Samacheer Kalvi Class 11 Physics Solutions Chapter 2 Kinematics

11th Physics Guide Kinematics Book Back Questions and Answers

Part - I:

I. Multiple choice questions:

 

Question 1. Which one of the following Cartesian coordinate systems is not followed in
(a)
(b)
(c)
(d)
Answer: (d) The system where the x, y, z axes are shown to form a left-handed coordinate system.
In simple words: A right-handed system means if you curl the fingers of your right hand from the x-axis to the y-axis, your thumb points along the z-axis. The option (d) does not follow this rule, making it a left-handed system.

🎯 Exam Tip: Remember the right-hand thumb rule for identifying right-handed coordinate systems. It's a fundamental concept in vector calculus and physics.

 

Question 2. Identify the unit vector in the following _________
(a) \( \hat{i} + \hat{j} \)
(b) \( \frac{\hat{i}}{\sqrt{2}} \)
(c) \( \hat{k} - \frac{\hat{i}}{\sqrt{2}} \)
(d) \( \frac{\hat{i} + \hat{j}}{\sqrt{2}} \)
Answer: (d) \( \frac{\hat{i} + \hat{j}}{\sqrt{2}} \)
In simple words: A unit vector is a vector that has a magnitude (length) of exactly 1. When you calculate the magnitude of option (d), you find it is equal to 1, unlike the other options.

🎯 Exam Tip: To check if a vector is a unit vector, always calculate its magnitude. If the magnitude is 1, it is a unit vector.

 

Question 3. Which one of the following quantities cannot be represented by a scalar?
(b) length
(c) momentum
(d) magnitude of the acceleration
Answer: (c) momentum
In simple words: Momentum is a quantity that has both size (how much) and direction (where it's going), so it needs to be described by a vector, not just a single number (scalar).

🎯 Exam Tip: Scalars have only magnitude (like mass or temperature), while vectors have both magnitude and direction (like velocity or force). Momentum needs direction, making it a vector quantity.

 

Question 4. Two objects of masses \( m_1 \) and \( m_2 \) fall from the heights \( h_1 \) and \( h_2 \) respectively. The ratio of the magnitude of their momenta when they hit the ground is _________ (AIPMT 2012)
(a) \( \frac{m_1}{m_2}\sqrt{\frac{h_1}{h_2}} \)
(b) \( \sqrt{\frac{m_1 h_1}{m_2 h_2}} \)
(c) \( \frac{m_1}{m_2}\sqrt{\frac{h_2}{h_1}} \)
(d) \( \frac{m_1}{m_2} \)
Answer: (a) \( \frac{m_1}{m_2}\sqrt{\frac{h_1}{h_2}} \)
In simple words: When objects fall, their speed just before hitting the ground depends on the height. We use this speed and their mass to find their momentum, then compare them to get this ratio. This ratio tells us how their momentum compares based on their mass and starting height.

🎯 Exam Tip: Remember to use the conservation of energy to find the velocity of the objects just before impact. Kinetic energy at impact equals potential energy at the start: \( \frac{1}{2}mv^2 = mgh \implies v = \sqrt{2gh} \). Momentum \( p = mv \). Then calculate the ratio of momenta.

 

Question 5. If a particle has negative velocity and negative acceleration, it speeds _________
(a) increases
(b) decreases
(c) remains the same
(d) zero
Answer: (a) increases
In simple words: If an object is moving in a certain direction (negative velocity) and something is pushing it or making it go faster in that same direction (negative acceleration), then its speed will increase.

🎯 Exam Tip: Speed increases when velocity and acceleration have the same sign (both positive or both negative). Speed decreases when they have opposite signs. Think of pushing a car backwards while it's already rolling backwards – it gets faster in that direction.

 

Question 6. If the velocity is \( \overline{V} = 2 \hat{i}+t^{2} \hat{j}-9 \hat{k} \) then the magnitude of acceleration at \( t = 0.5s \) is _________
(a) \( 1ms^{-2} \)
(b) \( 2 ms^{-2} \)
(c) zero
(d) \( -1ms^{-2} \)
Answer: (a) \( 1ms^{-2} \)
In simple words: Acceleration is how much the velocity changes over time. We calculate the acceleration from the given velocity equation and then find its total size at the specific time of 0.5 seconds.

🎯 Exam Tip: To find acceleration from velocity, differentiate the velocity vector with respect to time: \( \vec{a} = \frac{d\vec{V}}{dt} \). Then, substitute the given time to find the acceleration vector and calculate its magnitude using \( |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).

 

Question 7. If an object is dropped from the top of a building and it reaches the ground at \( t = 4s \), then the height of the building is (ignoring air resistance) (g = \( 9.8ms^{-2} \))
(a) 77.3m
(b) 78.4m
(c) 80.5m
(d) 79.2m
Answer: (b) 78.4m
In simple words: When an object is dropped, it starts with no speed and gravity makes it fall faster. We use a formula that connects how far it falls (height) with how long it takes and the pull of gravity.

🎯 Exam Tip: For an object dropped from rest, use the kinematic equation \( h = ut + \frac{1}{2}gt^2 \). Since it starts from rest, \( u = 0 \), so the equation simplifies to \( h = \frac{1}{2}gt^2 \). Substitute the given values of \( g \) and \( t \).

 

Question 8. A ball is projected vertically upwards with a velocity v. It comes back to the ground in time t. Which v-t graph shows the motion correctly? (NSEP 00-01)
(a)
(b)
(c)
(d)
Answer: (c)
In simple words: When a ball is thrown upwards, its velocity is positive at first and decreases linearly due to gravity until it becomes zero at the highest point. Then, it starts falling, so its velocity becomes negative and increases linearly in magnitude. The graph showing a straight line going from positive to negative velocity passing through zero is the correct representation.

🎯 Exam Tip: For vertical motion under gravity, the velocity-time graph is a straight line because acceleration due to gravity is constant. The slope of the v-t graph represents acceleration.

 

Question 9. If one object is dropped vertically downward and another object is thrown horizontally from the same height, then the ratio of vertical distance covered by both objects at any instant t is _________
(a) 1
(b) 2
(c) 4
(d) 0.5
Answer: (a) 1
In simple words: When an object is dropped, its vertical motion is only affected by gravity. If another object is thrown horizontally from the same height, its horizontal motion doesn't change its vertical motion. So, both objects fall the same vertical distance in the same amount of time.

🎯 Exam Tip: Remember that horizontal and vertical motions are independent of each other under gravity (ignoring air resistance). The vertical distance covered only depends on initial vertical velocity, acceleration due to gravity, and time.

 

Question 10. A bail is dropped from some height towards the ground. Which one of the following represents the correct motion of the ball?
(a)
(b)
(c)
(d)
Answer: (a)
In simple words: When a ball is dropped, it falls straight down. So, its x-position stays the same, and only its y-position changes downwards. The graph showing dots vertically aligned at a constant x-value is the correct one.

🎯 Exam Tip: Remember that gravity acts only in the vertical direction. If an object is simply dropped, it has no initial horizontal velocity and thus no horizontal movement.

 

Question 11. If a particle executes uniform circular motion in the XY plane in a clockwise direction, then the angular velocity is in _________
(a) +y direction
(b) +z direction
(c) -z direction
(d) -x direction
Answer: (c) -z direction
In simple words: For a circular motion, we use the right-hand rule. If you curl your fingers in the direction of the clockwise motion in the XY plane, your thumb will point downwards along the z-axis, which is the negative z-direction.

🎯 Exam Tip: Use the right-hand rule for angular velocity. Curl the fingers of your right hand in the direction of rotation. Your thumb will point in the direction of the angular velocity vector. Clockwise rotation in the XY plane means the angular velocity is in the negative z-direction.

 

Question 12. If a particle executes uniform circular motion, choose the correct statement (NEET 2016)
(a) The velocity and speed are constant.
(b) The acceleration and speed are constant.
(c) The velocity and acceleration are constant.
(d) The speed and magnitude of acceleration are constant
Answer: (d) The speed and magnitude of acceleration are constant
In simple words: In uniform circular motion, an object moves at a steady speed, but its direction changes all the time. This change in direction means its velocity is not constant, and because the velocity is changing, there must be an acceleration. The size of this acceleration stays the same.

🎯 Exam Tip: Uniform circular motion means constant speed, but changing velocity (due to changing direction). Changing velocity implies acceleration (centripetal acceleration), which is constant in magnitude but continuously changes direction, always pointing towards the center of the circle.

 

Question 13. If an object is thrown vertically up with the initial speed u from the ground, then the time taken by the object to return back to the ground is _________
(a) \( \frac{u^{2}}{2 g} \)
(b) \( \frac{u^{2}}{g} \)
(c) \( \frac { u }{ 2g } \)
(d) \( \frac { 2u }{ g } \)
Answer: (d) \( \frac { 2u }{ g } \)
In simple words: When an object is thrown up, it takes a certain time to reach its highest point, and then the same amount of time to fall back down to the start. The total time in the air is twice the time it takes to go up.

🎯 Exam Tip: The time to reach the maximum height is \( t_{up} = \frac{u}{g} \). Since the motion is symmetrical, the time to fall back down is also \( t_{down} = \frac{u}{g} \). Therefore, the total time of flight is \( T = t_{up} + t_{down} = \frac{u}{g} + \frac{u}{g} = \frac{2u}{g} \).

 

Question 14. Two objects are projected at angles 30° and 60° respectively with respect to the horizontal direction. The range of two objects are denoted as \( R_{30°} \) and \( R_{60°} \) Choose the correct relation from the following
(a) \( R_{30°} = R_{60°} \)
(b) \( R_{30°} = 4R_{60°} \)
(c) \( R_{30°} = R\frac { 60° }{ 2 } \)
(d) \( R_{30°} = 2R_{60°} \)
Answer: (a) \( R_{30°} = R_{60°} \)
In simple words: If you throw two objects with the same initial speed at angles that add up to 90 degrees (like 30 and 60), they will travel the same horizontal distance before landing.

🎯 Exam Tip: The range of a projectile launched with initial velocity \( u \) at an angle \( \theta \) is given by \( R = \frac{u^2 \sin(2\theta)}{g} \). For complementary angles \( \theta \) and \( (90° - \theta) \), the value of \( \sin(2\theta) \) is the same as \( \sin(2(90° - \theta)) = \sin(180° - 2\theta) = \sin(2\theta) \). Thus, their ranges are equal.

 

Question 15. An object is dropped in an unknown planet from a height of 50m, it reaches the ground in 2s. The acceleration due to gravity in this unknown planet is _________
(a) \( g = 20ms^{-2} \)
(b) \( g = 25ms^{-2} \)
(c) \( g = 15ms^{-2} \)
(d) \( g = 30ms^{-2} \)
Answer: (b) \( g = 25ms^{-2} \)
In simple words: We can find the acceleration of gravity on the new planet by looking at how far the object fell and how long it took. Using a basic physics formula, we can calculate the unknown gravity.

🎯 Exam Tip: Use the kinematic equation for free fall: \( h = ut + \frac{1}{2}gt^2 \). Since the object is dropped, its initial velocity \( u = 0 \). So, \( h = \frac{1}{2}gt^2 \). Rearrange to find \( g = \frac{2h}{t^2} \) and substitute the given values.

II. Short Answer Questions:

 

Question 1. Explain what is meant by the Cartesian coordinate system?
Answer: A Cartesian coordinate system is a way to describe where an object is at any moment in time. We use three perpendicular lines called axes (x, y, and z) to define its position in space. If the x, y, and z axes are drawn in a specific counter-clockwise order (x to y, then thumb along z), the system is called a right-handed Cartesian coordinate system.

In simple words: It's a system that uses x, y, and z lines to pinpoint exactly where something is. If you draw these lines in a special way (right-handed rule), it helps describe motion and forces correctly.

🎯 Exam Tip: When defining coordinate systems, remember to mention the three perpendicular axes and the concept of "right-handed" vs. "left-handed" as it affects vector operations.

 

Question 2. Define a vector. Give Example.
Answer: A vector is a quantity that is fully described by both its magnitude (size or amount) and its direction. For example, when we talk about a force, we need to know how strong it is (magnitude) and in which direction it is acting. Other examples include velocity and displacement. Geometrically, a vector can be shown as a directed line segment with an arrow indicating its direction.
In simple words: A vector is a measurement that needs both a size and a direction to be completely understood, like how hard and where you push something.

🎯 Exam Tip: Always specify both magnitude and direction when defining or using vectors. Providing relevant examples like force, velocity, or displacement helps illustrate the concept clearly.

 

Question 3. Define a Scalar. Give Examples.
Answer: A scalar is a physical quantity that can be completely described by its magnitude (size or numerical value) alone. It does not have a direction. For example, if you say the distance is 10 meters, you don't need to specify a direction. Other common examples include mass, temperature, speed, and energy. These quantities are simply numerical values.
In simple words: A scalar is a measurement that only tells you how much, not which way, like how hot it is or how heavy something is.

🎯 Exam Tip: The key difference between a scalar and a vector is the presence of direction. When asked for examples, choose quantities that intuitively do not require a direction to be fully understood.

 

Question 4. Write short note on the scalar product between two vectors.
Answer: The scalar product, also known as the dot product, of two vectors is found by multiplying their magnitudes and the cosine of the angle between them. The result is always a scalar quantity, meaning it has only magnitude and no direction. If \( \vec{A} \) and \( \vec{B} \) are two vectors with an angle \( \theta \) between them, their scalar product is given by \( \vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}| \cos \theta \). This can also be written as \( \vec{A} \cdot \vec{B} = AB \cos \theta \), where A and B are the magnitudes of \( \vec{A} \) and \( \vec{B} \). For example, the work done (W) by a force (\( \vec{F} \)) causing a displacement (\( \vec{r} \)) is a scalar product: \( W = \vec{F} \cdot \vec{r} \).
In simple words: The dot product of two vectors gives a simple number, not a new vector. You get it by multiplying their lengths and the cosine of the angle between them.

🎯 Exam Tip: When discussing scalar products, always state that the result is a scalar. Mention the formula \( \vec{A} \cdot \vec{B} = AB \cos \theta \) and give a common example like work done, as it clearly illustrates the scalar nature of the result.

 

Question 5. Write a Short note on vector product between two vectors.
Answer: The vector product, or cross product, of two vectors creates a new vector. The magnitude of this new vector is equal to the product of the magnitudes of the original two vectors and the sine of the angle between them. The direction of this product vector is always perpendicular to the plane that contains the two original vectors. We determine its direction using the right-hand screw rule or the right-hand thumb rule. So, if \( \vec{A} \) and \( \vec{B} \) are two vectors, their vector product \( \vec{A} \times \vec{B} \) is a vector \( \vec{C} \) defined by \( \vec{C} = \vec{A} \times \vec{B} = (AB \sin \theta) \hat{n} \), where \( \hat{n} \) is the unit vector perpendicular to the plane containing \( \vec{A} \) and \( \vec{B} \).
In simple words: The cross product of two vectors makes a new vector. Its size depends on the lengths of the first two vectors and the angle between them, and its direction is always straight out of the flat area where the first two vectors lie.

🎯 Exam Tip: For vector products, remember that the result is a vector, and its direction is determined by the right-hand rule. The formula for magnitude \( |\vec{A} \times \vec{B}| = AB \sin \theta \) is crucial.

 

Question 6. How do you deduce that two vectors are perpendicular?
Answer: We can deduce that two vectors are perpendicular if their vector product (cross product) has its maximum possible magnitude. The magnitude of the vector product of two vectors \( \vec{A} \) and \( \vec{B} \) is given by \( |\vec{A} \times \vec{B}| = AB \sin \theta \). This magnitude is maximum when \( \sin \theta = 1 \), which means the angle \( \theta \) between the vectors is \( 90^\circ \). Therefore, if \( |\vec{A} \times \vec{B}|_{\text{max}} = AB\hat{n} \), the two vectors are perpendicular.
In simple words: Two vectors are perpendicular if their cross product gives the biggest possible value, which happens when they are at a 90-degree angle to each other.

🎯 Exam Tip: Two vectors are perpendicular if their scalar product is zero (\( \vec{A} \cdot \vec{B} = 0 \)). They are perpendicular if their vector product has maximum magnitude (\( |\vec{A} \times \vec{B}| = AB \)). Both are valid conditions, choose the one most relevant to the context given in the question (here it explicitly refers to the vector product being maximum).

 

Question 7. Define Displacement and distance.
Answer: Distance is the actual total path length traveled by an object during its motion within a given time. It is a scalar quantity, always positive, and only considers how far the object moved along its path. Displacement, on the other hand, is the shortest straight-line distance between an object's initial position and its final position, along with the direction. It is a vector quantity and can be positive, negative, or zero.
In simple words: Distance is how much ground you cover; displacement is how far you are from where you started, and in what direction.

🎯 Exam Tip: Always emphasize that distance is a scalar and displacement is a vector. Highlight the "path length" versus "shortest path" distinction, and note that displacement can be zero even if distance is not (e.g., returning to the starting point).

 

Question 8. Define velocity and speed.
Answer: Velocity is defined as the rate at which an object changes its position, specifically considering both the rate of movement and the direction. It can also be defined as the rate of change of displacement. Velocity is a vector quantity. Speed, however, is simply the rate at which an object moves, without considering the direction. It is defined as the rate of change of distance and is a scalar quantity.

In simple words: Velocity tells you how fast something is going and in what direction, while speed just tells you how fast it's going.

🎯 Exam Tip: Highlight that velocity is a vector (requires direction) and speed is a scalar (only magnitude). Providing the formulas and units solidifies the definitions.

 

Question 9. Define acceleration.
Answer: Acceleration is defined as the rate of change of an object's velocity over time. Since velocity is a vector quantity, acceleration is also a vector quantity, meaning it has both magnitude and direction. If an object's velocity changes (either its speed, direction, or both), it is accelerating. The mathematical expression for acceleration is \( \vec{a} = \frac{d\vec{v}}{dt} \). Its unit is meters per second squared (\( ms^{-2} \)), and its dimensional formula is \( [LT^{-2}] \).
In simple words: Acceleration is how quickly an object's speed or direction changes. If something is speeding up, slowing down, or turning, it's accelerating.

🎯 Exam Tip: Emphasize that acceleration is a vector quantity. It's important to remember that a change in direction, even with constant speed, constitutes acceleration (e.g., in circular motion).

 

Question 10. What is the difference between velocity and average velocity?
Answer: Velocity describes the instantaneous rate of change of an object's position in a specific direction. It tells you the speed and direction at any given moment. Average velocity, however, is the ratio of the total displacement vector to the total time interval taken. It gives an overall measure of how much an object's position changed relative to the time, without detailing the instantaneous changes.

In simple words: Velocity is what your speedometer shows right now, plus the direction. Average velocity is like taking your total straight-line trip from start to finish and dividing by the total time.

🎯 Exam Tip: Differentiate between instantaneous velocity (at a specific moment) and average velocity (over a time interval). Both are vector quantities, but average velocity simplifies the motion. Average speed, in contrast, uses total distance traveled.

 

Question 11. Define a radian.
Answer: A radian is defined as the angle subtended at the center of a circle by an arc whose length is equal to the radius of the circle. This means if you take a part of the circle's edge (an arc) that has the same length as the distance from the center to the edge (radius), the angle formed at the center by that arc is one radian. We can express it as \( \theta = \frac{\text{arc length}}{\text{radius}} \), so for one radian, arc length equals radius: \( 1 \text{ rad} = \frac{r}{r} = 1 \).

In simple words: A radian is a way to measure angles where the angle size matches the arc length divided by the circle's radius. It's like saying if the curved part of a slice of pizza is as long as the straight edge, that's one radian.

🎯 Exam Tip: The definition of a radian connects arc length and radius. Make sure to specify that it is the angle at the *center* of the circle. Clearly show the relationship \( \theta = \frac{s}{r} \).

 

Question 12. Define angular displacement and angular velocity.
Answer:
1. Angular displacement: This is the angle an object rotates about its axis of rotation in a given amount of time. It tells us how much the object has turned from its starting position. It's a vector quantity, with direction given by the right-hand rule.
2. Angular velocity: This is the rate at which angular displacement changes over time. It measures how fast an object is rotating or revolving. It's also a vector quantity, pointing in the same direction as the angular displacement.
In simple words: Angular displacement is how much something has spun around, and angular velocity is how fast it is spinning.

🎯 Exam Tip: For both definitions, stress that they refer to rotational motion. Emphasize that angular velocity is the rate of change of angular displacement, similar to how linear velocity is the rate of change of linear displacement.

 

Question 13. What is non-uniform circular motion?
Answer: Non-uniform circular motion happens when an object moves in a circular path but its speed is not constant. This means the object covers different distances along the circle in equal amounts of time. In this type of motion, both the object's speed and its direction are continuously changing. An object experiencing this motion will have both centripetal and tangential acceleration.
In simple words: It's when something goes in a circle, but its speed changes as it moves, so it doesn't cover equal parts of the circle in the same amount of time.

🎯 Exam Tip: The key differentiator for non-uniform circular motion is that *speed* changes, not just direction. This leads to both centripetal and tangential acceleration being present.

 

Question 14. Write down the kinematic equations for angular motion.
Answer: The kinematic equations for angular motion describe how angular quantities like angular velocity, angular acceleration, angular displacement, and time are related when the angular acceleration is constant. These equations are:
1. \( \omega = \omega_0 + \alpha t \)
2. \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)
3. \( \omega^2 = \omega_0^2 + 2\alpha\theta \)
4. \( \theta = \left(\frac{\omega_0 + \omega}{2}\right) t \)
Here:
\( \omega_0 \) → initial angular velocity
\( \omega \) → final angular velocity
\( \alpha \) → angular acceleration
\( \theta \) → angular displacement
\( t \) → time interval
These equations are similar to the linear kinematic equations but adapted for rotational movement.
In simple words: These are special math rules that help us figure out how fast something spins, how much it turns, and how long it takes, when its spinning speed changes steadily.

🎯 Exam Tip: These equations are analogous to linear kinematic equations (\( v = u + at \), \( s = ut + \frac{1}{2}at^2 \), \( v^2 = u^2 + 2as \), \( s = \left(\frac{u+v}{2}\right)t \)). Clearly define all variables used in the equations.

 

Question 15. Write down the expression for angle made by resultant acceleration and radius vector in the non-uniform circular motion.
Answer: In non-uniform circular motion, a particle experiences two types of acceleration: centripetal acceleration (\( a_c \)) which points towards the center of the circle, and tangential acceleration (\( a_t \)) which is tangent to the circle. The overall (resultant) acceleration (\( a_R \)) is the vector sum of these two components. The magnitude of the resultant acceleration is \( a_R = \sqrt{a_t^2 + a_c^2} \). The angle \( \phi \) (often represented as \( \theta \)) that this resultant acceleration makes with the radius vector (which is in the direction of \( a_c \)) is given by \( \tan \phi = \frac{a_t}{a_c} \).

In simple words: When something goes in a circle and changes speed, its total push (acceleration) isn't just towards the center. It has a part that makes it speed up or slow down (tangential) and a part that keeps it in the circle (centripetal). The angle of the total push compared to the radius shows how these two parts combine.

🎯 Exam Tip: Clearly distinguish between centripetal (\( v^2/r \)) and tangential (\( d|v|/dt \)) accelerations. The resultant acceleration is their vector sum, and its angle with the radius is found using the tangent function.

III. Long Answer Questions:

 

Question 1. Explain in detail the triangle law of addition.
Answer: The triangle law of vector addition is a method to find the resultant (single combined effect) of two vectors. It states that if two vectors are represented in both magnitude and direction by the two adjacent sides of a triangle taken in the same order, then their resultant vector is represented in both magnitude and direction by the third side of the triangle taken in the reverse order.
Let's say we have two vectors, \( \vec{A} \) and \( \vec{B} \). We place the tail of vector \( \vec{B} \) at the head of vector \( \vec{A} \). The resultant vector \( \vec{R} \) is then drawn from the tail of \( \vec{A} \) to the head of \( \vec{B} \).

The magnitude of the resultant vector \( \vec{R} \) can be found using the Law of Cosines. If \( \theta \) is the angle between vectors \( \vec{A} \) and \( \vec{B} \), then the magnitude of the resultant \( R \) is given by: \[ R = \sqrt{A^2 + B^2 + 2AB \cos \theta} \] The direction of the resultant vector \( \vec{R} \) (let's say it makes an angle \( \alpha \) with vector \( \vec{A} \)) can be found using the Law of Sines: \[ \tan \alpha = \frac{B \sin \theta}{A + B \cos \theta} \] This law is fundamental for combining forces, velocities, or displacements in physics.
In simple words: To add two arrows (vectors), draw the first one, then draw the second one starting from the end of the first. The total arrow, from the start of the first to the end of the second, is the combined result.

🎯 Exam Tip: Clearly state the triangle law. A well-drawn diagram is essential, showing the two vectors in order and the resultant in reverse order. Include the formulas for both magnitude and direction of the resultant.

 

Question 1. Explain in detail the triangle law of addition.
Answer: The triangle law helps us add two vectors. If we draw two vectors, A and B, as two sides of a triangle, one after the other, then the third side of the triangle shows their total effect or resultant. This third side is drawn in the opposite order to complete the triangle. This law is very useful in physics for combining forces or velocities to find their overall effect.
Let us consider two vectors \( \vec{A} \) and \( \vec{B} \) as shown in the figure.
Applied as follows: \( \vec{A} \) and \( \vec{B} \) are represented as the two adjacent sides of a triangle taken in the same order. The resultant is given by the third side of the triangle taken in reverse order.

\( \vec{OQ} = \vec{R} = \vec{OP} + \vec{PQ} \)
Magnitude of the resultant vector:
Let's say the angle between the two vectors is \( \theta \). Using geometry, specifically the sine and cosine rules in triangle ABN, we can find BN and AN in terms of B and \( \theta \).
From figure:
From \( \triangle ABN \), \( \sin \theta = \frac{BN}{AB} \implies BN = B \sin \theta \)
\( \cos \theta = \frac{AN}{AB} \implies AN = B \cos \theta \)
Now, using the Pythagorean theorem in triangle OBN, we can relate its sides.
From \( \triangle OBN \), \( OB^2 = ON^2 + BN^2 \)
\( OB^2 = (OA + AN)^2 + BN^2 \)
\( R^2 = (A + B \cos \theta)^2 + (B \sin \theta)^2 \)
\( R^2 = A^2 + 2AB \cos \theta + B^2 \cos^2 \theta + B^2 \sin^2 \theta \)
\( R^2 = A^2 + B^2 (\cos^2 \theta + \sin^2 \theta) + 2AB \cos \theta \)
\( R^2 = A^2 + B^2 + 2AB \cos \theta \)
\( \therefore R = \sqrt{A^2 + B^2 + 2AB \cos \theta} \)
The direction of the resultant vector:
To find the direction of this total vector \( \vec{R} \), we look at the angle \( \alpha \) it makes with vector \( \vec{A} \). Using the tangent function in triangle OBN, we can calculate this angle.
If \( \vec{R} \) makes an angle \( \alpha \) with \( \vec{A} \) then
From \( \triangle OBN \), \( \tan \alpha = \frac{BN}{ON} = \frac{BN}{OA+AN} \)
\( \tan \alpha = \frac{B \sin \theta}{A + B \cos \theta} \)
\( \therefore \alpha = \tan^{-1} \left(\frac{B \sin \theta}{A+B \cos\theta}\right) \)
In simple words: The triangle law of vector addition says that if you draw two vectors one after another like two sides of a triangle, the third side, drawn to close the triangle, gives their combined effect. It helps us find both how strong the combined effect is and in what direction it acts.

🎯 Exam Tip: Remember to clearly label all vectors and angles in your diagrams. Showing the steps for both magnitude and direction is crucial for full marks.

 

Question 2. Discuss the properties of scalar and vector products.
Answer:The scalar product, also called the dot product, tells us how much two vectors point in the same direction. We find it by multiplying the lengths of the two vectors by the cosine of the angle between them. If \( \vec{A} \) and \( \vec{B} \) are two vectors with an angle \( \theta \) between them, their dot product is \( \vec{A}.\vec{B} = AB \cos \theta \). Here, A and B are the sizes of vectors \( \vec{A} \) and \( \vec{B} \). The dot product is always a scalar (just a number), not a vector. Scalar products are used to calculate work done by a force, while vector products are used to calculate torque.
Properties of scalar product:
1. The scalar product \( \overline{A}.\overline{B} \) is always a scalar quantity. It is positive if the angle between the vectors is acute (0° < \( \theta \) < 90°) and negative if the angle between them is obtuse (90° < \( \theta \) < 180°).
2. The scalar product is commutative: \( \overline{A}.\overline{B} = \overline{B}.\overline{A} \).
3. The scalar product obeys the distributive law: \( \overline{A}.(\overline{B} + \overline{C}) = \overline{A}.\overline{B} + \overline{A}.\overline{C} \).
4. The angle between the vectors can be found using: \( \theta = \cos^{-1}\left(\frac{\overline{A} \cdot \overline{B}}{AB}\right) \).
5. The scalar product of two vectors will be maximum when \( \cos \theta = 1 \), which means \( \theta = 0^{\circ} \) (when they are parallel).
6. The scalar product of two vectors will be minimum when \( \cos \theta = -1 \), which means \( \theta = 180^{\circ} \) (when they are anti-parallel). So, \( (\overline{A}.\overline{B})_{min} = -AB \).
7. If two vectors \( \overline{A} \) & \( \overline{B} \) are perpendicular to each other, then \( \overline{A}.\overline{B} = 0 \). This is because \( \cos 90^{\circ} = 0 \), so vectors A & B are mutually orthogonal.
8. The scalar product of a vector with itself is called a self or dot product: \( (\overline{A})^2 = \overline{A}.\overline{A} = AA \cos 0^{\circ} = A^2 \). Here \( \theta = 0^{\circ} \).
The magnitude or norm of the vector \( \overline{A} \) is \( |A| = A = \sqrt{\overline{A} \cdot \overline{A}} = A \).
9. In the case of orthogonal unit vectors, \( \hat{n}.\hat{n} = 1 \times 1 \cos 0^{\circ} = 1 \). For example, \( \hat{i}.\hat{i} = \hat{j}.\hat{j} = \hat{k}.\hat{k} = 1 \).
10. For orthogonal unit vectors \( \hat{i}, \hat{j}, \hat{k} \), their dot products are: \( \hat{i}.\hat{j} = \hat{j}.\hat{k} = \hat{k}.\hat{j} = 1.1 \cos 90^{\circ} = 0 \).
11. In terms of components, the scalar product of \( \vec{A} \) and \( \vec{B} \) can be written as:
\( \vec{A}.\vec{B}= (A_x\hat{i}+A_y\hat{j}+A_z\hat{k}).(B_x\hat{i}+B_y\hat{j}+B_z\hat{k}) \)
\( = A_x B_x + A_y B_y + A_z B_z \)
The magnitude of \( \vec{A} \) = \( A = \sqrt{A_x^2 + A_y^2 + A_z^2} \)
The magnitude of \( \vec{B} \) = \( B = \sqrt{B_x^2 + B_y^2 + B_z^2} \)
Properties of cross product:
The vector product, or cross product, gives us a new vector that is at right angles to both of the original vectors. Its size is found by multiplying the lengths of the two vectors by the sine of the angle between them. To figure out its direction, we use the right-hand screw rule or the right-hand thumb rule. This product is always a vector.
Formula: \( \vec{A} \times \vec{B} = AB \sin \theta \hat{n} \) (where \( \hat{n} \) is a unit vector perpendicular to both \( \vec{A} \) and \( \vec{B} \)).
1. The vector product of any two vectors is always another vector. Its direction is perpendicular to the plane containing these two vectors. So, it is orthogonal to \( \overline{A} \) & \( \overline{B} \), even if \( \overline{A} \) & \( \overline{B} \) are not mutually orthogonal themselves.
2. The vector product is not commutative: \( \overline{A} \times \overline{B} = -\overline{B} \times \overline{A} \). This means \( \overline{A} \times \overline{B} \neq \vec{B} \times \vec{A} \). The magnitudes \( | \overline{A} \times \overline{B} | \) and \( | \overline{B} \times \overline{A} | \) are equal, but their directions are opposite.
3. The vector product of two vectors is maximum when \( \sin \theta = 1 \), i.e., \( \theta = 90^{\circ} \) (when \( \overline{A} \) and \( \overline{B} \) are orthogonal to each other). The maximum cross product is \( (\overline{A} \times \overline{B} )_{max} = AB \hat{n} \).
4. The vector product of two non-zero vectors is minimum if \( |\sin\theta| = 0 \), i.e., \( \theta = 0^{\circ} \) or \( 180^{\circ} \). The minimum cross product is \( (\overline{A} \times \overline{B} )_{min} = 0 \). The vector product of two non-zero vectors is zero if they are either parallel or anti-parallel.
5. The self cross product (product of a vector with itself) is a null vector: \( \overline{A} \times \overline{A} = AA \sin 0^{\circ} = 0 \).
6. The self-vector product of the unit vector is zero, e.g., \( \hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0 \).
7. In the case of orthogonal unit vectors \( \hat{i}, \hat{j}, \hat{k} \), according to the right-hand corkscrew rule: \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{k} \times \hat{i} = \hat{j} \).
The source states: \( \hat{i}\times \hat{j}=-\hat{k}, \hat{k}\times \hat{j}=−\hat{i}, \hat{i}\times \hat{k} = −\hat{j} \) also since cross product is not commutative.

8. In terms of components, the cross product of \( \vec{A} \) and \( \vec{B} \) can be written as:
\( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} \)
\( = \hat{i}(A_yB_z-A_zB_y)+\hat{j}(A_zB_x-A_xB_z)+\hat{k}(A_xB_y-A_yB_x) \)
9. If two vectors \( \overline{A} \) & \( \overline{B} \) form adjacent sides of a parallelogram, then the magnitude of \( |\overline{A} \times \overline{B}| \) will give the area of the parallelogram.
10. Since one can divide a parallelogram into two equal triangles, the area of the triangle is \( \frac{1}{2} |\overline{A} \times \overline{B}| \).
In simple words: Scalar product gives a simple number, telling us how much two forces or movements act in the same direction. Vector product gives a new direction, showing the turning effect or twist that two forces can create.

🎯 Exam Tip: When discussing properties, it's helpful to provide a brief example or application for each type of product to show a deeper understanding.

 

Question 3. Derive the kinematic equations of motion for constant acceleration.
Answer: Let's think about an object moving straight with a steady acceleration, which we call 'a'. Its starting speed at time zero is 'u', and its speed after some time 't' is 'v'. These three equations are fundamental in classical mechanics and allow us to predict the motion of objects under constant acceleration.
(i) Velocity-time relation:
We know that acceleration \( a \) is the rate of change of velocity.
\( a = \frac{dv}{dt} \)
\( dv = a dt \)
To solve this, we will integrate both sides of the equation from initial velocity \( u \) to final velocity \( v \) and from initial time 0 to final time \( t \).
\( \int_u^v dv = \int_0^t a dt \)
\( [V]_u^v = a[t]_0^t \)
\( v - u = at \)
\( v = u + at \)
(ii) Displacement-time relation:
Velocity is how fast the object's position changes over time. We can express it as the change in displacement divided by the change in time.
\( v = \frac{ds}{dt} \)
\( ds = v dt \)
From the velocity-time relation, we know \( v = u + at \). Substitute this into the displacement equation:
\( ds = (u + at)dt \)
\( ds = u dt + at dt \)
Now, we integrate both sides to find the displacement, from initial displacement 0 to final displacement \( s \) and from time 0 to \( t \).
\( \int_0^s ds = \int_0^t u dt + \int_0^t at dt \)
\( [s]_0^s = u[t]_0^t + a\left[\frac{t^2}{2}\right]_0^t \)
\( s = ut + \frac{1}{2}at^2 \)
(iii) Velocity-displacement relation:
We start with the definition of acceleration and velocity:
\( a = \frac{dv}{dt} \) and \( v = \frac{ds}{dt} \)
We can write \( a = \frac{dv}{ds} \times \frac{ds}{dt} \)
\( a = v \frac{dv}{ds} \)
Rearrange this to integrate:
\( a ds = v dv \)
Now, integrate both sides from initial displacement 0 to final displacement \( s \) and from initial velocity \( u \) to final velocity \( v \).
\( \int_0^s a ds = \int_u^v v dv \)
\( a[s]_0^s = \left[\frac{v^2}{2}\right]_u^v \)
\( as = \frac{v^2}{2} - \frac{u^2}{2} \)
\( 2as = v^2 - u^2 \)
\( v^2 - u^2 = 2as \)
We can also find another useful equation by combining the first kinematic equation, \( v = u + at \), with the displacement equation.
From \( v = u + at \), we get \( t = \frac{v-u}{a} \).
Substitute this into \( s = ut + \frac{1}{2}at^2 \):
\( s = u\left(\frac{v-u}{a}\right) + \frac{1}{2}a\left(\frac{v-u}{a}\right)^2 \)
\( s = \frac{uv-u^2}{a} + \frac{1}{2}a\frac{(v-u)^2}{a^2} \)
\( s = \frac{uv-u^2}{a} + \frac{(v-u)^2}{2a} \)
\( s = \frac{2(uv-u^2) + (v^2-2uv+u^2)}{2a} \)
\( s = \frac{2uv-2u^2 + v^2-2uv+u^2}{2a} \)
\( s = \frac{v^2-u^2}{2a} \)
\( v^2 - u^2 = 2as \)
The kinematic equations are:
\( v = u + at \)
\( s = ut + \frac{1}{2}at^2 \)
\( v^2 - u^2 = 2as \)
\( s = \left(\frac{u+v}{2}\right)t \)
In simple words: Kinematic equations are like special formulas that help us understand how things move. If something is speeding up or slowing down at a steady rate, these equations let us find its final speed, how far it went, or how long it took, just by knowing a few details about its start.

🎯 Exam Tip: Clearly state the assumptions (constant acceleration, straight line motion) at the beginning of your derivation. Make sure to present the final three equations concisely.

 

Question 4. Derive the equations of motion for a particle (a) falling vertically (b) projected vertically.
Answer: We'll look at the equations for objects moving straight up or down. These equations simplify complex motion into predictable patterns, crucial for understanding everyday phenomena like a dropped ball or a thrown object.
(a) For a body falling vertically from a height 'h':
First, imagine an object of mass 'm' falling from height 'h'. We ignore air resistance, and say that going downwards is the positive y-axis direction. The Earth's gravity pulls the object down with a constant acceleration 'g'. So, the acceleration 'a' is equal to 'g'.

So, consider \( a = a_y = g \)
Case - 1: If an object is thrown downwards with a starting speed 'u'.
\( v = u + gt \)
\( y = ut + \frac{1}{2}gt^2 \)
\( v^2 - u^2 = 2gy \)
Case - 2: If the object simply starts falling from rest (its initial speed 'u' is zero).
\( u = 0 \)
\( v = gt \)
\( y = \frac{1}{2}gt^2 \)
\( v^2 = 2gy \)
(b) For a body projected vertically:
Next, consider an object of mass 'm' thrown straight upwards with a starting speed 'u'. Again, we ignore air friction. If upwards is the positive y-axis direction, then gravity acts downwards, making the acceleration '\(-g\)'.
\( a = -g \)
The kinematic equations of motion are:
\( v = u - gt \)
\( s = ut - \frac{1}{2}gt^2 \)
\( v^2 - u^2 = -2gs \)

In simple words: When an object moves straight up or down, special formulas help us find its speed and position. If it falls down, gravity helps it speed up. If it's thrown up, gravity makes it slow down and then fall back.

🎯 Exam Tip: Clearly distinguish between positive and negative directions for displacement, velocity, and acceleration based on your chosen coordinate system. Remember that 'g' is positive when falling and negative when projected upwards if 'up' is positive.

 

Question 5. Derive the equations of motion, range, and maximum height reached by a particle thrown at an oblique angle \( \theta \) with respect to the horizontal direction.
Answer: Let's consider an object launched with a starting speed 'u' at an angle \( \theta \) above the horizontal ground. This type of motion is called projectile motion. Understanding these concepts is key for sports like basketball or javelin throw, where trajectory planning is essential.

Then the initial velocity 'u' is broken down into two parts (resolved into two components):
\( u_x = u \cos \theta \) horizontally
\( u_y = u \sin \theta \) vertically
At its highest point, the object stops moving upwards for a brief moment, so its vertical speed \( u_y \) becomes zero. This happens because gravity constantly pulls it downwards, slowing its upward motion.
The horizontal speed of the object, \( u_x = u \cos \theta \), stays the same throughout its flight because there's no force (like air resistance) pushing it sideways (assuming negligible air resistance).

**1. Equation for trajectory (Path of the projectile):**
The horizontal distance traveled by the projectile in a time 't' is given by \( S_x = u_x t + \frac{1}{2} a_x t^2 \).
Here \( S_x = x \), \( u_x = u \cos \theta \), and \( a_x = 0 \) (since there is no horizontal acceleration).
\( \therefore x = (u \cos \theta) t \) (1)
From this, we can find the time \( t = \frac{x}{u \cos \theta} \) (2)
The vertical distance traveled by the projectile in the same time 't' is \( S_y = U_y t + \frac{1}{2} a_y t^2 \).
Here \( S_y = y \), \( U_y = u \sin \theta \), and \( a_y = -g \) (acceleration due to gravity acting downwards).
\( \therefore y = (u \sin \theta) t - \frac{1}{2} g t^2 \) (3)
Substituting the value of \( t \) from equation (2) into equation (3):
\( y = (u \sin \theta) \left(\frac{x}{u \cos \theta}\right) - \frac{1}{2} g \left(\frac{x}{u \cos \theta}\right)^2 \)
\( y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \)
This equation shows that the path an object follows when thrown in the air is shaped like an inverted parabola.

**2. Expression for Maximum height (\( h_{max} \)):**
The maximum height is the highest vertical point the object reaches from its starting position. We can find this height using the kinematic equations. At the maximum height, the vertical velocity (\( v_y \)) of the projectile is zero.
We use the equation: \( v_y^2 = u_y^2 + 2a_y s \)
Here, \( v_y = 0 \), \( u_y = u \sin \theta \), \( a_y = -g \), and \( s = h_{max} \).
\( 0^2 = (u \sin \theta)^2 + 2(-g) h_{max} \)
\( 0 = u^2 \sin^2 \theta - 2g h_{max} \)
\( 2g h_{max} = u^2 \sin^2 \theta \)
\( h_{max} = \frac{u^2 \sin^2 \theta}{2g} \)

**3. Expression for Horizontal Range (R):**
The horizontal range is how far the object travels horizontally from where it was thrown to where it lands back on the same level. The time of flight (\( t_f \)) is the total time the object spends in the air, from when it's launched until it lands.
First, let's find the time of flight (\( t_f \)). When the projectile lands back on the ground, its vertical displacement (\( S_y \)) is zero.
Using the equation: \( S_y = u_y t_f + \frac{1}{2} a_y t_f^2 \)
Here, \( S_y = 0 \), \( u_y = u \sin \theta \), and \( a_y = -g \).
\( 0 = (u \sin \theta) t_f - \frac{1}{2} g t_f^2 \)
Factor out \( t_f \):
\( t_f \left(u \sin \theta - \frac{1}{2} g t_f\right) = 0 \)
This gives two solutions: \( t_f = 0 \) (the starting point) or \( u \sin \theta - \frac{1}{2} g t_f = 0 \).
\( u \sin \theta = \frac{1}{2} g t_f \)
\( t_f = \frac{2u \sin \theta}{g} \)
Now, we can find the horizontal range \( R \). The horizontal range is the horizontal distance covered during the time of flight.
\( R = u_x t_f \)
\( R = (u \cos \theta) \left(\frac{2u \sin \theta}{g}\right) \)
\( R = \frac{u^2 (2 \sin \theta \cos \theta)}{g} \)
Using the trigonometric identity \( 2 \sin \theta \cos \theta = \sin 2\theta \):
\( R = \frac{u^2 \sin 2\theta}{g} \)
In simple words: When you throw a ball, it goes up and then comes down in a curved path. We can use math to figure out how high it will go, how far it will land, and for how long it stays in the air. These calculations help us understand how thrown objects move.

🎯 Exam Tip: Be careful with signs (positive/negative) for vertical motion, especially for acceleration due to gravity. Clearly label initial velocity components and use correct kinematic equations for horizontal and vertical motion separately.

 

Question 6. Derive the expression for centripetal acceleration.
Answer: When an object moves in a circle at a steady speed (uniform circular motion), its velocity is always changing direction, even if its speed stays the same. This change in direction means there's an acceleration. This acceleration always points towards the center of the circle and is called centripetal acceleration. Centripetal acceleration is essential for keeping satellites in orbit and cars turning on a curve without skidding.
**Expression for Centripetal Acceleration:**
We can find the formula for centripetal acceleration using simple geometry. We look at how the object's position and velocity vectors change over a very short time, \( \Delta t \), with a small angle \( \theta \).
For uniform circular motion, the magnitudes of position vectors are equal \( r = |\overline{r_1}| = |\overline{r_2}| \), and the magnitudes of velocity vectors are equal \( v = |\overline{v_1}| = |\overline{v_2}| \).
If the particle moves from position vector \( \overline{r_1} \) to \( \overline{r_2} \), the displacement is given by \( \overrightarrow{\Delta r} = \overline{r_2} – \overline{r_1} \).

And the change in velocity from \( \overline{v_1} \) to \( \overline{v_2} \) is given by \( \Delta \overline{v} = \overline{v_2} - \overline{v_1} \).

The magnitudes of the displacement \( \Delta r \) and \( \Delta v \) are related by:
\( \frac{|\Delta r|}{r} = \frac{|\Delta v|}{v} = \theta \) (for small angles)

Here, the negative sign indicates that \( \Delta v \) points radially inwards, towards the center of the circle. This is why centripetal acceleration is always directed inwards.
From the relation: \( \Delta v = -v \left(\frac{\Delta r}{r}\right) \)
Acceleration \( a \) is defined as \( a = \frac{\Delta v}{\Delta t} \).
\( \implies a = \frac{-v}{r} \frac{\Delta r}{\Delta t} \)
As \( \Delta t \rightarrow 0 \), \( \frac{\Delta r}{\Delta t} \) becomes the speed \( v \).
\( \implies a = \frac{-v}{r} (v) \)
\( \implies a = -v^2/r \)
The magnitude of centripetal acceleration is \( a_c = \frac{v^2}{r} \).
For uniform circular motion, we can also link linear speed 'v' to angular speed '\(\omega\)' using the radius 'r' of the circle. Angular speed tells us how fast the angle changes.
We know that \( v = r\omega \).
Substituting this into the centripetal acceleration formula:
\( a_c = \frac{(r\omega)^2}{r} \)
\( a_c = \frac{r^2\omega^2}{r} \)
\( a_c = \omega^2r \)
In simple words: When something goes in a circle, even if its speed is steady, its direction keeps changing. This change in direction means it is accelerating, and this acceleration always points to the center of the circle, pulling the object inwards. This is called centripetal acceleration.

🎯 Exam Tip: Clearly define uniform circular motion and explain why acceleration occurs even with constant speed. Remember the two forms of the formula for centripetal acceleration (using linear speed and angular speed).

 

Question 7. Derive the expression for total acceleration in the non-uniform circular motion.
Answer: In non-uniform circular motion, an object's velocity changes in both its speed and its direction. This means the object experiences two types of acceleration: centripetal acceleration, which changes direction, and tangential acceleration, which changes speed. This combined acceleration helps describe the complex motion of a car speeding up while turning a corner.

When a particle executes non-uniform circular motion, its velocity changes both in magnitude and direction. Therefore, the particle will have two components of acceleration:
1. **Centripetal acceleration (\( a_c \))**: This component is responsible for changing the direction of the velocity. It always points towards the center of the circular path and has a magnitude of \( a_c = \frac{v^2}{r} \).
2. **Tangential acceleration (\( a_t \))**: This component is responsible for changing the magnitude (speed) of the velocity. It acts along the tangent to the circular path and has a magnitude of \( a_t = \frac{dv}{dt} \).

The total (resultant) acceleration (\( a_R \)) of the particle is the vector sum of the centripetal acceleration and the tangential acceleration. Since these two components are perpendicular to each other, we can find the magnitude of the resultant acceleration using the Pythagorean theorem.

The magnitude of the resultant acceleration is:
\( a_R = \sqrt{a_c^2 + a_t^2} \)
Substituting the expressions for \( a_c \) and \( a_t \):
\( a_R = \sqrt{\left(\frac{v^2}{r}\right)^2 + \left(\frac{dv}{dt}\right)^2} \)
This expression gives the total acceleration of a particle undergoing non-uniform circular motion.
In simple words: When something moves in a circle but also changes its speed, it has two accelerations. One pulls it towards the center (centripetal), and the other pushes it along the circle (tangential). The total acceleration is a mix of these two.

🎯 Exam Tip: Clearly define both centripetal and tangential acceleration components. Emphasize that they are perpendicular, allowing the use of the Pythagorean theorem for the resultant magnitude.

IV. Exercises:

 

Question 1. The position vector particle has a length of 1m and makes 30° with the x-axis what are the lengths of x and y components of the position vector?
Answer: The position vector has a length (magnitude) of 1 meter. It forms an angle of 30 degrees with the x-axis. We need to find its x and y parts. The x-component is found by multiplying the length by the cosine of the angle. So, \( I_x = 1 \times \cos(30^\circ) = \frac{\sqrt{3}}{2} \) meters. This means how far it extends along the x-direction. The y-component is found by multiplying the length by the sine of the angle. So, \( I_y = 1 \times \sin(30^\circ) = \frac{1}{2} \) meter. This shows how far it extends along the y-direction. We use trigonometry to break down a single vector into its horizontal and vertical effects.
In simple words: For a vector of length 1m at 30° to the x-axis, the x-part is \( 1 \times \cos(30^\circ) \) and the y-part is \( 1 \times \sin(30^\circ) \).

🎯 Exam Tip: Remember that the x-component of a vector is calculated using cosine, and the y-component using sine, when the angle is measured from the positive x-axis.

 

Question 2. A particle has its position moved from \(\left|\bar{r}_{1}\right| = 3\hat{i} + 4\hat{j}\) to \(\left|\bar{r}_{2}\right| = \hat{i}+ 2\hat{j}\) calculate the displacement vector \((\Delta \vec{r})\) and draw the \(\left|\bar{r}_{1}\right|\), \(\left|\bar{r}_{2}\right|\) and \((\Delta \vec{r})\) vector in a two dimensional Cartesian co-ordinate system.
Answer: The particle's first position vector is \( \vec{r_1} = 3\hat{i} + 4\hat{j} \). This means it is 3 units in the x-direction and 4 units in the y-direction from the origin. The second position vector is \( \vec{r_2} = \hat{i} + 2\hat{j} \), meaning 1 unit in x and 2 units in y. To find the displacement vector, we subtract the initial position from the final position.
\( \Delta \vec{r} = \vec{r_2} - \vec{r_1} \)
\( \Delta \vec{r} = (\hat{i} + 2\hat{j}) - (3\hat{i} + 4\hat{j}) \)
\( \Delta \vec{r} = (1-3)\hat{i} + (2-4)\hat{j} \)
\( \Delta \vec{r} = -2\hat{i} - 2\hat{j} \) This result indicates the overall change in the particle's position, showing it moved 2 units in the negative x-direction and 2 units in the negative y-direction.
In simple words: To find how much a particle's position changed, subtract its starting position vector from its ending position vector. This new vector is called the displacement.

🎯 Exam Tip: Always subtract the initial position vector from the final position vector to correctly calculate displacement, ensuring vector components are handled separately.

 

Question 3. Calculate the average velocity of the particle whose position vector changes from \(\left|\bar{r}_{1}\right| = 5\hat{i} + 6\hat{j}\) to \(\left|\bar{r}_{2}\right| = 2\hat{i} + 3\hat{j}\) in a time 5 seconds.
Answer: First, we find the change in position (displacement) by subtracting the initial position vector from the final position vector. Initial position: \( \vec{r_1} = 5\hat{i} + 6\hat{j} \) Final position: \( \vec{r_2} = 2\hat{i} + 3\hat{j} \) Time taken: \( \Delta t = 5 \) seconds. The displacement is \( \Delta \vec{r} = \vec{r_2} - \vec{r_1} \).
\( \Delta \vec{r} = (2\hat{i} + 3\hat{j}) - (5\hat{i} + 6\hat{j}) \)
\( \Delta \vec{r} = (2-5)\hat{i} + (3-6)\hat{j} \)
\( \Delta \vec{r} = -3\hat{i} - 3\hat{j} \) Next, we calculate the average velocity by dividing the displacement by the time taken.
Average velocity \( \vec{V}_{ave} = \frac{\Delta \vec{r}}{\Delta t} \)
\( \vec{V}_{ave} = \frac{-3\hat{i} - 3\hat{j}}{5} \)
\( \vec{V}_{ave} = -\frac{3}{5}(\hat{i} + \hat{j}) \) This means the particle moved in a direction of negative x and negative y, with an average speed related to the magnitude of this vector.
In simple words: Average velocity is found by dividing the total change in position (displacement) by the total time taken.

🎯 Exam Tip: Remember that average velocity is a vector quantity, so it has both magnitude and direction, and you must correctly calculate both the displacement vector and divide by the scalar time.

 

Question 4. Convert the vector \(\overline{r} = 3\hat{i} + 2\hat{j}\) into a unit vector.
Answer: To convert a vector into a unit vector, we need to divide the vector by its own magnitude. The given vector is \( \vec{r} = 3\hat{i} + 2\hat{j} \). First, find the magnitude of the vector:
\( |\vec{r}| = \sqrt{(\text{x-component})^2 + (\text{y-component})^2} \)
\( |\vec{r}| = \sqrt{3^2 + 2^2} \)
\( |\vec{r}| = \sqrt{9 + 4} \)
\( |\vec{r}| = \sqrt{13} \) Now, divide the vector by its magnitude to get the unit vector \( \hat{r} \):
\( \hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{3\hat{i} + 2\hat{j}}{\sqrt{13}} \) A unit vector always has a magnitude of 1 and points in the same direction as the original vector, simplifying direction representation.
In simple words: A unit vector shows the direction of a vector and has a length of 1. You get it by dividing the vector by its own length.

🎯 Exam Tip: Always calculate the magnitude of the vector correctly before dividing, as an error in magnitude will lead to an incorrect unit vector.

 

Question 5. Find the vector product of two given vectors \(\overline{A} = 4\hat{i} – 2\hat{j} + \hat{k}\) and \(\overline{B} = 5\hat{i} + 3\hat{j} – 4\hat{k}\).
Answer: To find the vector product (or cross product) of two vectors \( \vec{A} \) and \( \vec{B} \), we set up a determinant using their components. Given vectors are:
\( \vec{A} = 4\hat{i} - 2\hat{j} + \hat{k} \)
\( \vec{B} = 5\hat{i} + 3\hat{j} - 4\hat{k} \) We arrange the components in a 3x3 determinant: \[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -2 & 1 \\ 5 & 3 & -4 \end{vmatrix} \] Now, we expand the determinant:
\( = \hat{i}((-2) \times -4 - 1 \times 3) \)
\( - \hat{j}(4 \times -4 - 1 \times 5) \)
\( + \hat{k}(4 \times 3 - (-2) \times 5) \)
\( = \hat{i}(8 - 3) - \hat{j}(-16 - 5) + \hat{k}(12 - (-10)) \)
\( = 5\hat{i} - \hat{j}(-21) + \hat{k}(12 + 10) \)
\( = 5\hat{i} + 21\hat{j} + 22\hat{k} \) The vector product is a new vector that is perpendicular to both original vectors, and its magnitude represents the area of the parallelogram formed by them.
In simple words: The vector product of two vectors is found using a special calculation called a determinant, which gives you a new vector that is at right angles to both original vectors.

🎯 Exam Tip: Be careful with the signs when expanding the determinant for the cross product, especially with the minus sign for the \( \hat{j} \) component.

 

Question 6. An object at an angle such that the horizontal range is 4 times the maximum height. What is the angle of projection of the object?
Answer: We are given that the horizontal range (R) of a projectile is four times its maximum height (\( H_{max} \)). We need to find the angle at which the object was launched. The formula for horizontal range is \( R = \frac{u^2 \sin(2\theta)}{g} \). The formula for maximum height is \( H_{max} = \frac{u^2 \sin^2(\theta)}{2g} \). Using the given condition \( R = 4 H_{max} \):
\( \frac{u^2 \sin(2\theta)}{g} = 4 \times \frac{u^2 \sin^2(\theta)}{2g} \)
\( \implies \frac{u^2 \sin(2\theta)}{g} = \frac{2 u^2 \sin^2(\theta)}{g} \) We can cancel \( \frac{u^2}{g} \) from both sides.
\( \implies \sin(2\theta) = 2 \sin^2(\theta) \) Using the trigonometric identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\( \implies 2 \sin(\theta) \cos(\theta) = 2 \sin^2(\theta) \) Divide both sides by \( 2 \sin(\theta) \) (assuming \( \theta \) is not 0 or 180 degrees, where there would be no projectile motion).
\( \implies \cos(\theta) = \sin(\theta) \)
\( \implies \frac{\sin(\theta)}{\cos(\theta)} = 1 \)
\( \implies \tan(\theta) = 1 \) The angle whose tangent is 1 is \( 45^\circ \). So, \( \theta = 45^\circ \). This angle is special because it balances the horizontal and vertical motions to achieve the greatest range for a given initial speed.
In simple words: When the range of a launched object is four times its highest point, the object was thrown at a 45-degree angle.

🎯 Exam Tip: Always remember the standard formulas for projectile range and maximum height, and know the key trigonometric identities like \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \) to simplify equations.

 

Question 7. The following graphs represent velocity-time graph. Identify what kind of motion a particle undergoes in each graph.
Answer: Let's analyze each velocity-time graph to understand the type of motion it shows.
(a) This graph shows a straight line starting from the origin and sloping upwards. This means the object starts from rest (velocity = 0 at time = 0) and its velocity increases steadily. So, it represents motion with **constant positive acceleration**.
(b) This graph shows a horizontal straight line above the time axis. This means the velocity is not changing over time. So, it represents motion with **constant velocity (zero acceleration)**. The object moves at the same speed in the same direction.
(c) This graph shows a straight line sloping upwards but it does not start from the origin. This indicates the object starts with some initial velocity (not zero) and its velocity increases at a constant rate. So, it represents motion with **constant positive acceleration**, starting from an initial velocity.
(d) This graph shows a curve that becomes steeper over time. This means the rate at which velocity is changing is increasing. So, it represents motion with **increasing acceleration**. The object is speeding up faster and faster.
In simple words: The shape of a velocity-time graph tells you if an object is moving at a steady speed, speeding up, slowing down, or changing its acceleration.

🎯 Exam Tip: The slope of a velocity-time graph represents acceleration, and the area under the graph represents displacement. A straight line indicates constant acceleration, while a curve indicates varying acceleration.

 

Question 8. The following velocity-time graph represents a particle moving in the positive x-direction. Analyse its motion from o to 7s calculate the displacement covered and distance traveled by the particle from 0 to 2s.
Answer: Let's analyze the motion of the particle based on the given velocity-time graph from 0 to 7 seconds._m/s 0 1 2 3 4 5 6 7 1 2 -1 -2 B C D E A
**Analysis of Motion:** * **From 0s to 1s (Segment O to A):** The velocity changes from 0 m/s to -2 m/s. This means the particle accelerates in the negative x-direction. The acceleration is \( \frac{-2 - 0}{1 - 0} = -2 \text{ m/s}^2 \). * **From 1s to 2s (Segment A to B):** The velocity changes from -2 m/s to 1 m/s. The particle is slowing down in the negative direction, then momentarily stopping, and finally speeding up in the positive direction. The acceleration is \( \frac{1 - (-2)}{2 - 1} = 3 \text{ m/s}^2 \). * **From 2s to 5s (Segment B to C):** The velocity remains constant at 1 m/s. This means the acceleration is zero. The particle moves at a steady speed in the positive x-direction. * **From 5s to 6s (Segment C to D):** The velocity changes from 1 m/s to 0 m/s. The particle is decelerating (slowing down) in the positive x-direction until it stops. The acceleration is \( \frac{0 - 1}{6 - 5} = -1 \text{ m/s}^2 \). * **From 6s to 7s (Segment D to E):** The velocity is zero. The particle is at rest. **Displacement and Distance from 0 to 2s:** Displacement is the total signed area under the velocity-time graph. Distance is the total absolute area. Area from 0s to 1s (Triangle below axis): Base = 1s, Height = -2 m/s.
Area\(_{0-1} = \frac{1}{2} \times 1 \times (-2) = -1 \text{ m}\). To find the area from 1s to 2s, we first find the time \( t_0 \) when velocity is zero. The line connects (1, -2) and (2, 1). Using the equation of a line \( v - v_1 = m(t - t_1) \):
\( v - (-2) = \frac{1 - (-2)}{2 - 1}(t - 1) \)
\( v + 2 = 3(t - 1) \)
\( v = 3t - 5 \) Setting \( v = 0 \): \( 0 = 3t_0 - 5 \implies t_0 = \frac{5}{3} \text{ s} \). Area from 1s to \( \frac{5}{3} \)s (Triangle below axis): Base = \( \frac{5}{3} - 1 = \frac{2}{3} \)s, Height = -2 m/s.
Area\(_{1-5/3} = \frac{1}{2} \times \frac{2}{3} \times (-2) = -\frac{2}{3} \text{ m}\). Area from \( \frac{5}{3} \)s to 2s (Triangle above axis): Base = \( 2 - \frac{5}{3} = \frac{1}{3} \)s, Height = 1 m/s.
Area\(_{5/3-2} = \frac{1}{2} \times \frac{1}{3} \times 1 = \frac{1}{6} \text{ m}\). Total Displacement (0 to 2s):
\( \Delta x = \text{Area}_{0-1} + \text{Area}_{1-5/3} + \text{Area}_{5/3-2} = -1 - \frac{2}{3} + \frac{1}{6} = -\frac{6}{6} - \frac{4}{6} + \frac{1}{6} = -\frac{9}{6} = -1.5 \text{ m}\). Total Distance Traveled (0 to 2s):
\( D = |\text{Area}_{0-1}| + |\text{Area}_{1-5/3}| + |\text{Area}_{5/3-2}| = |-1| + |-\frac{2}{3}| + |\frac{1}{6}| \)
\( D = 1 + \frac{2}{3} + \frac{1}{6} = \frac{6+4+1}{6} = \frac{11}{6} \approx 1.83 \text{ m}\).
In simple words: To understand motion from a velocity-time graph, look at the slope for acceleration and the area for displacement. Displacement considers direction, while distance is the total path covered.

🎯 Exam Tip: When calculating displacement and distance from a velocity-time graph, remember that areas below the time axis contribute negatively to displacement but positively to total distance.

 

Question 9. A particle is projected at an angle of \( \theta \) with respect to the horizontal direction. Match the following for the above motion.
(a) \( v_x \) - decreases and increases
(b) \( v_y \) – remains constant
(c) Acceleration – varies
(d) Position vector – remains downwards
Answer: In projectile motion, when a particle is launched at an angle to the horizontal, these are its characteristics:
(a) The horizontal component of velocity, \( v_x \), **remains constant** throughout the motion (ignoring air resistance). This is because there is no horizontal force acting on the projectile.
(b) The vertical component of velocity, \( v_y \), **decreases and increases**. It decreases as the projectile moves upwards, becomes zero at the highest point, and then increases (in magnitude) as it falls downwards.
(c) The acceleration, \( \vec{a} \), **remains constant and points downwards**. This acceleration is due to gravity (g).
(d) The position vector, \( \vec{r} \), **varies**. It continuously changes as the particle moves through its trajectory.
In simple words: When an object is thrown, its horizontal speed stays the same, its vertical speed goes up and down, its acceleration is always constant and downwards, and its position changes all the time.

🎯 Exam Tip: For projectile motion (ignoring air resistance), acceleration is always 'g' downwards, and the horizontal velocity component is constant, which are critical concepts.

 

Question 10. A water fountain on the ground sprinkles water all around it. If the speed of the water coming out of the fountains is V. Calculate the total area around the fountain that gets wet.
Answer: When a water fountain sprinkles water, the water streams act like projectiles. The maximum distance the water can reach from the fountain is the maximum range of a projectile. If the initial speed of the water is V, the maximum horizontal range (\( R_{max} \)) for a projectile happens when the launch angle is 45 degrees. The formula for this maximum range is:
\( R_{max} = \frac{V^2}{g} \) Here, 'g' is the acceleration due to gravity. This maximum range defines the radius of the circular area that gets wet around the fountain. So, the radius of the wet area is \( r = \frac{V^2}{g} \). The total area covered is the area of a circle with this radius:
Area \( A = \pi r^2 \) Substitute the expression for \( r \):
\( A = \pi \left(\frac{V^2}{g}\right)^2 \)
\( A = \pi \frac{V^4}{g^2} \) This calculation helps us understand the reach of a fountain based on its water speed, showing a wider area for higher initial speeds.
In simple words: The largest circle a fountain can wet has a radius equal to the maximum distance the water can travel, which depends on its speed and gravity. The total wet area is then calculated using this radius.

🎯 Exam Tip: Remember that the maximum range for a projectile (and thus the radius of the wet area) is achieved at a launch angle of 45 degrees, and its formula is \( R_{max} = \frac{u^2}{g} \).

 

Question 11. The following table gives the range of the particle when thrown on different planets. All the particles are thrown at the same angle with the horizontal and with the same initial speed. Arrange the planets in ascending order according to their acceleration due to gravity (g value)
Answer: We are comparing projectile motion on different planets. We know the initial speed (\( u \)) and launch angle (\( \theta \)) are the same for all tests. The formula for the horizontal range (R) of a projectile is:
\( R = \frac{u^2 \sin(2\theta)}{g} \) Since \( u \) and \( \theta \) are constant, the term \( u^2 \sin(2\theta) \) is also constant. Let's call this constant C. So, we can write \( R = \frac{C}{g} \). This means that the acceleration due to gravity (g) is inversely proportional to the range (R). In simpler terms, if the range is small, gravity is strong, and if the range is large, gravity is weak. We need to arrange the planets by their 'g' values from smallest to largest (ascending order). This means we need to arrange them by their 'R' values from largest to smallest (descending order). Here are the ranges given:

Ordering the ranges from largest to smallest: Mercury (95m) > Mars (90m) > Earth (75m) > Jupiter (50m) Therefore, ordering the planets by their 'g' values from smallest to largest is: Mercury, Mars, Earth, Jupiter. A planet with a smaller range has stronger gravity.
In simple words: Since the range of a thrown object depends on gravity, planets where the object travels further have weaker gravity, and planets where it travels less far have stronger gravity. We ordered planets by gravity from weakest to strongest.

🎯 Exam Tip: When dealing with inverse proportionality, a larger value of one quantity corresponds to a smaller value of the other, and vice versa. Always clarify the order (ascending or descending) requested in the question.

 

Question 12. The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B. Then the angle between A and B is
(a) 30°
(b) 45°
(c) 150°
(d) 120°
Answer: The correct answer is **(c) 150°** Let \( \vec{A} \) and \( \vec{B} \) be two vectors. Their resultant vector is \( \vec{R} = \vec{A} + \vec{B} \). We are given two conditions: 1. The resultant vector \( \vec{R} \) is perpendicular to vector \( \vec{A} \). This means the angle (\( \alpha \)) between \( \vec{R} \) and \( \vec{A} \) is \( 90^\circ \). The formula for the angle \( \alpha \) is \( \tan(\alpha) = \frac{B \sin(\theta)}{A + B \cos(\theta)} \), where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \). Since \( \alpha = 90^\circ \), \( \tan(90^\circ) \) is infinitely large, which means the denominator must be zero:
\( A + B \cos(\theta) = 0 \)
\( \implies \cos(\theta) = -\frac{A}{B} \) (Equation 1) 2. The magnitude of the resultant \( |\vec{R}| \) is half the magnitude of vector \( \vec{B} \). So, \( R = \frac{B}{2} \). The magnitude of the resultant is also given by \( R = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \). Substitute \( R = \frac{B}{2} \) into this equation and square both sides:
\( \left(\frac{B}{2}\right)^2 = A^2 + B^2 + 2AB \cos(\theta) \)
\( \implies \frac{B^2}{4} = A^2 + B^2 + 2AB \cos(\theta) \) (Equation 2) Now, substitute \( \cos(\theta) = -\frac{A}{B} \) from Equation 1 into Equation 2:
\( \frac{B^2}{4} = A^2 + B^2 + 2AB \left(-\frac{A}{B}\right) \)
\( \implies \frac{B^2}{4} = A^2 + B^2 - 2A^2 \)
\( \implies \frac{B^2}{4} = B^2 - A^2 \) Rearranging to find A:
\( A^2 = B^2 - \frac{B^2}{4} \)
\( \implies A^2 = \frac{3B^2}{4} \)
\( \implies A = \frac{\sqrt{3}}{2} B \) (Equation 3) Finally, substitute this value of A back into Equation 1:
\( \cos(\theta) = -\frac{\frac{\sqrt{3}}{2} B}{B} \)
\( \implies \cos(\theta) = -\frac{\sqrt{3}}{2} \) The angle \( \theta \) for which the cosine is \( -\frac{\sqrt{3}}{2} \) is \( 150^\circ \). This type of problem often involves using the properties of vector addition and trigonometry to find unknown angles or magnitudes.
In simple words: When the combined effect of two forces is at a right angle to one force, and its strength is half of the other force, the angle between the two original forces is 150 degrees.

🎯 Exam Tip: For vector problems involving resultants and specific angle conditions, it's crucial to use both the magnitude and direction formulas for vector addition and be proficient with trigonometric identities.

 

Question 13. Compare the components for the following vector equations.
(a) \( T\hat{j} – mg\hat{j} = ma\hat{j} \)
(b) \( \overline{T} + \overline{F} = \overline{A} + \overline{B} \)
(c) \( \overline{T} – \overline{F} = \overline{A} – \overline{B} \)
(d) \( T\hat{j} + mg\hat{j} = ma\hat{j} \)
Answer: To compare components in vector equations, we break each vector into its x, y, and z parts (components). Then, we equate the corresponding components on both sides of the equation.
(a) For the equation \( T\hat{j} – mg\hat{j} = ma\hat{j} \): This equation only has a y-component. By matching the terms that have \( \hat{j} \), we get:
\( T - mg = ma \)
(b) For the equation \( \vec{T} + \vec{F} = \vec{A} + \vec{B} \): Let \( \vec{T} = T_x\hat{i} + T_y\hat{j} + T_z\hat{k} \), \( \vec{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{k} \), and so on for \( \vec{A} \) and \( \vec{B} \). Equating the components: - x-components: \( T_x + F_x = A_x + B_x \) - y-components: \( T_y + F_y = A_y + B_y \) - z-components: \( T_z + F_z = A_z + B_z \)
(c) For the equation \( \vec{T} – \vec{F} = \vec{A} – \vec{B} \): Equating the components: - x-components: \( T_x - F_x = A_x - B_x \) - y-components: \( T_y - F_y = A_y - B_y \) - z-components: \( T_z - F_z = A_z - B_z \)
(d) For the equation \( T\hat{j} + mg\hat{j} = ma\hat{j} \): This equation also only has a y-component. By matching the terms with \( \hat{j} \), we get:
\( T + mg = ma \) Breaking down vectors into components is a key step in solving physics problems in multiple dimensions.
In simple words: To solve vector equations, you look at the parts of each vector (x, y, and z components) separately and make sure they are equal on both sides of the equation.

🎯 Exam Tip: Always ensure consistency in component representation (e.g., all x-components on one side, all y-components on another) to avoid errors when comparing vector equations.

 

Question 14. Calculate the area of the triangle for which two of its sides are given by the vectors \(\overline{A} = 5\hat{i} – 3\hat{j}\) and \(\overline{B} = 4\hat{i} + 6\hat{j}\).
Answer: To find the area of a triangle when two of its sides are given as vectors, we first calculate the cross product of these two vectors. Then, we find the magnitude of this cross product and divide it by two. The given vectors are:
\( \vec{A} = 5\hat{i} - 3\hat{j} \)
\( \vec{B} = 4\hat{i} + 6\hat{j} \) We can write these in 3D form with a zero z-component: \( \vec{A} = 5\hat{i} - 3\hat{j} + 0\hat{k} \) and \( \vec{B} = 4\hat{i} + 6\hat{j} + 0\hat{k} \). The cross product \( \vec{A} \times \vec{B} \) is: \[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -3 & 0 \\ 4 & 6 & 0 \end{vmatrix} \]
\( = \hat{i}((-3 \times 0) - (0 \times 6)) - \hat{j}((5 \times 0) - (0 \times 4)) + \hat{k}((5 \times 6) - (-3 \times 4)) \)
\( = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}(30 - (-12)) \)
\( = 0\hat{i} - 0\hat{j} + \hat{k}(30 + 12) \)
\( = 42\hat{k} \) The magnitude of this cross product is \( |\vec{A} \times \vec{B}| = |42\hat{k}| = 42 \). The area of the triangle is half of this magnitude:
Area \( = \frac{1}{2} \times 42 = 21 \text{ square meters} \). The cross product provides a vector whose magnitude is the area of the parallelogram formed by the two vectors.
In simple words: To find the area of a triangle from two vectors, first calculate their cross product, then find the length of that resulting vector, and finally divide that length by two.

🎯 Exam Tip: Remember that the magnitude of the cross product of two vectors gives the area of the parallelogram formed by them, so for a triangle, it's half of that value.

 

Question 15. If the earth completes one revolution in 24 hours, what is the angular displacement made by the earth in one hour? Express your answer in both radian and degree.
Answer: The Earth completes one full revolution in 24 hours. A full revolution is equal to an angular displacement of \( 360^\circ \) or \( 2\pi \) radians. We need to find the angular displacement for just one hour. To find the angular displacement in degrees:
Angular displacement \( = \frac{\text{Total degrees in one revolution}}{\text{Total hours in one revolution}} \times \text{Hours needed} \)
Angular displacement \( = \frac{360^\circ}{24 \text{ hours}} \times 1 \text{ hour} = 15^\circ \) To find the angular displacement in radians:
Angular displacement \( = \frac{\text{Total radians in one revolution}}{\text{Total hours in one revolution}} \times 1 \text{ hour} = \frac{2\pi \text{ radians}}{24 \text{ hours}} \times 1 \text{ hour} = \frac{\pi}{12} \text{ radians} \). This means the Earth rotates by a small amount each hour, which is key for understanding time zones and how different parts of the planet experience daylight.
In simple words: The Earth turns 360 degrees, or \( 2\pi \) radians, in 24 hours. So, in just one hour, it turns 15 degrees or \( \frac{\pi}{12} \) radians.

🎯 Exam Tip: Always remember the conversion between degrees and radians: \( 180^\circ = \pi \) radians. This allows you to easily convert angular measurements between the two units.

 

Question 16. An object is thrown with initial speed of 5ms\(^{-1}\) with an angle of projection of 30°. What is the height and range reached by the particle?
Answer: First, let's find the maximum height the object reaches. We use the formula for maximum height in projectile motion: \( h_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g} \). Here, the initial speed \( u \) is 5 m/s, the projection angle \( \theta \) is 30°, and the acceleration due to gravity \( g \) is 9.8 m/s\(^2\). Putting these values into the formula, we calculate \( h_{\text{max}} = \frac{(5 \, \text{m/s})^2 \times (\sin 30^\circ)^2}{2 \times 9.8 \, \text{m/s}^2} = \frac{25 \times (0.5)^2}{19.6} = \frac{6.25}{19.6} \approx 0.318 \, \text{m} \). So, the object reaches a maximum height of about 0.318 meters. This calculation helps us understand the peak elevation of the object's path.
In simple words: To find how high it goes, we use a special height formula with its speed and angle. The object will reach about 0.318 meters high.

🎯 Exam Tip: Remember to correctly use the square of the sine function in the height formula, not just sine squared after calculating the angle. Ensure 'g' is consistently 9.8 m/s² unless otherwise specified.

 

Question 16. An object is thrown with initial speed of 5ms\(^{-1}\) with an angle of projection of 30°. What is the height and range reached by the particle?
Answer: (Continued) Next, we calculate the horizontal range, which is how far the object travels horizontally before landing. The formula for horizontal range is \( R = \frac{u^2 \sin 2\theta}{g} \). Using the same initial speed \( u = 5 \, \text{m/s} \), angle \( \theta = 30^\circ \), and \( g = 9.8 \, \text{m/s}^2 \): we get \( R = \frac{(5 \, \text{m/s})^2 \times \sin (2 \times 30^\circ)}{9.8 \, \text{m/s}^2} = \frac{25 \times \sin 60^\circ}{9.8} \). Since \( \sin 60^\circ \approx 0.866 \), the range is \( R = \frac{25 \times 0.866}{9.8} = \frac{21.65}{9.8} \approx 2.21 \, \text{m} \). So, the object travels about 2.21 meters horizontally. Understanding both height and range helps to fully describe the projectile's motion.
In simple words: Then, we find how far it travels sideways, which is called the range. Using a different formula with the speed and angle, we find it travels about 2.21 meters horizontally.

🎯 Exam Tip: Distinguish between the formulas for maximum height (involving \( \sin^2 \theta \)) and horizontal range (involving \( \sin 2\theta \)) to avoid common calculation errors.

 

Question 17. A football player hits the ball with a speed 20m/s with angle 30° with respect to the horizontal directions. The goal post is at a distance of 40 m from him. Find out whether the ball reaches the goal post.
Answer: To see if the ball reaches the goal post, we need to calculate its horizontal range. The horizontal range \( R \) is given by the formula \( R = \frac{u^2 \sin 2\theta}{g} \). The initial speed \( u \) of the ball is 20 m/s, the projection angle \( \theta \) is 30°, and acceleration due to gravity \( g \) is 9.8 m/s\(^2\). Plugging in the values, we get \( R = \frac{(20 \, \text{m/s})^2 \sin (2 \times 30^\circ)}{9.8 \, \text{m/s}^2} = \frac{400 \sin 60^\circ}{9.8} \). Since \( \sin 60^\circ \approx 0.866 \), the range is \( R = \frac{400 \times 0.866}{9.8} = \frac{346.4}{9.8} \approx 35.35 \, \text{m} \). The calculated range is approximately 35.35 meters. Because 35.35 meters is less than the 40-meter distance to the goal post, the ball will not reach the goal post. This shows that the initial speed and angle are critical for achieving the desired range.

🎯 Exam Tip: Always compare the calculated range with the target distance to determine if a projectile reaches its intended mark. Clearly state your conclusion based on the comparison.

 

Question 18. If an object is thrown horizontally with an initial speed 10 ms\(^{-1}\) from the top of a building of height 100 m. What is the horizontal distance covered by the particle?
Answer: First, we need to find the time it takes for the object to fall to the ground. For horizontal projection, the time of flight \( T \) is determined by the height of the building and gravity, using the formula \( T = \sqrt{\frac{2h}{g}} \). Here, the height \( h \) is 100 m and acceleration due to gravity \( g \) is 9.8 m/s\(^2\). So, \( T = \sqrt{\frac{2 \times 100}{9.8}} = \sqrt{\frac{200}{9.8}} = \sqrt{20.408} \approx 4.518 \, \text{s} \). Once we have the time of flight, the horizontal distance covered \( x \) is found by multiplying the horizontal initial speed \( u_x \) by the time \( T \). The initial horizontal speed \( u_x \) is 10 m/s. Therefore, \( x = u_x \times T = 10 \, \text{m/s} \times 4.518 \, \text{s} = 45.18 \, \text{m} \). The object will travel 45.18 meters horizontally. This calculation helps predict where the object will land.

🎯 Exam Tip: For horizontally projected objects, remember that the horizontal velocity remains constant, and the time of flight depends only on the vertical drop and gravity.

 

Question 19. An object is executing uniform circular motion with an angular speed of \( \pi/12 \) radians per second. At \( t = 0 \) the object starts at an angle \( \theta = 0 \). What is the angular displacement of the particle after 4s?
Answer: We are given the angular speed \( \omega = \pi/12 \, \text{rad/s} \) and the time \( t = 4 \, \text{s} \). To find the angular displacement \( \theta \), we use the formula \( \theta = \omega t \). So, \( \theta = (\pi/12 \, \text{rad/s}) \times (4 \, \text{s}) = \frac{4\pi}{12} = \frac{\pi}{3} \, \text{radians} \). To convert this to degrees, we know that \( \pi \) radians equals 180°. Therefore, \( \theta = \frac{180^\circ}{3} = 60^\circ \). The angular displacement after 4 seconds is \( \frac{\pi}{3} \) radians or 60 degrees. This demonstrates how to calculate the angular distance covered in circular motion.

🎯 Exam Tip: Remember to use radians for angular speed and displacement calculations unless the question specifically asks for degrees, and be careful with unit conversions between radians and degrees.

 

Question 20. Consider the x-axis as representing east, the y-axis as north, and the z-axis as vertically upwards. Give the vector representing each of the following points.
Answer: To represent points as vectors, we use \( \hat{i} \) for the x-direction (East), \( \hat{j} \) for the y-direction (North), and \( \hat{k} \) for the z-direction (Up). This helps to locate positions in 3D space. We will apply this to the specific scenarios presented.

🎯 Exam Tip: Always clearly define your coordinate system axes and their corresponding unit vectors (e.g., \( \hat{i}, \hat{j}, \hat{k} \)) before representing any vector quantity.

(a) 5m northeast and 2m up.
Answer: Northeast means 45° between East (x-axis) and North (y-axis). The magnitude in the XY-plane is 5m. So, the x-component is \( 5 \cos 45^\circ \) and the y-component is \( 5 \sin 45^\circ \). The upward component is 2m in the z-direction.
So, the vector is \( 5 \cos 45^\circ \hat{i} + 5 \sin 45^\circ \hat{j} + 2 \hat{k} \).
Since \( \cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}} \), the vector is \( \frac{5}{\sqrt{2}} \hat{i} + \frac{5}{\sqrt{2}} \hat{j} + 2 \hat{k} = \frac{5(\hat{i} + \hat{j})}{\sqrt{2}} + 2 \hat{k} \). This represents the position in three-dimensional space.
In simple words: For 5m northeast and 2m up, we split the 5m into equal East and North parts (each \( \frac{5}{\sqrt{2}} \) m), then add 2m for the Up direction.

🎯 Exam Tip: For "northeast," remember that it implies equal components along the positive x and y axes, so the angle with the x-axis is 45°.

(b) 4m southeast and 3m up.
Answer: Southeast means 45° from the East (positive x-axis) towards the South (negative y-axis). The magnitude in the XY-plane is 4m. So, the x-component is \( 4 \cos 45^\circ \) and the y-component is \( -4 \sin 45^\circ \). The upward component is 3m in the z-direction.
So, the vector is \( 4 \cos 45^\circ \hat{i} - 4 \sin 45^\circ \hat{j} + 3 \hat{k} \).
Since \( \cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}} \), the vector is \( \frac{4}{\sqrt{2}} \hat{i} - \frac{4}{\sqrt{2}} \hat{j} + 3 \hat{k} = \frac{4(\hat{i} - \hat{j})}{\sqrt{2}} + 3 \hat{k} \). This vector accurately locates the point.
In simple words: For 4m southeast and 3m up, we split the 4m into equal East and South parts (East is positive, South is negative), then add 3m for the Up direction.

🎯 Exam Tip: For directions like "southeast" or "northwest," pay attention to the signs of the components in the x and y directions, as they indicate the quadrant.

(c) 2m northwest and 4m up.
Answer: Northwest means 45° from the North (positive y-axis) towards the West (negative x-axis), or 135° from the positive x-axis. The magnitude in the XY-plane is 2m. So, the x-component is \( -2 \cos 45^\circ \) and the y-component is \( 2 \sin 45^\circ \). The upward component is 4m in the z-direction.
So, the vector is \( -2 \cos 45^\circ \hat{i} + 2 \sin 45^\circ \hat{j} + 4 \hat{k} \).
Since \( \cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}} \), the vector is \( -\frac{2}{\sqrt{2}} \hat{i} + \frac{2}{\sqrt{2}} \hat{j} + 4 \hat{k} = \frac{2(-\hat{i} + \hat{j})}{\sqrt{2}} + 4 \hat{k} \). We can simplify this to \( (-\hat{i} + \hat{j})\sqrt{2} + 4 \hat{k} \). This represents the location precisely.
In simple words: For 2m northwest and 4m up, we split the 2m into equal West (negative) and North (positive) parts, then add 4m for the Up direction.

🎯 Exam Tip: When dealing with directions like "northwest," it's crucial to correctly assign negative signs to components pointing in the negative x or y directions.

 

Question 21. The moon is orbiting the earth approximately once in 27 days. What is the angle transversed by the moon per day?
Answer: The moon completes one full orbit around the Earth in 27 days. One full orbit is an angular displacement of 360° or \( 2\pi \) radians. To find the angle covered per day, we divide the total angle by the total number of days.
Angle per day \( = \frac{\text{Total Angle}}{\text{Total Days}} = \frac{360^\circ}{27} \approx 13.3^\circ \).
In radians, the angle per day \( = \frac{2\pi \, \text{radians}}{27} \approx 0.23 \, \text{radians} \). So, the moon moves about 13.3 degrees or 0.23 radians each day. This shows its average daily angular movement.

🎯 Exam Tip: When calculating angular displacement per unit time, ensure that the total angle (360° or \( 2\pi \) radians for a full circle) is divided by the correct time period.

 

Question 22. An object of mass m has an angular acceleration \( \alpha = 0.2 \, \text{rad/s}^2 \). What is the angular displacement covered by the object after 3 seconds? (Assume that the object started with angle zero with zero angular velocity)
Answer: We are given the angular acceleration \( \alpha = 0.2 \, \text{rad/s}^2 \), time \( t = 3 \, \text{s} \), initial angular position \( \theta_0 = 0 \), and initial angular velocity \( \omega_0 = 0 \). To find the angular displacement \( \theta \), we use the kinematic equation for angular motion: \( \theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2 \). Plugging in the given values, we get \( \theta = 0 + (0 \times 3) + \frac{1}{2} (0.2 \, \text{rad/s}^2) (3 \, \text{s})^2 \).
\( \theta = \frac{1}{2} \times 0.2 \times 9 = 0.1 \times 9 = 0.9 \, \text{radians} \).
To convert this to degrees, we multiply by \( \frac{180^\circ}{\pi} \). So, \( \theta = 0.9 \times \frac{180^\circ}{3.14159} \approx 51.57^\circ \). The angular displacement is 0.9 radians, which is approximately 51.6 degrees. This formula is similar to the linear displacement equation, but for rotational motion.

🎯 Exam Tip: When solving angular motion problems, always ensure initial conditions (like starting angular velocity and position) are correctly incorporated into the kinematic equations.

11th Physics Guide Kinematics Additional Important Questions and Answers

I. Multiple Choice Questions:

 

Question 1. A particle moves in a circle of radius R from A to B as in the figure. The distance and displacement covered.
(a) \( \frac{\pi R}{3}, R \)
(b) \( \frac{\pi^2 R}{3}, R \)
(c) \( \frac{\pi R}{3}, R\sqrt{2} \)
(d) \( \frac{\pi^2 R}{3}, R\sqrt{2} \)
Answer: (a) \( \frac{\pi R}{3}, R \)
In simple words: The distance covered is the arc length, which is \( R \times (\text{angle in radians}) \). Since 60° is \( \pi/3 \) radians, the distance is \( \frac{\pi R}{3} \). The displacement is the straight line connecting points A and B. In an isosceles triangle with two sides R and an angle of 60° between them, the third side is also R, making it an equilateral triangle. So, the displacement is R.

🎯 Exam Tip: For motion along a circular arc, remember that distance is the arc length, while displacement is the straight-line distance between the start and end points.

 

Question 2. The branch of mechanics which deals with the motion of objects without taking force into account is -
(a) kinetics
(b) dynamics
(c) kinematics
(d) statics
Answer: (c) kinematics
In simple words: Kinematics is the study of how things move, focusing on speed, velocity, and acceleration, but it doesn't care about the forces that cause the movement.

🎯 Exam Tip: Understand the key distinction: kinematics describes motion, dynamics relates motion to forces, and statics deals with forces on objects at rest.

 

Question 3. A particle moves in a straight line from A to B with speed \( v_1 \) and then from B to A with speed \( v_2 \). The average velocity and average speed are
(a) \( \frac{2 v_1 v_2}{v_1 + v_2}, 0 \)
(b) \( 0, \frac{2 v_1 v_2}{v_1 + v_2} \)
(c) \( 0, 0 \)
(d) \( \frac{2 v_1 v_2}{v_1 + v_2}, \frac{v_1 v_2}{v_1 + v_2} \)
Answer: (b) \( 0, \frac{2 v_1 v_2}{v_1 + v_2} \)
In simple words: Average velocity is total displacement divided by total time. Since the particle returns to its starting point (A), its total displacement is zero, so average velocity is zero. Average speed is total distance divided by total time. For travel at two different speeds over the same distance, the average speed is calculated as \( \frac{2 v_1 v_2}{v_1 + v_2} \).

🎯 Exam Tip: Always remember that average velocity depends on displacement, while average speed depends on total distance covered. If an object returns to its starting point, its displacement is zero.

 

Question 4. A particle is moving in a straight line under constant acceleration. It travels 15m in the 3rd second and 31m in the 7th second. The initial velocity and acceleration are
(a) 5 m/s, 4 m/s\(^2\)
(b) 4 m/s, 5 m/s\(^2\)
(c) 4 m/s, 4 m/s\(^2\)
(d) 5 m/s, 5 m/s\(^2\)
Answer: (a) 5 m/s, 4 m/s\(^2\)
In simple words: We use a formula for distance traveled in the nth second, \( S_n = u + a(n - \frac{1}{2}) \). Setting up equations for the 3rd and 7th seconds and solving them gives the initial velocity \( u \) as 5 m/s and acceleration \( a \) as 4 m/s\(^2\).

🎯 Exam Tip: The formula for distance covered in the nth second, \( S_n = u + a(n - \frac{1}{2}) \), is very useful for problems involving constant acceleration over specific time intervals.

 

Question 5. If an object is moving in a straight line then the motion is known as -
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer: (a) linear motion
In simple words: When an object moves along a straight path, we call that linear motion. It's the simplest type of movement.

🎯 Exam Tip: Linear motion is defined by movement along a straight path, in contrast to circular, curvilinear (curved path), or rotational motion.

 

Question 6. A car is moving at a constant speed of 15 m/s. Suddenly the driver sees an obstacle on the road and takes 0.4 s to apply the brake. The brake causes a deceleration of 5 m/s². The distance traveled by car before it stops
(a) 6 m
(b) 22.5 m
(c) 28.5 m
(d) 16.2 m
Answer: (c) 28.5 m
In simple words: First, the car travels for 0.4 seconds at 15 m/s (6 m) before the brakes are applied. Then, using equations of motion, we find the distance traveled while decelerating at 5 m/s\(^2\) until it stops. Adding these distances gives the total.

🎯 Exam Tip: This is a two-part problem: first, calculate the distance covered during the reaction time (constant velocity), then the distance covered during deceleration (uniformly decelerated motion), and finally sum them up.

 

Question 7. A car accelerates from rest at a constant rate \( \alpha \) for some time after which it decelerates at a constant rate \( \beta \) to come to rest. If the total time lapses in 't' seconds, then the maximum velocity reached is
(a) \( \frac{\alpha \beta t}{\alpha + \beta} \)
(b) \( \frac{\alpha^2 \beta t^2}{\alpha + \beta} \)
(c) \( \frac{\alpha \beta t^2}{\alpha + \beta} \)
(d) \( \frac{\alpha \beta t^2}{2(\alpha + \beta)} \)
Answer: (a) \( \frac{\alpha \beta t}{\alpha + \beta} \)
In simple words: Imagine the car speeding up and then slowing down. If it takes a total time \( t \), its top speed depends on both how fast it can speed up \( \alpha \) and how fast it can slow down \( \beta \). The formula provided tells us this maximum speed.

🎯 Exam Tip: For problems involving acceleration and deceleration over a total time, drawing a velocity-time graph (a triangle) can often simplify the calculation of maximum velocity.

 

Question 8. Spinning of the earth about its own axis is known as -
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer: (d) rotational motion
In simple words: When something spins around its own center, like the Earth turning, that's called rotational motion.

🎯 Exam Tip: Differentiate between rotational motion (an object spinning about an internal axis) and circular motion (an object moving in a circle around an external point).

 

Question 9. A particle is thrown vertically up with a speed of 40m/s, The velocity at half of the maximum height
(a) 20 m/s
(b) \( 20\sqrt{2} \) m/s
(c) 10 m/s
(d) \( 10\sqrt{2} \) m/s
Answer: (b) \( 20\sqrt{2} \) m/s
In simple words: If you throw a ball up at 40 m/s, to find its speed when it's halfway to its highest point, we use physics formulas. The speed will be \( 20\sqrt{2} \) m/s.

🎯 Exam Tip: Use the kinematic equation \( v^2 = u^2 + 2as \) for vertical motion. Remember that at maximum height, the final velocity is zero, and the acceleration is \( -g \).

 

Question 10. The ratio of the numerical values of the average velocity and the average speed of the body is always
(a) unity
(b) unity or less
(c) unity or more
(d) less than unity
Answer: (b) unity or less
In simple words: Average velocity is based on displacement (shortest path), while average speed is based on total distance traveled. Distance is always equal to or greater than displacement, so the average velocity can be equal to or less than the average speed.

🎯 Exam Tip: Always remember that the magnitude of displacement can never be greater than the total distance traveled, which directly implies the relationship between average velocity and average speed.

 

Question 11. The motion of a satellite around the earth is an example for -
(a) circular motion
(b) rotational motion
(c) curvilinear motion
(d) spinning
Answer: (a) circular motion
In simple words: A satellite moves in a nearly round path around the Earth, which is a type of circular motion.

🎯 Exam Tip: Circular motion describes an object moving in a circle around an external point, as opposed to rotational motion where an object spins on its own axis.

 

Question 12. One car moving on a straight road covers one-third of the distance with 20 km/h and the rest with 60 km/h. The average speed is
(a) 40 km/h
(b) 80 km/h
(c) \( 46 \frac{2}{3} \) km/hr
(d) 36 km/h
Answer: (d) 36 km/h
In simple words: To find the average speed, we need to divide the total distance by the total time. If the total distance is \( D \), then one-third is \( D/3 \) and two-thirds is \( 2D/3 \). Calculate the time for each part using \( \text{time} = \frac{\text{distance}}{\text{speed}} \), then sum them up. The average speed is \( \frac{D}{\text{Total Time}} \).

🎯 Exam Tip: When dealing with average speed over different segments of a journey, always calculate the total distance and total time separately, then divide them.

 

Question 13. A 150m long train is moving with a uniform velocity of 45 km/h. The time taken by the train to cross a bridge of length 850m is
(a) 56s
(b) 68s
(c) 80s
(d) 92s
Answer: (c) 80s
In simple words: When a train crosses a bridge, the total distance it needs to cover is its own length plus the bridge's length. Convert the train's speed to m/s, then use \( \text{time} = \frac{\text{total distance}}{\text{speed}} \) to find the time taken.

🎯 Exam Tip: Remember to convert all units to be consistent (e.g., meters and seconds) before performing calculations. For a train crossing a bridge, the total distance is the sum of the train's length and the bridge's length.

 

Question 14. A particle moves in a straight line with constant acceleration. It changes its velocity from 10 m/s to 20 m/s while passing through a distance of 135 m in 't' seconds. The value of t is
(a) 6s
(b) 9s
(c) 10s
(d) 1.8s
Answer: (b) 9s
In simple words: We know the initial velocity (10 m/s), final velocity (20 m/s), and distance (135 m). First, find the acceleration using \( v^2 = u^2 + 2as \). Then, use \( v = u + at \) to find the time \( t \).

🎯 Exam Tip: When given initial velocity, final velocity, and distance, use the equation \( v^2 = u^2 + 2as \) to find acceleration, which then allows you to find time using \( v = u + at \).

 

Question 15. If a ball is thrown vertically upwards with a speed u the distance covered during the last 't' seconds of its ascent is
(a) \( \frac{1}{2} gt^2 \)
(b) \( ut - \frac{1}{2} gt^2 \)
(c) \( (u - gt)t \)
(d) \( ut \)
Answer: (a) \( \frac{1}{2} gt^2 \)
In simple words: When a ball is thrown upwards, during the last 't' seconds of its upward journey, its motion is like falling from rest under gravity. So, the distance covered is the same as if it started falling from rest for 't' seconds, which is \( \frac{1}{2} gt^2 \).

🎯 Exam Tip: The motion during the last 't' seconds of ascent is symmetrical to the motion during the first 't' seconds of descent from the peak height, simplifying the distance calculation.

 

Question 16. A particle moves along a straight line such that its displacement 's' at any time 't' is given by \( s = t^3 - 6t^2 + 3t + 4 \) meters, t being in second. The velocity when acceleration is zero is
Answer: First, we find the velocity \( v \) by differentiating the displacement \( s \) with respect to time \( t \): \( v = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 3t + 4) = 3t^2 - 12t + 3 \). Next, we find the acceleration \( a \) by differentiating the velocity \( v \) with respect to time \( t \): \( a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 3) = 6t - 12 \). To find when acceleration is zero, we set \( a = 0 \): \( 6t - 12 = 0 \implies 6t = 12 \implies t = 2 \, \text{s} \). Finally, we substitute \( t = 2 \, \text{s} \) back into the velocity equation: \( v = 3(2)^2 - 12(2) + 3 = 3(4) - 24 + 3 = 12 - 24 + 3 = -9 \, \text{m/s} \). So, the velocity when acceleration is zero is -9 m/s. This shows the interplay between displacement, velocity, and acceleration using calculus.

🎯 Exam Tip: Remember that velocity is the first derivative of displacement with respect to time, and acceleration is the second derivative of displacement (or first derivative of velocity) with respect to time.

 

Question 17. Which one of the following is not a scalar?
(a) Volume
(b) angular momentum
(c) Relative density
(d) time
Answer: (b) angular momentum
In simple words: A scalar quantity only has a size (magnitude), while a vector quantity has both size and direction. Angular momentum has both a size and a direction, making it a vector, not a scalar.

🎯 Exam Tip: Understand the difference between scalar quantities (magnitude only, like time, volume, relative density) and vector quantities (magnitude and direction, like angular momentum, velocity, force).

 

Question 18. Vector is having -
(a) only magnitude
(b) only direction
(c) both magnitude and direction
(d) either magnitude or direction
Answer: (c) both magnitude and direction
In simple words: A vector is a type of measurement that tells us both how much there is (its size) and which way it's going (its direction).

🎯 Exam Tip: Clearly distinguish between scalar quantities (magnitude only) and vector quantities (magnitude and direction) as this is fundamental to physics.

 

Question 19. The displacement-time graph of a moving particle is shown below. The instant velocity of the particle is negative at the point
(a) D
(b) F
(c) C
(d) E
Answer: (d) E
In simple words: On a displacement-time graph, the velocity is shown by the slope of the line. If the line is going downwards (decreasing displacement), the slope is negative, meaning the velocity is negative. Point E is on a downward sloping part of the curve.

🎯 Exam Tip: For displacement-time graphs, a positive slope indicates positive velocity, a negative slope indicates negative velocity, and a zero slope indicates zero velocity.

 

Question 20. If two vectors are having equal magnitude and the same direction is known as –
(a) equal vectors
(b) col-linear vectors
(c) parallel vectors
(d) on it vector
Answer: (a) equal vectors
In simple words: When two arrows (vectors) are exactly the same length and point in the same way, they are called equal vectors. They both represent the same amount and direction of something.

🎯 Exam Tip: Remember that for vectors, both magnitude (size) and direction are crucial. Equal vectors must match in both aspects.

 

Question 21. The velocity-time graph of a body moving in a straight line is shown below

Which are of the following represents its acceleration-time graph?

Answer: (a)
In simple words: The first graph (velocity-time) shows that velocity stays the same over time, which means there is no change in speed or direction. If velocity is constant, then acceleration must be zero. So, the correct acceleration-time graph is a straight line along the time axis, indicating zero acceleration.

🎯 Exam Tip: Remember that the slope of a velocity-time graph gives acceleration. A horizontal line on a v-t graph means zero slope, hence zero acceleration.

 

Question 22. Indicate which of the following graph represents the one-dimensional motion of particle?

Answer: (b)
In simple words: For one-dimensional motion, a particle can only move forward or backward along a single line. This means its velocity can only have one value (positive or negative) for any given time. Graph (b) shows that for any time 't', there is only one value of velocity 'v'. The other graphs show multiple velocity values for a single time, which is not possible in one-dimensional motion.

🎯 Exam Tip: In one-dimensional motion, an object cannot be at two different positions or have two different velocities at the exact same moment in time. The curve on a position-time or velocity-time graph must pass the vertical line test.

 

Question 23. The variation of velocity of a particle with time moving along a straight line is illustrated in the following figure. The distance travelled by the particle in 4s is _____.

Answer: (b) 55m
In simple words: To find the distance traveled from a velocity-time graph, you need to calculate the area under the graph. Break the shape into simple rectangles and triangles, then add their areas. From 0 to 2 seconds, it's a triangle and a rectangle. From 2 to 3 seconds, it's a rectangle. From 3 to 4 seconds, it's a trapezoid. Add all these areas to get the total distance.

🎯 Exam Tip: For velocity-time graphs, the area under the curve represents the displacement. If the curve goes below the x-axis, that area contributes negatively to displacement, but for total distance, you sum the absolute values of all areas.

 

Question 24. An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement (s), velocity (v) graph of this object is _____.

Answer: (b)
In simple words: When an object moves with constant acceleration in the same direction as its motion, its velocity steadily increases. This is shown by a straight line with a positive slope on a velocity-time graph, starting from zero. The displacement-time graph for this motion would be a curve bending upwards, showing that the object covers more distance in equal time intervals. Graph (b) correctly shows both increasing velocity (positive slope on v-t) and continuously increasing displacement (upward curving s-t graph).

🎯 Exam Tip: Remember that a straight line on a v-t graph indicates constant acceleration. If the line has a positive slope, acceleration is positive; if it has a negative slope, acceleration is negative.

 

Question 25. A unit vector is used to specify –
(a) only magnitude
(b) only direction
(c) either magnitude (or) direction
(d) absolute value
Answer: (b) only direction
In simple words: A unit vector is like a special tiny arrow that only tells you the direction of something. It has a length of exactly one, so its size doesn't change, only its pointing direction matters.

🎯 Exam Tip: Unit vectors are fundamental for describing directions in physics, especially when dealing with vector components or specifying the orientation of a force or velocity.

 

Question 26. A vector is not changed if _____.
(a) It is rotated through an arbitrary angle
(b) It is multiplied by an arbitrary scalar
(c) It is cross multiplied by a unit vector
(d) It is parallel to itself.
Answer: (d) It is parallel to itself.
In simple words: A vector stays exactly the same if you just move it without changing its length or the direction it points. Moving a vector parallel to itself means you're just sliding it without rotating it or making it longer or shorter.

🎯 Exam Tip: Understanding vector properties is key: a vector is defined by its magnitude and direction; only operations that preserve both will leave the vector unchanged.

 

Question 27. Two forces each of magnitude 'F' have a resultant of the same magnitude. The angle between two forces _____.
(a) 45°
(b) 120°
(c) 150°
(d) 60°
Answer: (b) 120°
In simple words: If you have two forces of the same strength and their combined effect is also that same strength, then the angle between them must be 120 degrees. This is a special case in vector addition.

🎯 Exam Tip: This is a common result derived from the law of cosines for vector addition: \( R^2 = A^2 + B^2 + 2AB \cos\theta \). If \( R=A=B=F \), then \( F^2 = F^2 + F^2 + 2F^2 \cos\theta \), which simplifies to \( \cos\theta = -1/2 \), so \( \theta = 120^\circ \).

 

Question 28. The magnitude of a vector can not be-
(a) positive
(b) negative
(c) zero
Answer: (b) negative
In simple words: The magnitude of a vector is its length or size, which is always a positive number or zero. It cannot be a negative value, just like a distance cannot be negative.

🎯 Exam Tip: Magnitude is a scalar quantity, representing the "how much" of a vector, and by definition, it is always non-negative. Zero magnitude means a null vector.

 

Question 29. Six vectors \( \vec{a} \) through \( \vec{f} \) have magnitudes and directions as indicated in figure. Which of the following statement is true?

(a) \( \overline{b} + \overline{e} = \overline{f} \)
(b) \( 3\hat{i} – 2\hat{j} + \hat{k} \hat{b} + \hat{c} = \hat{f} \)
(c) \( \hat{d} + \hat{c} = \hat{f} \)
(d) \( \hat{d} + \hat{e} = \hat{f} \)
Answer: (d) \( \hat{d} + \hat{e} = \hat{f} \)
In simple words: If you place vectors one after another, the sum (resultant) is the vector that goes from the start of the first to the end of the last. Looking at the diagram, if you put vector \( \hat{d} \) and then vector \( \hat{e} \) right after it, the final path is exactly the same as vector \( \hat{f} \). This shows that \( \hat{d} + \hat{e} = \hat{f} \).

🎯 Exam Tip: To find the sum of two vectors graphically, use the triangle rule or parallelogram rule of vector addition. The resultant vector connects the tail of the first vector to the head of the second.

 

Question 30. A force of 3 N and 4 N are acting perpendicular to an object, the resultant force is-
(a) 9 N
(b) 16 N
(c) 5 N
(d) 7 N
Answer: (c) 5 N
In simple words: When two forces push on something at a right angle (like the sides of a square), you can find their total combined push using the Pythagorean theorem. So, for a 3 N force and a 4 N force at 90 degrees, the total force is 5 N.

🎯 Exam Tip: When two vectors (like forces) are perpendicular, their resultant magnitude can be found using \( R = \sqrt{F_1^2 + F_2^2} \). This is a direct application of the Pythagorean theorem.

 

Question 31. The figure shows ABCDEF as regular hexagon. What is the value of \( \overline{AC} + \overline{AD} + \overline{AE} + \overline{AF} \)?

(a) \( \overline{A0} \)
(b) \( 2 \overline{A0} \)
(c) \( 4 \overline{A0} \)
(d) \( 6 \overline{A0} \)
Answer: (d) \( 6 \overline{A0} \)
In simple words: In a regular hexagon, if you draw lines from one corner (A) to all other corners and the center (O), you can combine these vectors. The sum of all vectors from A to the other points (C, D, E, F) is equal to six times the vector from A to the center (O). This is because of the symmetry of the hexagon.

🎯 Exam Tip: For regular polygons, the vector sum from one vertex to all other vertices (including the center, usually labeled O) often simplifies due to symmetry. Remember that \( \vec{AD} = 2\vec{AO} \) and other pairs sum to \( 2\vec{AO} \).

 

Question 32. One of the two rectangular components of a force is 20N. And it makes an angle of 30° with the force. The magnitude of the other component is _____.
(a) \( 20/\sqrt{3} \)
(b) \( 10/\sqrt{3} \)
(c) \( 15/\sqrt{3} \)
(d) \( 40/\sqrt{3} \)
Answer: (a) \( 20/\sqrt{3} \)
In simple words: If a force is broken down into two parts that are at right angles to each other (rectangular components), and one part is 20 N acting at a 30-degree angle to the main force, then the other part can be found using trigonometry. You'll use the sine function for the perpendicular component.

🎯 Exam Tip: When resolving a force \( F \) into two perpendicular components \( F_x \) and \( F_y \) with an angle \( \theta \) to the x-axis, remember that \( F_x = F \cos\theta \) and \( F_y = F \sin\theta \). You can find \( F \) first using the given component, then calculate the other component.

 

Question 33. The angle between \( (\overrightarrow{\mathrm{A}} + \overrightarrow{\mathrm{B}}) \) and \( (\overrightarrow{\mathrm{A}} – \overrightarrow{\mathrm{B}}) \) can be –
(a) only 0°
(b) only 90°
(c) between 0° and 90°
(d) between 0° and 180°
Answer: (d) between 0° and 180°
In simple words: The angle between the sum of two vectors and their difference can be any angle from 0 degrees (when they point in the same direction) up to 180 degrees (when they point in opposite directions). It just depends on what the original two vectors are.

🎯 Exam Tip: The dot product of two vectors \( \vec{X} \) and \( \vec{Y} \) is \( |\vec{X}||\vec{Y}|\cos\theta \). By using \( \vec{X} = \vec{A} + \vec{B} \) and \( \vec{Y} = \vec{A} - \vec{B} \), we can find \( (\vec{A} + \vec{B}) \cdot (\vec{A} - \vec{B}) = A^2 - B^2 \). The angle depends on whether \( A^2 > B^2 \), \( A^2 < B^2 \), or \( A^2 = B^2 \).

 

Question 34. If the sum of two unit vectors is a unit vector the magnitude of the difference is _____.
(a) \( \sqrt{2} \)
(b) \( \sqrt{3} \)
(c) \( 1/\sqrt{2} \)
(d) \( \sqrt{5} \)
Answer: (b) \( \sqrt{3} \)
In simple words: Imagine two arrows, each one unit long. If you add them together and the resulting arrow is also one unit long, it means the original two arrows were at a 120-degree angle to each other. If you then find the difference between those same two arrows, its length will be \( \sqrt{3} \) units. This happens when the vectors form an equilateral triangle with the resultant.

🎯 Exam Tip: This is a classic vector problem. If \( |\vec{A}|=|\vec{B}|=|\vec{A}+\vec{B}|=1 \), then the angle between \( \vec{A} \) and \( \vec{B} \) is 120°. The magnitude of the difference \( |\vec{A}-\vec{B}| = \sqrt{1^2+1^2-2(1)(1)\cos(120^\circ)} = \sqrt{1+1-2(-1/2)} = \sqrt{1+1+1} = \sqrt{3} \).

 

Question 35. If P = mV then the direction of P along-
(a) m
(b) v
(c) both (a) and (b)
(d) neither m nor v
Answer: (b) v
In simple words: The equation P = mV shows that momentum (P) is found by multiplying mass (m) by velocity (v). Since mass is just a number and doesn't have a direction, the direction of momentum will always be the same as the direction of velocity.

🎯 Exam Tip: Momentum is a vector quantity, and its direction is always identical to the velocity vector. Mass is a scalar and only affects the magnitude.

 

Question 36. If \( \overline{A} = 2\hat{i} + \hat{j} – \hat{k} \), \( \overline{B} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \overline{C} = 6\hat{i} – 2\hat{j} – 6\hat{k} \) then angle between \( \overline{A} + \overline{B} \) and \( \overline{C} \) will be _____.
(a) 30°
(b) 45°
(c) 60°
Answer: (d) 90°
In simple words: First, you add vectors A and B to get a new vector. Then, you find the angle between this new vector and vector C. If their dot product is zero, it means they are at a 90-degree angle to each other.

🎯 Exam Tip: To find the angle between two vectors \( \vec{X} \) and \( \vec{Y} \), use the dot product formula: \( \vec{X} \cdot \vec{Y} = |\vec{X}||\vec{Y}|\cos\theta \). If the dot product is zero, the vectors are perpendicular, and \( \theta = 90^\circ \).

 

Question 37. The scalar product \( \overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}} \) is equal to-
(a) \( \overrightarrow{\mathrm{A}} +\overrightarrow{\mathrm{B}} \)
(b) \( \overrightarrow{\mathrm{A}}. \overrightarrow{\mathrm{B}} \)
(c) AB sin 0
(d) \( (\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}} \)
Answer: (b) \( \overrightarrow{\mathrm{A}}. \overrightarrow{\mathrm{B}} \)
In simple words: The scalar product, also called the dot product, of two vectors A and B is written as \( \overrightarrow{A} \cdot \overrightarrow{B} \). It calculates a single number (a scalar) that tells you how much one vector points in the direction of the other.

🎯 Exam Tip: Remember that the scalar product (dot product) of two vectors \( \vec{A} \) and \( \vec{B} \) is defined as \( AB\cos\theta \), where \( \theta \) is the angle between them. This operation always results in a scalar quantity.

 

Question 38. If \( \overline{A} \times \overline{B} = \overline{C} \) then which of the following statement is wrong?
(a) \( \overline{C} \perp \overline{A} \)
(b) \( \overline{B} \perp \overline{B} \)
(c) \( \overline{C} \pm (\overline{A} + \overline{B}) \)
(d) \( \overline{C} \pm (\overline{A} \times \overline{B}) \)
Answer: (b) \( \overline{B} \perp \overline{B} \)
In simple words: The cross product of two vectors (A and B) gives a new vector (C) that is at a right angle (perpendicular) to both A and B. This means C is perpendicular to the plane containing A and B. The statement that B is perpendicular to B itself is wrong because a vector is always parallel to itself.

🎯 Exam Tip: The cross product \( \vec{A} \times \vec{B} \) produces a vector \( \vec{C} \) that is orthogonal to both \( \vec{A} \) and \( \vec{B} \). A vector can never be perpendicular to itself unless it is a zero vector, which is not the general case.

 

Question 39. The scalar product of two vectors will be minimum. When \( \theta \) is equal to –
(a) 0°
(b) 45°
(c) 180°
(d) 60°
Answer: (c) 180°
In simple words: The scalar product (dot product) of two vectors is smallest when the vectors point in exact opposite directions. At this angle (180 degrees), the cosine value is -1, making the product a negative maximum, or the most negative value possible, hence the minimum.

🎯 Exam Tip: The scalar product is \( \vec{A} \cdot \vec{B} = AB\cos\theta \). Since A and B are positive magnitudes, the minimum value occurs when \( \cos\theta \) is minimum, which is -1 (at \( \theta = 180^\circ \)).

 

Question 40. If \( |\overline{A} \times \overline{B}| \), then value of \( |\overline{A} \times \overline{B}| \) is _____.
(a) \( A^2 + B^2 + \frac{AB}{\sqrt{3}} \)^{1/2}
(b) A+B
(c) \( (A^2 + B^2 + \sqrt{3}AB)^{1/2} \)
(d) \( (A^2 + B^2 + AB)^{1/2} \)
Answer: (d) \( (A^2 + B^2 + AB)^{1/2} \)
In simple words: The question seems to be asking for the magnitude of \( |\vec{A} + \vec{B}| \) rather than \( |\vec{A} \times \vec{B}| \). The magnitude of the sum of two vectors A and B, if the angle between them is 120 degrees, can be related to the expression \( (A^2 + B^2 - AB)^{1/2} \). The given options suggest a typo in the question and are instead related to the magnitude of the resultant vector or other operations. Assuming the question intends to ask for magnitude of resultant: \( |\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB\cos\theta} \). If \( \theta = 60^\circ \), then \( \cos 60^\circ = 1/2 \), so \( |\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + AB} \).

🎯 Exam Tip: Pay close attention to the operation symbol. \( |\vec{A} \times \vec{B}| \) is the magnitude of the cross product \( (AB\sin\theta) \), while \( |\vec{A} + \vec{B}| \) is the magnitude of the resultant vector, typically found using the law of cosines \( \sqrt{A^2 + B^2 + 2AB\cos\theta} \).

 

Question 41. The angle between vectors \( \overline{A} \) and \( \overline{B} \) is A. The value of the triple product \( \overline{A} ( \overline{A} \times \overline{B} ) \) is _____.
(a) A² B
(b) zero
(c) A² B \( \sin\theta \)
(d) A² B \( \cos\theta \)
Answer: (b) zero
In simple words: The cross product \( \overline{A} \times \overline{B} \) gives a new vector that is always at a right angle to \( \overline{A} \). When you then take the dot product of \( \overline{A} \) with this new vector, the result is zero. This is because the dot product of two vectors at a right angle to each other is always zero.

🎯 Exam Tip: Remember the property that the vector \( (\vec{A} \times \vec{B}) \) is always perpendicular to \( \vec{A} \). Therefore, the dot product \( \vec{A} \cdot (\vec{A} \times \vec{B}) \) will always be zero, as \( \cos 90^\circ = 0 \).

 

Question 42. Two adjacent sides of a parallelogram are represented by the two vectors \( \hat{i} + 2\hat{j} + 3\hat{k} \) and \( 3\hat{i} – 2\hat{j} + \hat{k} \). The area parallelogram _____.
(a) 8
(b) \( 8\sqrt{3} \)
Answer: (b) \( 8\sqrt{3} \)
In simple words: To find the area of a parallelogram formed by two vectors, you need to calculate the magnitude of their cross product. First, find the cross product of the two given vectors. Then, find the length (magnitude) of the resulting vector. This length will be the area of the parallelogram.

🎯 Exam Tip: The area of a parallelogram with adjacent sides given by vectors \( \vec{A} \) and \( \vec{B} \) is \( |\vec{A} \times \vec{B}| \). The cross product can be computed using a determinant: \( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} \).

 

Question 43. If \( \overrightarrow{\mathrm{A}} \) and \( \overrightarrow{\mathrm{B}} \) are two vectors, which are acting along x, y respectively, then \( \overrightarrow{\mathrm{A}} \) and \( \overrightarrow{\mathrm{B}} \) lies along-
(a) x
(b) y
(c) z
(d) none
Answer: (c) z
In simple words: If you have two vectors, A along the x-axis and B along the y-axis, and you take their cross product \( \vec{A} \times \vec{B} \), the resulting vector will always be along the z-axis. This is a rule of vector cross products in a 3D coordinate system.

🎯 Exam Tip: Remember the right-hand rule for cross products. If \( \vec{A} \) is along the x-axis and \( \vec{B} \) is along the y-axis, then \( \vec{A} \times \vec{B} \) will be along the positive z-axis. If the question implies the plane containing A and B, then it's the XY plane.

 

Question 44. Galileo writes that for angles of the projectile (45 + \( \theta \)) and (45 – \( \theta \)) the horizontal ranges described by the projectile are in the ratio of (if \( \theta \le 45 \))
(a) 2:1
(b) 1:2
(c) 1:1
(d) 2:3
Answer: (c) 1:1
In simple words: Galileo found that if you throw an object with the same speed at two different angles that are equally above and below 45 degrees (like 30 degrees and 60 degrees, or 40 degrees and 50 degrees), they will land at the same horizontal distance. So, the ratio of their horizontal ranges will always be 1:1.

🎯 Exam Tip: The horizontal range of a projectile is given by \( R = \frac{u^2 \sin(2\alpha)}{g} \). For angles \( (45^\circ + \theta) \) and \( (45^\circ - \theta) \), the term \( \sin(2\alpha) \) will be \( \sin(90^\circ + 2\theta) = \cos(2\theta) \) and \( \sin(90^\circ - 2\theta) = \cos(2\theta) \) respectively. Since \( \cos(2\theta) \) is the same for both, their ranges are equal, resulting in a 1:1 ratio.

 

Question 45. A projectile is thrown into the air so as to have the minimum possible range. The projection point as the origin the Coordinates of the point where the velocity of the projectile is minimum are _____.
Answer: (b) 100,50
In simple words: For a projectile, the velocity is at its lowest point when the object reaches the highest point in its path. At this peak, only the horizontal velocity component remains. The coordinates of this point for a minimum range are specific to the given conditions and calculation.

🎯 Exam Tip: The minimum velocity for a projectile occurs at the highest point of its trajectory, where the vertical component of velocity is zero. The horizontal component of velocity remains constant throughout the flight.

 

Question 46. \( \overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}} \) is equal to –
(a) \( \overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}} \)
(b) \( \overrightarrow{\mathrm{A}} + \overrightarrow{\mathrm{B}} \)
(c) \( -\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}} \)
(d) \( \overrightarrow{\mathrm{A}} – \overrightarrow{\mathrm{B}} \)
Answer: (c) \( -\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}} \)
In simple words: The cross product of two vectors is not commutative, meaning the order matters. If you swap the order of the vectors in a cross product, the direction of the resulting vector reverses. So, \( \vec{A} \times \vec{B} \) is the opposite of \( \vec{B} \times \vec{A} \).

🎯 Exam Tip: Remember the anticommutative property of the cross product: \( \vec{A} \times \vec{B} = - (\vec{B} \times \vec{A}) \). This means swapping the order changes the direction of the resultant vector.

 

Question 47. The vector product of any two vectors gives a –
(a) vector
(b) scalar
(c) tensor
(d) col-linear
Answer: (a) vector
In simple words: When you perform a vector product (also known as a cross product) between any two vectors, the answer you get is always another vector. This new vector has both a size (magnitude) and a direction.

🎯 Exam Tip: The key distinction between dot product and cross product is their result: a dot product yields a scalar, while a cross product yields a vector. This resultant vector is perpendicular to the plane containing the original two vectors.

 

Question 48. A 150 m long train is moving the north at a speed of 10 m/s. A parrot flying towards the south with a speed of 5 m/s crosses the train. The time taken would be _____.
Answer: (d) 10s
In simple words: When the train and parrot are moving towards each other, their speeds add up to give a relative speed. The total distance the parrot needs to "cross" is the length of the train. Divide the train's length by the combined speed to find the time taken.

🎯 Exam Tip: For relative motion problems, if two objects move in opposite directions, their relative speed is the sum of their individual speeds. The time taken to cross each other is the total distance (usually the length of one object if the other is a point, or sum of lengths if both have length) divided by this relative speed.

 

Question 49. A boat is moving with a velocity of 3i+4j with respect to the ground. The water in the river is moving with a velocity of -3i-4j with respect to the ground. The relative velocity of the boat with respect to water _____.
(a) 8j
(b) -6i -8j
(c) 6i + 8j
(d) \( 5\sqrt{2} \)
Answer: (c) 6i + 8j
In simple words: To find how fast the boat moves compared to the water, you subtract the water's velocity from the boat's velocity relative to the ground. Since the water is moving in the exact opposite direction to the boat, their speeds add up when seen from the water's perspective.

🎯 Exam Tip: The relative velocity of object A with respect to object B is given by \( \vec{V}_{AB} = \vec{V}_A - \vec{V}_B \). Be careful with vector subtraction, subtracting a negative vector component means adding its positive counterpart.

 

Question 50. The vector product of two non-zero vectors will be minimum when \( \theta \) is equal to-
(a) 0°
(b) 180°
(e) both (a) and (b)
(d) neither (a) nor (b)
Answer: (e) both (a) and (b)
In simple words: The vector product (cross product) of two vectors becomes the smallest possible value (zero) when the vectors are either pointing in the exact same direction (0 degrees) or in exact opposite directions (180 degrees). In both cases, they are parallel, so there is no area between them in 3D space, which is what the cross product measures.

🎯 Exam Tip: The magnitude of the cross product is \( |\vec{A} \times \vec{B}| = AB\sin\theta \). This value is minimum (zero) when \( \sin\theta = 0 \), which occurs at \( \theta = 0^\circ \) (parallel) and \( \theta = 180^\circ \) (anti-parallel). In both cases, the vectors are collinear.

 

II. Long Answer Questions:

 

Question 1. What are the different types of motion? State one example for each & explain.
Answer: The various kinds of motion are:
a) Linear motion: An object moves in a straight line. For example, an athlete running on a straight track. The object's path is a single straight dimension.
b) Circular motion: An object travels along a circular path. For instance, the motion of a satellite around the Earth. The object maintains a constant distance from a central point.
c) Rotational motion: An object spins around an axis. During this motion, every point on the object follows a circular path around that axis. An example is the Earth spinning on its own axis. This is often an internal motion of a rigid body.
d) Vibratory motion: An object or particle moves back and forth repeatedly around a fixed point. This is also known as oscillatory motion. A guitar string vibrating is a good example. This is a rhythmic, repetitive motion.
In simple words: Motion can be straight (like a runner), in a circle (like a satellite), spinning around (like the Earth), or wiggling back and forth (like a guitar string). Each type describes how something moves.

🎯 Exam Tip: When defining motion types, clearly state the characteristic path (straight line, circle, etc.) and provide a distinct, easy-to-understand example for each to score full marks.

 

Question 2. How will you differentiate motion in one dimension, two dimensions, and in three dimensions?
Answer: To distinguish between one, two, and three-dimensional motion, we look at how many coordinates are needed to describe the object's position at any time:
Motion in one dimension: This happens when a particle moves along a single straight line. Only one coordinate (like x, y, or z) is needed to show its position. For example, an object falling freely under gravity near the Earth's surface moves only up or down.
Motion in two dimensions: In this case, a particle moves along a curved path in a flat surface (plane). Two coordinates (like x and y) are needed to describe its position. For instance, the motion of a coin on a carrom board moves across a flat surface.
Motion in three dimensions: A particle moves through usual three-dimensional space. All three coordinates (x, y, and z) are needed to describe its position. A bird flying in the sky is an example, as it can move in any direction (up-down, left-right, forward-backward).
In simple words: One-dimensional motion is like moving only forward or backward. Two-dimensional motion is like moving on a flat floor. Three-dimensional motion is like a bird flying anywhere in the air. The number of directions it can change tells you the dimension.

🎯 Exam Tip: For clear differentiation, specify the number of coordinates required to define position in each dimension and provide illustrative examples that distinctly fit each category.

 

Question 3. State and define different types of vectors.
Answer: Here are the various kinds of vectors:
1. Equal vectors: Two vectors, say \( \vec{A} \) and \( \vec{B} \), are considered equal if they have the same length (magnitude) and point in the same direction. They also represent the same physical quantity.

(a) Collinear vectors: These are vectors that act along the same line. The angle between them can be either 0° (if they are in the same direction) or 180° (if they are in opposite directions).
(i) Parallel vectors – If two vectors \( \vec{A} \) and \( \vec{B} \) act in the same direction along the same line or in parallel lines, the angle between them is zero.

(ii) Antiparallel vectors: Two vectors \( \vec{A} \) and \( \vec{B} \) are said to be antiparallel if they are in opposite directions along the same line or in parallel lines. The angle between them is 180°.

2. Unit vector: This is a vector divided by its own magnitude. The unit vector of \( \vec{A} \) is represented as \( \hat{A} \). Its magnitude is always 1. \( \hat{A} = \frac{\vec{A}}{|\vec{A}|} \) or \( \hat{A} = \frac{\vec{A}}{A} \). This is useful for showing direction without caring about length.
3. Orthogonal unit vectors: These are three unit vectors, \( \hat{i}, \hat{j}, \hat{k} \), that point along the positive x-axis, positive y-axis, and positive z-axis, respectively. They are all directly perpendicular to each other, with the angle between any two of them being 90°. For example, \( \hat{i}, \hat{j}, \hat{k} \) are orthogonal vectors. Two vectors that are perpendicular to each other are called orthogonal vectors.

In simple words: Vectors come in different types. Equal vectors are the same size and point the same way. Collinear vectors are on the same line (parallel or opposite). Unit vectors are tiny arrows with a length of one, only showing direction. Orthogonal unit vectors are three special unit vectors that point along the x, y, and z axes, forming perfect right angles with each other.

🎯 Exam Tip: When defining vector types, ensure you include both magnitude and direction for clear understanding. For unit vectors, emphasize their magnitude of one and their role in indicating direction.

 

Question 3. Define a Scalar. Give Examples.
Answer: A scalar is a physical quantity that can be fully described only by its magnitude. It does not have any direction. For example, distance, mass, temperature, speed, and energy are all scalar quantities. Understanding scalars helps distinguish them from vectors which have both magnitude and direction.
In simple words: A scalar is a quantity that only tells you "how much" (its size), not "which way." For example, 5 kilograms is a scalar; it just tells you the amount of mass.

🎯 Exam Tip: Remember that scalar quantities are often distinguished from vector quantities by their lack of direction. Identifying them correctly is fundamental in physics.

 

Question 4. Write short note on the scalar product between two vectors.
Answer: The scalar product, also known as the dot product, of two vectors is found by multiplying their magnitudes and the cosine of the angle between them. If you have two vectors, \( \vec{A} \) and \( \vec{B} \), with an angle \( \theta \) between them, their scalar product is given by \( \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \). The result of a scalar product is always a scalar quantity, meaning it has magnitude but no direction. A common real-world example is work done, which is the scalar product of force and displacement, indicating the energy transferred.
In simple words: The scalar product (or dot product) of two vectors combines their strengths and how much they point in the same direction. It always gives a single number, not a direction.

🎯 Exam Tip: The scalar product tells you how much one vector "aligns" with another. If vectors are perpendicular, their scalar product is zero because \( \cos 90^\circ = 0 \).

 

Question 5. Write a Short note on vector product between two vectors.
Answer: The vector product, also known as the cross product, of two vectors results in another vector. This new vector has a magnitude equal to the product of the magnitudes of the two original vectors and the sine of the angle between them. Its direction is perpendicular to the plane containing the original two vectors, determined by the right-hand screw rule or right-hand thumb rule. For two vectors \( \vec{A} \) and \( \vec{B} \), their vector product \( \vec{C} \) is written as \( \vec{C} = \vec{A} \times \vec{B} = (AB \sin \theta) \hat{n} \), where \( \hat{n} \) is the unit vector perpendicular to the plane of \( \vec{A} \) and \( \vec{B} \). This product is widely used to calculate torque or magnetic force, which are rotational effects.
In simple words: The vector product (or cross product) of two vectors gives you a new vector. The size of this new vector depends on the size of the first two and how much they are at right angles to each other. Its direction is always straight out of the plane they both sit in.

🎯 Exam Tip: The right-hand rule is crucial for determining the direction of the resultant vector in a cross product. Visualize it carefully to avoid errors.

 

Question 6. How do you deduce that two vectors are perpendicular?
Answer: Two vectors are considered perpendicular if their vector product (cross product) has its maximum magnitude. This occurs when the angle \( \theta \) between them is \( 90^\circ \), because \( \sin 90^\circ = 1 \). When \( \theta = 90^\circ \), the magnitude of the cross product \( |\vec{A} \times \vec{B}| \) equals \( AB \). Conversely, if the dot product of two non-zero vectors is zero, they are also perpendicular, as \( \cos 90^\circ = 0 \). Both methods confirm orthogonality, but the cross product yields maximum magnitude for perpendicular vectors, while the dot product yields zero.
In simple words: Two vectors are perpendicular if their cross product is the largest possible value for their given sizes, or if their dot product is zero.

🎯 Exam Tip: To quickly check if two vectors are perpendicular, calculate their dot product. If the result is zero, they are perpendicular.

 

Question 7. Define Displacement and distance.
Answer:
**Distance** is the actual total length of the path traveled by an object during its motion, regardless of direction. It is a scalar quantity and is always positive. For example, if you walk around a block, the total meters you covered is your distance. Distance is crucial for calculating energy usage or fuel consumption.
**Displacement** is the shortest straight-line distance between an object's initial and final positions, along with its direction. It is a vector quantity and can be positive, negative, or zero. For instance, if you walk around a block and return to your starting point, your displacement is zero. Displacement is important in understanding net change in position and velocity calculations.
In simple words: Distance is how far you actually moved, no matter the path. Displacement is only how far you are from where you started, in a straight line, and includes direction.

🎯 Exam Tip: Always remember that distance is a scalar (magnitude only) and displacement is a vector (magnitude and direction). This distinction is fundamental in kinematics.

 

Question 8. Define velocity and speed.
Answer:
**Velocity** is the rate at which an object changes its position with respect to time, including the direction of motion. It is defined as the rate of change of the position vector or the rate of change of displacement. Velocity is a vector quantity, meaning it has both magnitude and direction. Its unit is meters per second (ms\(^{-1}\)) and its dimensional formula is \( [LT^{-1}] \). A car moving at 60 km/h north has a specific velocity.
**Speed** is the rate at which an object covers distance. It is defined as the rate of change of distance. Speed is a scalar quantity, meaning it only has magnitude and no direction. Its unit is also meters per second (ms\(^{-1}\)) and its dimensional formula is \( [LT^{-1}] \). A car moving at 60 km/h has a specific speed, regardless of its direction. Even if a car's speed is constant, its velocity can change if its direction changes.
In simple words: Speed tells you how fast something is moving. Velocity tells you how fast it is moving AND in what direction.

🎯 Exam Tip: The key difference lies in direction: speed is a scalar, while velocity is a vector. This distinction is vital for understanding motion accurately.

 

Question 9. Define acceleration.
Answer: Acceleration is defined as the rate at which an object's velocity changes over time. Since velocity is a vector quantity, acceleration is also a vector quantity, possessing both magnitude and direction. It can involve a change in speed, a change in direction, or both. The formula for acceleration is \( \vec{a} = \frac{d \vec{v}}{d t} \). Its standard unit is meters per second squared (ms\(^{-2}\)), and its dimensional formula is \( [LT^{-2}] \). For example, a car speeding up, slowing down, or turning a corner is undergoing acceleration.
In simple words: Acceleration is how quickly an object's speed or direction changes. If something speeds up, slows down, or turns, it is accelerating.

🎯 Exam Tip: Remember that acceleration doesn't just mean speeding up; slowing down (deceleration) or changing direction at a constant speed are also forms of acceleration.

 

Question 10. What is the difference between velocity and average velocity?
Answer:

Velocity refers to the instantaneous rate of change of an object's position, including its direction, at a specific moment. Average velocity, on the other hand, considers the total displacement over a specific time interval, providing an overall measure of motion. For example, a car's speedometer shows its instantaneous speed, while calculating its average speed for a trip requires dividing total distance by total time.
In simple words: Velocity is how fast you are going right now and in what direction. Average velocity is your total change in position divided by the total time it took.

🎯 Exam Tip: Always distinguish between instantaneous velocity and average velocity. Instantaneous velocity describes motion at a specific point, while average velocity describes the overall motion over an interval.

 

Question 11. Define a radian.
Answer: A radian is a unit of angle measurement. One radian is defined as the angle subtended at the center of a circle by an arc whose length is equal to the radius of the circle. This means that if you wrap a string around a circle, and the length of that string segment is equal to the circle's radius, the angle it forms at the center is exactly one radian. Radians simplify many mathematical and physical formulas, especially those involving circular motion and waves.
In simple words: A radian is a way to measure angles. Imagine a circle: if you take a piece of its edge that is exactly the same length as the circle's arm (radius), the angle this piece makes at the center is one radian.

🎯 Exam Tip: Remember that \( 2\pi \) radians is equivalent to \( 360^\circ \). This conversion factor is crucial when switching between radians and degrees.

 

Question 12. Define angular displacement and angular velocity.
Answer:
1. **Angular displacement:** This is the angle described by a particle rotating about an axis in a given amount of time. It measures how much an object has turned around a central point or axis. It is a vector quantity, often measured in radians.
2. **Angular velocity:** This is the rate at which the angular displacement changes over time. It measures how fast an object is rotating or revolving. It is also a vector quantity, commonly measured in radians per second (rad/s). Angular velocity is vital for understanding spinning objects like a merry-go-round.
In simple words: Angular displacement is how much something has rotated or turned. Angular velocity is how fast it is rotating or turning.

🎯 Exam Tip: Both angular displacement and angular velocity are vector quantities, meaning their direction (clockwise or counter-clockwise) is important. Use the right-hand rule to determine direction.

 

Question 13. What is non-uniform circular motion?
Answer: Non-uniform circular motion occurs when an object moves in a circular path but its speed is not constant. In this type of motion, the object covers unequal distances along the circular path in equal time intervals. Both the speed and the direction of the object's motion are continuously changing. A car going around a curve while speeding up or slowing down is an example of non-uniform circular motion, where there's both centripetal and tangential acceleration.
In simple words: Non-uniform circular motion is when something moves in a circle but its speed keeps changing, not staying the same.

🎯 Exam Tip: In non-uniform circular motion, there is both centripetal (towards the center) and tangential (along the circle's edge) acceleration, causing both speed and direction to change.

 

Question 14. Write down the kinematic equations for angular motion.
Answer: The kinematic equations for angular motion describe the relationship between angular displacement, initial angular velocity, final angular velocity, angular acceleration, and time interval. These equations are:
\( \omega = \omega_0 + \alpha t \)
\( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)
\( \omega^2 = \omega_0^2 + 2\alpha\theta \)
\( \theta = \left(\frac{\omega_0 + \omega}{2}\right) t \)
Where:
\( \omega_0 \) represents the initial angular velocity,
\( \omega \) represents the final angular velocity,
\( \alpha \) represents the angular acceleration,
\( \theta \) represents the angular displacement, and
\( t \) represents the time interval.
These equations are similar to linear kinematic equations but adapted for rotational movement, providing a powerful tool to analyze spinning systems.
In simple words: These are formulas that help us figure out how much something spins, how fast it spins, how quickly it changes its spin speed, and for how long. They are like rules for spinning objects.

🎯 Exam Tip: These angular kinematic equations are analogous to linear kinematic equations; learning them together can help you remember them better.

 

Question 15. Write down the expression for angle made by resultant acceleration and radius vector in the non-uniform circular motion.
Answer: In non-uniform circular motion, a particle experiences both centripetal acceleration (towards the center of the circle) and tangential acceleration (along the tangent to the circular path). The resultant acceleration is the vector sum of these two components. If \( a_R \) is the centripetal acceleration and \( a_t \) is the tangential acceleration, the magnitude of the resultant acceleration is \( a = \sqrt{a_R^2 + a_t^2} \). The angle \( \theta \) that this resultant acceleration makes with the radius vector (which points towards the center) can be found using the formula:
\( \tan \theta = \frac{a_t}{a_R} \)
\[ \text{Here, } a_R = \frac{V^2}{r} \]
This angle \( \theta \) helps to fully describe the direction of the total acceleration vector at any point in non-uniform circular motion. This combination of accelerations causes both a change in speed and direction.
In simple words: When something moves in a circle but its speed changes, it has two kinds of push: one towards the middle (centripetal) and one along the side (tangential). The total push makes an angle with the line pointing to the middle, which can be found by dividing the side-push by the middle-push.

🎯 Exam Tip: Remember that tangential acceleration changes the speed, while centripetal acceleration changes the direction of velocity. The angle gives the overall direction of the net acceleration.

 

III. Long Answer Questions:

 

Question 1. Explain in detail the triangle law of addition.
Answer: The triangle law of vector addition is a graphical method for finding the resultant of two vectors. It states that if two vectors are represented in both magnitude and direction by the two adjacent sides of a triangle taken in the same order, then their resultant (or sum) is given by the third side of the triangle, taken in the opposite order.
Consider two vectors, \( \vec{A} \) and \( \vec{B} \), as shown in the figure below:

To find the resultant \( \vec{R} \) of \( \vec{A} \) and \( \vec{B} \), we place the tail of \( \vec{B} \) at the head of \( \vec{A} \). The vector drawn from the tail of \( \vec{A} \) to the head of \( \vec{B} \) then represents the resultant vector \( \vec{R} = \vec{A} + \vec{B} \). The length of this resultant vector gives its magnitude, and its direction points from the start of \( \vec{A} \) to the end of \( \vec{B} \). This law is a simple way to visualize and calculate the sum of forces or velocities.
The magnitude of the resultant vector can be found using the Law of Cosines. If \( \theta \) is the angle between vectors \( \vec{A} \) and \( \vec{B} \), then the magnitude of the resultant \( R \) is given by:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos \theta} \]
The direction of the resultant vector, making an angle \( \alpha \) with vector \( \vec{A} \), can be found using the Law of Sines:
\[ \tan \alpha = \frac{B \sin \theta}{A + B \cos \theta} \]
This law is foundational in physics for combining vector quantities like forces, velocities, and displacements.
In simple words: The triangle law helps add two push-or-pull arrows (vectors). You place the second arrow's start at the first arrow's end. Then, a new arrow drawn from the very start of the first to the very end of the second is the total (resultant) arrow.

🎯 Exam Tip: When applying the triangle law, ensure the vectors are placed "head to tail." The resultant vector closes the triangle and points from the tail of the first to the head of the second.

 

Question 2. Discuss the properties of scalar and vector products.
Answer: The properties of scalar (dot) and vector (cross) products are fundamental in understanding how vectors combine in physics.

**Properties of Scalar Product (Dot Product):**
1. The scalar product of two vectors, \( \vec{A} \cdot \vec{B} \), is always a scalar quantity. It is positive if the angle \( \theta \) between the vectors is acute (\( 0^\circ < \theta < 90^\circ \)) and negative if the angle is obtuse (\( 90^\circ < \theta < 180^\circ \)).
2. The scalar product is **commutative**, meaning the order of multiplication does not change the result: \( \vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A} \).
3. The scalar product obeys the **distributive law**: \( \vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C} \).
4. The angle \( \theta \) between two vectors can be found using: \( \theta = \cos^{-1} \left( \frac{\vec{A} \cdot \vec{B}}{AB} \right) \).
5. The scalar product of two non-zero vectors is **maximum** when they are parallel (\( \theta = 0^\circ \), \( \cos 0^\circ = 1 \)), giving \( \vec{A} \cdot \vec{B} = AB \).
6. The scalar product is **minimum** when they are anti-parallel (\( \theta = 180^\circ \), \( \cos 180^\circ = -1 \)), giving \( \vec{A} \cdot \vec{B} = -AB \).
7. If two non-zero vectors are **perpendicular** (\( \theta = 90^\circ \), \( \cos 90^\circ = 0 \)), their scalar product is zero: \( \vec{A} \cdot \vec{B} = 0 \).
8. The scalar product of a vector with itself is its squared magnitude: \( \vec{A} \cdot \vec{A} = A^2 \).
9. For orthogonal unit vectors (\( \hat{i}, \hat{j}, \hat{k} \)): \( \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1 \) and \( \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0 \).
10. In terms of components, if \( \vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k} \) and \( \vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \), then \( \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z \).

**Properties of Vector Product (Cross Product):**
1. The vector product of two vectors, \( \vec{A} \times \vec{B} \), is always another vector. Its direction is perpendicular to the plane containing \( \vec{A} \) and \( \vec{B} \).
2. The vector product is **not commutative**; changing the order reverses the direction of the resultant vector: \( \vec{A} \times \vec{B} = -(\vec{B} \times \vec{A}) \). The magnitudes are equal, but directions are opposite.
3. The vector product of two non-zero vectors is **maximum** when they are orthogonal (\( \theta = 90^\circ \), \( \sin 90^\circ = 1 \)), giving \( |\vec{A} \times \vec{B}| = AB \).
4. The vector product of two non-zero vectors is **zero** if they are parallel or anti-parallel (\( \theta = 0^\circ \) or \( \theta = 180^\circ \), \( \sin 0^\circ = \sin 180^\circ = 0 \)).
5. The cross product of a vector with itself is a **null vector**: \( \vec{A} \times \vec{A} = 0 \).
6. For orthogonal unit vectors, the cross product follows the right-hand corkscrew rule: \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{k} \times \hat{i} = \hat{j} \). Also, \( \hat{j} \times \hat{i} = -\hat{k} \), etc.
7. In terms of components, \( \vec{A} \times \vec{B} \) can be calculated as the determinant of a matrix:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} = \hat{i}(A_y B_z - A_z B_y) - \hat{j}(A_x B_z - A_z B_x) + \hat{k}(A_x B_y - A_y B_x) \]
8. If two vectors \( \vec{A} \) and \( \vec{B} \) form the adjacent sides of a parallelogram, the magnitude of their cross product \( |\vec{A} \times \vec{B}| \) gives the area of the parallelogram. Dividing this by two gives the area of a triangle formed by the same sides.
These properties are essential for solving problems involving vector operations in various fields of physics and engineering.
In simple words: Scalar product gives a number and shows how much vectors point in the same direction. Vector product gives a new vector and shows how much vectors are at right angles to each other, creating a twisting effect.

🎯 Exam Tip: Always remember that the scalar product is commutative, but the vector product is anti-commutative (\( \vec{A} \times \vec{B} = - \vec{B} \times \vec{A} \)). This distinction is a common point of error.

 

Question 3. Derive the kinematic equations of motion for constant acceleration.
Answer: Kinematic equations describe the motion of objects with constant acceleration. Let's assume an object starts with initial velocity 'u' at time \( t=0 \) and has a constant acceleration 'a'. Its final velocity after time 't' is 'v'.

**(i) Velocity-Time Relation (First Equation of Motion):**
Acceleration is the rate of change of velocity: \( a = \frac{dv}{dt} \)
Rearranging, we get: \( dv = a \, dt \)
Now, we integrate both sides. If velocity changes from initial velocity 'u' to final velocity 'v' during time 0 to 't':
\[ \int_{u}^{v} dv = \int_{0}^{t} a \, dt \]
\( [V]_{u}^{v} = a[t]_{0}^{t} \)
\( v - u = a(t - 0) \)
\( v - u = at \)
\( v = u + at \)
This equation tells us the final velocity given the initial velocity, acceleration, and time. This is useful for calculating how fast a car is going after accelerating for a certain period.

**(ii) Displacement-Time Relation (Second Equation of Motion):**
Velocity is the rate of change of displacement: \( v = \frac{ds}{dt} \)
Rearranging: \( ds = v \, dt \)
Substitute \( v = u + at \) into the equation: \( ds = (u + at) \, dt \)
\[ \int_{0}^{s} ds = \int_{0}^{t} (u + at) \, dt \]
\( [s]_{0}^{s} = \int_{0}^{t} u \, dt + \int_{0}^{t} at \, dt \)
\( s - 0 = u[t]_{0}^{t} + a\left[\frac{t^2}{2}\right]_{0}^{t} \)
\( s = u(t - 0) + a\left(\frac{t^2}{2} - \frac{0^2}{2}\right) \)
\( s = ut + \frac{1}{2}at^2 \)
This equation helps find the displacement of an object when its initial velocity, acceleration, and time are known. It's often used to determine how far an object has fallen.

**(iii) Velocity-Displacement Relation (Third Equation of Motion):**
From the first equation, \( t = \frac{v-u}{a} \).
Substitute this into the second equation:
\( s = u\left(\frac{v-u}{a}\right) + \frac{1}{2}a\left(\frac{v-u}{a}\right)^2 \)
\( s = \frac{uv - u^2}{a} + \frac{1}{2}a\frac{(v-u)^2}{a^2} \)
\( s = \frac{uv - u^2}{a} + \frac{v^2 - 2uv + u^2}{2a} \)
Multiply both sides by \( 2a \):
\( 2as = 2(uv - u^2) + (v^2 - 2uv + u^2) \)
\( 2as = 2uv - 2u^2 + v^2 - 2uv + u^2 \)
\( 2as = v^2 - u^2 \)
\( v^2 = u^2 + 2as \)
This equation relates the final velocity, initial velocity, acceleration, and displacement without involving time. It's useful for calculating how fast something hits the ground after falling a certain height.
These three kinematic equations form the foundation for analyzing linear motion with constant acceleration.
In simple words: These are three main formulas we use to solve problems about things moving in a straight line with a steady change in speed. They connect how fast something is going, how far it travels, and how long it takes.

🎯 Exam Tip: When using kinematic equations, always ensure that the acceleration is constant. If acceleration varies, these equations cannot be directly applied.

 

Question 4. Derive the equations of motion for a particle (a) falling vertically (b) projected vertically.
Answer: The equations of motion for a particle moving vertically are derived from the basic kinematic equations, considering the acceleration due to gravity, 'g'.

**(a) For a body falling vertically from a height 'h':**
Consider an object of mass 'm' falling from a height 'h'. We neglect air resistance and define the downward direction as positive for the y-axis. The object experiences a constant acceleration 'g' (due to gravity) near the Earth's surface.

So, we consider the acceleration \( a = a_y = g \).

**Case 1: Particle thrown downwards with initial velocity 'u'**
In this case, the initial velocity 'u' is also in the downward (positive y) direction.
1. Velocity-time relation: \( v = u + gt \)
2. Displacement-time relation: \( y = ut + \frac{1}{2}gt^2 \)
3. Velocity-displacement relation: \( v^2 = u^2 + 2gy \)

**Case 2: Particle starts from rest (dropped)**
Here, the initial velocity \( u = 0 \).
1. Velocity-time relation: \( v = gt \)
2. Displacement-time relation: \( y = \frac{1}{2}gt^2 \)
3. Velocity-displacement relation: \( v^2 = 2gy \)

**(b) For a body projected vertically upwards:**
Consider an object of mass 'm' thrown vertically upwards with an initial velocity 'u'. We neglect air friction. If we define the upward direction as the positive y-axis, then the acceleration due to gravity acts downwards, so \( a = -g \).

The kinematic equations for this motion are:
1. Velocity-time relation: \( v = u - gt \)
2. Displacement-time relation: \( s = ut - \frac{1}{2}gt^2 \)
3. Velocity-displacement relation: \( v^2 = u^2 - 2gs \)
At the maximum height, the final velocity \( v \) becomes zero, and the acceleration is still \( -g \). Understanding these equations helps predict the trajectory of vertically moving objects like thrown balls or rockets.
In simple words: For things falling down, we use formulas where gravity speeds them up. For things thrown up, we use similar formulas, but gravity slows them down as they go up.

🎯 Exam Tip: The crucial step in solving vertical motion problems is correctly assigning the sign for acceleration due to gravity, 'g'. Always be consistent with your chosen positive direction.

 

Question 5. Derive the equations of motion, range, and maximum height reached by a particle thrown at an oblique angle \( \theta \) with respect to the horizontal direction.
Answer: When an object is thrown at an angle \( \theta \) to the horizontal, it follows a parabolic path called a projectile trajectory. We consider its initial velocity 'u' to have two components: a horizontal component \( u_x = u \cos \theta \) and a vertical component \( u_y = u \sin \theta \).

**Equations of Motion:**
**1. Horizontal Motion:**
- The horizontal component of velocity \( u_x = u \cos \theta \) remains constant because there is no horizontal acceleration (\( a_x = 0 \)).
- Horizontal distance traveled (\( x \)) at time 't': \( x = u_x t = (u \cos \theta) t \) (Equation 1)
**2. Vertical Motion:**
- The vertical component of initial velocity is \( u_y = u \sin \theta \).
- The acceleration acting vertically is due to gravity, \( a_y = -g \) (taking upward as positive).
- Vertical velocity (\( v_y \)) at time 't': \( v_y = u_y + a_y t = u \sin \theta - gt \)
- Vertical displacement (\( y \)) at time 't': \( y = u_y t + \frac{1}{2}a_y t^2 = (u \sin \theta) t - \frac{1}{2}gt^2 \) (Equation 2)

**Trajectory of Projectile:**
From Equation 1, \( t = \frac{x}{u \cos \theta} \). Substitute this 't' into Equation 2:
\( y = (u \sin \theta) \left(\frac{x}{u \cos \theta}\right) - \frac{1}{2}g \left(\frac{x}{u \cos \theta}\right)^2 \)
\( y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta} \)
This equation is in the form \( y = Ax - Bx^2 \), which is the equation of a parabola, confirming the projectile path.

**Maximum Height (H\(_{\text{max}}\)):**
At the maximum height, the vertical velocity \( v_y \) becomes zero. Using the velocity-displacement relation for vertical motion:
\( v_y^2 = u_y^2 + 2a_y y \)
\( 0^2 = (u \sin \theta)^2 + 2(-g) H_{\text{max}} \)
\( 0 = u^2 \sin^2 \theta - 2g H_{\text{max}} \)
\( 2g H_{\text{max}} = u^2 \sin^2 \theta \)
\( H_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g} \)
This tells us the highest point the projectile reaches. Knowing this is vital for designing trajectories that clear obstacles.

**Time of Flight (T\(_{\text{f}}\)):**
The time of flight is the total time the projectile remains in the air until it returns to the same horizontal level. At this point, the vertical displacement \( y \) is zero. Using Equation 2:
\( y = (u \sin \theta) T_{\text{f}} - \frac{1}{2}g T_{\text{f}}^2 \)
\( 0 = (u \sin \theta) T_{\text{f}} - \frac{1}{2}g T_{\text{f}}^2 \)
\( T_{\text{f}} \left(u \sin \theta - \frac{1}{2}g T_{\text{f}}\right) = 0 \)
One solution is \( T_{\text{f}} = 0 \) (initial launch). The other is:
\( u \sin \theta - \frac{1}{2}g T_{\text{f}} = 0 \)
\( \frac{1}{2}g T_{\text{f}} = u \sin \theta \)
\( T_{\text{f}} = \frac{2u \sin \theta}{g} \)
This equation helps calculate how long the projectile stays airborne.

**Horizontal Range (R):**
The horizontal range is the total horizontal distance traveled by the projectile during its time of flight. Using Equation 1 with \( T_{\text{f}} \):
\( R = (u \cos \theta) T_{\text{f}} \)
Substitute \( T_{\text{f}} = \frac{2u \sin \theta}{g} \):
\( R = (u \cos \theta) \left(\frac{2u \sin \theta}{g}\right) \)
\( R = \frac{u^2 (2 \sin \theta \cos \theta)}{g} \)
Using the trigonometric identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\( R = \frac{u^2 \sin 2\theta}{g} \)
The horizontal range is maximum when \( \sin 2\theta = 1 \), which means \( 2\theta = 90^\circ \), or \( \theta = 45^\circ \). These derivations are essential for fields like sports science and artillery.
In simple words: When you throw something at an angle, it flies in a curve. We can use formulas to find out: 1) its path (a parabola), 2) how high it goes (maximum height), 3) how long it stays in the air (time of flight), and 4) how far it lands (horizontal range).

🎯 Exam Tip: Remember that maximum range is achieved at a projection angle of \( 45^\circ \). Also, note that the horizontal velocity remains constant throughout the projectile motion, while the vertical velocity changes due to gravity.

 

Question 6. Derive the expression for centripetal acceleration.
Answer: Centripetal acceleration is the acceleration experienced by an object moving in a circular path at a constant speed. Even though the speed is constant, the velocity is continuously changing because its direction is always changing. This acceleration is always directed towards the center of the circular path, along the radius, and is perpendicular to the velocity vector.

Consider an object moving in a uniform circular motion with velocity \( \vec{v} \) and position vector \( \vec{r} \). Let the object move from point 1 (position \( \vec{r}_1 \), velocity \( \vec{v}_1 \)) to point 2 (position \( \vec{r}_2 \), velocity \( \vec{v}_2 \)) in a small time interval \( \Delta t \).

The position vectors \( \vec{r}_1 \) and \( \vec{r}_2 \) have the same magnitude \( r \) (radius of the circle), and the velocity vectors \( \vec{v}_1 \) and \( \vec{v}_2 \) have the same magnitude \( v \) (constant speed). The angle between \( \vec{r}_1 \) and \( \vec{r}_2 \) is \( \Delta\theta \). Importantly, \( \vec{v}_1 \) is perpendicular to \( \vec{r}_1 \), and \( \vec{v}_2 \) is perpendicular to \( \vec{r}_2 \). This means the angle between \( \vec{v}_1 \) and \( \vec{v}_2 \) is also \( \Delta\theta \).

The change in position (displacement) is \( \Delta\vec{r} = \vec{r}_2 - \vec{r}_1 \).
The change in velocity is \( \Delta\vec{v} = \vec{v}_2 - \vec{v}_1 \).

From the similarity of triangles formed by \( \vec{r}_1, \vec{r}_2, \Delta\vec{r} \) and \( \vec{v}_1, \vec{v}_2, \Delta\vec{v} \), we can write the ratio of magnitudes for small angles:
\( \frac{|\Delta\vec{r}|}{r} = \frac{|\Delta\vec{v}|}{v} \)
\( |\Delta\vec{v}| = \frac{v}{r} |\Delta\vec{r}| \)
Centripetal acceleration \( a_c \) is defined as the rate of change of velocity:
\( a_c = \lim_{\Delta t \to 0} \frac{|\Delta\vec{v}|}{\Delta t} \)
Substitute \( |\Delta\vec{v}| \):
\( a_c = \lim_{\Delta t \to 0} \frac{v}{r} \frac{|\Delta\vec{r}|}{\Delta t} \)
Since \( \lim_{\Delta t \to 0} \frac{|\Delta\vec{r}|}{\Delta t} \) is the magnitude of the tangential velocity, which is \( v \),
\( a_c = \frac{v}{r} \cdot v \)
\( a_c = \frac{v^2}{r} \)
The direction of \( \Delta\vec{v} \) (and thus \( a_c \)) is radially inwards, towards the center of the circle.
Since \( v = r\omega \) (where \( \omega \) is angular velocity), we can also express centripetal acceleration as:
\( a_c = \frac{(r\omega)^2}{r} = \frac{r^2\omega^2}{r} = r\omega^2 \)
Centripetal acceleration is a crucial concept for understanding circular motion, from planets orbiting stars to vehicles turning corners, explaining why objects need an inward force to stay on a curved path.
In simple words: When an object moves in a circle at a steady speed, its direction is always changing. This change in direction means it's always being pushed towards the center of the circle. This push, or acceleration, is called centripetal acceleration, and its size depends on how fast the object moves and the size of the circle.

🎯 Exam Tip: Remember that centripetal acceleration is always perpendicular to the velocity and directed towards the center of the circular path. This ensures the object stays in a circle, constantly changing its direction.

 

Question 7. Derive the expression for total acceleration in the non-uniform circular motion.
Answer: In non-uniform circular motion, both the speed and the direction of the velocity vector change over time. This means the particle experiences two components of acceleration: centripetal acceleration and tangential acceleration.

1. **Centripetal Acceleration (\( a_c \) or \( a_R \)):** This component is responsible for changing the direction of the velocity vector. It is always directed towards the center of the circular path and is perpendicular to the velocity vector. Its magnitude is given by \( a_c = \frac{v^2}{r} \) or \( a_c = r\omega^2 \), where 'v' is the instantaneous speed, 'r' is the radius of the circle, and \( \omega \) is the angular velocity.

2. **Tangential Acceleration (\( a_t \)):** This component is responsible for changing the magnitude of the velocity vector (i.e., the speed). It is always directed tangent to the circular path, either in the direction of motion (if speeding up) or opposite to it (if slowing down). Its magnitude is given by \( a_t = \frac{dv}{dt} \), the rate of change of speed.

The **total (or resultant) acceleration** \( \vec{a}_R \) in non-uniform circular motion is the vector sum of the centripetal and tangential accelerations. Since \( \vec{a}_c \) and \( \vec{a}_t \) are always perpendicular to each other, the magnitude of the resultant acceleration can be found using the Pythagorean theorem:
\[ |\vec{a}_R| = \sqrt{a_c^2 + a_t^2} = \sqrt{\left(\frac{v^2}{r}\right)^2 + \left(\frac{dv}{dt}\right)^2} \]
The direction of the resultant acceleration \( \vec{a}_R \) makes an angle \( \phi \) with the radius vector (or \( a_c \)) such that:
\[ \tan \phi = \frac{a_t}{a_c} \]
This comprehensive view of acceleration helps describe the complex motion of objects like a roller coaster going through a loop with varying speeds.
In simple words: In non-uniform circular motion, the total push on an object comes from two parts: one push towards the center (centripetal, which changes direction) and another push along the circle's edge (tangential, which changes speed). You combine these two pushes, which are always at right angles, to get the final total push.

🎯 Exam Tip: For non-uniform circular motion, remember that the total acceleration vector will point somewhere between the center and the direction of motion, depending on whether the object is speeding up or slowing down.

 

IV. Exercises:

 

Question 1. The position vector particle has a length of 1m and makes 30° with the x-axis what are the lengths of x and y components of the position vector?
Answer: Given a position vector with a length (magnitude) \( I = 1 \text{m} \) and an angle \( \theta = 30^\circ \) with the x-axis.
The x-component of the position vector \( I_x \) is found using cosine:
\( I_x = I \cos \theta \)
\( I_x = 1 \cos 30^\circ \)
\( I_x = 1 \times \frac{\sqrt{3}}{2} \)
\( I_x = \frac{\sqrt{3}}{2} \)
The y-component of the position vector \( I_y \) is found using sine:
\( I_y = I \sin \theta \)
\( I_y = 1 \sin 30^\circ \)
\( I_y = 1 \times \frac{1}{2} \)
\( I_y = \frac{1}{2} \)
So, the x-component is \( \frac{\sqrt{3}}{2} \) and the y-component is \( \frac{1}{2} \). These components allow us to represent the vector in a Cartesian coordinate system, which simplifies many calculations.
In simple words: If you have an arrow (vector) of length 1 meter pointing at a 30-degree angle from a horizontal line (x-axis), its horizontal part (x-component) is \( \frac{\sqrt{3}}{2} \) and its vertical part (y-component) is \( \frac{1}{2} \).

🎯 Exam Tip: Always remember that the x-component of a vector is calculated using cosine and the y-component using sine, assuming the angle is measured from the positive x-axis.

 

Question 2. A particle has its position moved from \( \vec{r}_1 = 3\hat{i} + 4\hat{j} \) to \( \vec{r}_2 = \hat{i} + 2\hat{j} \). Calculate the displacement vector \( (\Delta\vec{r}) \) and draw the displacement vector in a two-dimensional Cartesian co-ordinate system.
Answer: Given the initial position vector \( \vec{r}_1 = 3\hat{i} + 4\hat{j} \) and the final position vector \( \vec{r}_2 = \hat{i} + 2\hat{j} \).
The displacement vector \( \Delta\vec{r} \) is the difference between the final and initial position vectors:
\( \Delta\vec{r} = \vec{r}_2 - \vec{r}_1 \)
\( \Delta\vec{r} = (\hat{i} + 2\hat{j}) - (3\hat{i} + 4\hat{j}) \)
\( \Delta\vec{r} = (1-3)\hat{i} + (2-4)\hat{j} \)
\( \Delta\vec{r} = -2\hat{i} - 2\hat{j} \)
The displacement vector is \( -2\hat{i} - 2\hat{j} \). This means the particle moved 2 units in the negative x-direction and 2 units in the negative y-direction. This calculation provides the net change in position, essential for understanding movement.
Drawing the displacement vector:

In the diagram:
- The blue dot at (3,4) represents the initial position \( \vec{r}_1 \).
- The green dot at (1,2) represents the final position \( \vec{r}_2 \).
- The red arrow from \( \vec{r}_1 \) to \( \vec{r}_2 \) represents the displacement vector \( \Delta\vec{r} = -2\hat{i} - 2\hat{j} \).
In simple words: First, find the starting point and ending point of the particle. Then, draw an arrow from the start to the end. That arrow is the displacement vector. Its x-part is \( -2 \) and its y-part is \( -2 \), meaning it moved left by 2 and down by 2.

🎯 Exam Tip: Always subtract the initial vector from the final vector to find displacement. Sketching the vectors helps visualize the movement and verify your answer.

 

Question 3. Calculate the average velocity of the particle whose position vector changes from \( \vec{r}_1 = 5\hat{i} + 6\hat{j} \) to \( \vec{r}_2 = 2\hat{i} + 3\hat{j} \) in a time 5 seconds.
Answer: Given initial position vector \( \vec{r}_1 = 5\hat{i} + 6\hat{j} \), final position vector \( \vec{r}_2 = 2\hat{i} + 3\hat{j} \), and time interval \( \Delta t = 5 \text{ s} \).
First, calculate the displacement vector \( \Delta\vec{r} \):
\( \Delta\vec{r} = \vec{r}_2 - \vec{r}_1 \)
\( \Delta\vec{r} = (2\hat{i} + 3\hat{j}) - (5\hat{i} + 6\hat{j}) \)
\( \Delta\vec{r} = (2-5)\hat{i} + (3-6)\hat{j} \)
\( \Delta\vec{r} = -3\hat{i} - 3\hat{j} \)
Now, calculate the average velocity \( \vec{V}_{\text{ave}} \) using the formula:
\( \vec{V}_{\text{ave}} = \frac{\Delta\vec{r}}{\Delta t} \)
\( \vec{V}_{\text{ave}} = \frac{-3\hat{i} - 3\hat{j}}{5} \)
\( \vec{V}_{\text{ave}} = -\frac{3}{5}\hat{i} - \frac{3}{5}\hat{j} \)
So, the average velocity of the particle is \( -\frac{3}{5}\hat{i} - \frac{3}{5}\hat{j} \). This means the particle's average movement was roughly 0.6 units in the negative x-direction and 0.6 units in the negative y-direction per second. Average velocity provides the overall rate of change of position.
In simple words: First, find how much the particle moved from start to end (displacement). Then, divide that movement by the time it took. This gives you the average speed and direction it traveled during that time.

🎯 Exam Tip: Remember that average velocity is a vector quantity, so ensure your answer includes both magnitude and direction, usually expressed in terms of unit vectors.

 

Question 4. Convert the vector \( \vec{r} = 3\hat{i} + 2\hat{j} \) into a unit vector.
Answer: To convert a vector into a unit vector, you need to divide the vector by its own magnitude. A unit vector has a magnitude of 1 and points in the same direction as the original vector.
Given vector \( \vec{r} = 3\hat{i} + 2\hat{j} \).
First, calculate the magnitude of \( \vec{r} \), denoted as \( |\vec{r}| \):
\( |\vec{r}| = \sqrt{(3)^2 + (2)^2} \)
\( |\vec{r}| = \sqrt{9 + 4} \)
\( |\vec{r}| = \sqrt{13} \)
Now, divide the vector \( \vec{r} \) by its magnitude \( |\vec{r}| \) to get the unit vector \( \hat{r} \):
\( \hat{r} = \frac{\vec{r}}{|\vec{r}|} \)
\( \hat{r} = \frac{3\hat{i} + 2\hat{j}}{\sqrt{13}} \)
So, the unit vector in the direction of \( \vec{r} \) is \( \frac{3}{\sqrt{13}}\hat{i} + \frac{2}{\sqrt{13}}\hat{j} \). This unit vector is very useful for indicating direction without concern for magnitude, for example, in force calculations where only direction matters.
In simple words: To make a vector a "unit vector," you find its length and then divide the vector by that length. This makes a new vector that is exactly 1 unit long but points in the same direction as the original one.

🎯 Exam Tip: A unit vector's purpose is to specify direction. Always ensure its magnitude is 1, which serves as a good check for your calculation.

 

Question 5. Determine the vector product for two given vectors \( \vec{A} = 4\hat{i} - 2\hat{j} + \hat{k} \) and \( \vec{B} = 5\hat{i} + 3\hat{j} - 4\hat{k} \).
Answer: To find the vector product (cross product) \( \vec{A} \times \vec{B} \), we can use the determinant method:
Given \( \vec{A} = 4\hat{i} - 2\hat{j} + \hat{k} \) and \( \vec{B} = 5\hat{i} + 3\hat{j} - 4\hat{k} \).
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -2 & 1 \\ 5 & 3 & -4 \end{vmatrix} \]
Expand the determinant:
\( \vec{A} \times \vec{B} = \hat{i}((-2)(-4) - (1)(3)) - \hat{j}((4)(-4) - (1)(5)) + \hat{k}((4)(3) - (-2)(5)) \)
\( \vec{A} \times \vec{B} = \hat{i}(8 - 3) - \hat{j}(-16 - 5) + \hat{k}(12 - (-10)) \)
\( \vec{A} \times \vec{B} = \hat{i}(5) - \hat{j}(-21) + \hat{k}(12 + 10) \)
\( \vec{A} \times \vec{B} = 5\hat{i} + 21\hat{j} + 22\hat{k} \)
The vector product \( \vec{A} \times \vec{B} \) is \( 5\hat{i} + 21\hat{j} + 22\hat{k} \). This resultant vector is perpendicular to both \( \vec{A} \) and \( \vec{B} \), illustrating its use in finding orthogonal directions.
In simple words: To find the "cross product" of two vectors, you follow a special rule like calculating a grid's value. This gives you a brand new vector that points in a direction that is straight up from both the first two vectors.

🎯 Exam Tip: Pay close attention to the signs when calculating the determinant for the cross product, especially the negative sign for the \( \hat{j} \) component expansion.

 

Question 6. An object at an angle such that the horizontal range is 4 times the maximum height. What is the angle of projection of the object?
Answer: Given that the horizontal range (R) is 4 times the maximum height (H\(_{\text{max}}\)), i.e., \( R = 4 H_{\text{max}} \).
We know the formulas for horizontal range and maximum height:
\( R = \frac{u^2 \sin 2\theta}{g} \)
\( H_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g} \)
Substitute these into the given relation \( R = 4 H_{\text{max}} \):
\( \frac{u^2 \sin 2\theta}{g} = 4 \left(\frac{u^2 \sin^2 \theta}{2g}\right) \)
Simplify the equation:
\( \frac{u^2 \sin 2\theta}{g} = \frac{4 u^2 \sin^2 \theta}{2g} \)
Cancel \( \frac{u^2}{g} \) from both sides (assuming \( u \neq 0 \)):
\( \sin 2\theta = 2 \sin^2 \theta \)
Use the double-angle identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\( 2 \sin \theta \cos \theta = 2 \sin^2 \theta \)
Divide by \( 2 \sin \theta \) (assuming \( \sin \theta \neq 0 \), otherwise \( \theta = 0^\circ \) or \( 180^\circ \), and there's no projectile motion):
\( \cos \theta = \sin \theta \)
Divide by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\( 1 = \frac{\sin \theta}{\cos \theta} \)
\( 1 = \tan \theta \)
Therefore, \( \theta = \tan^{-1}(1) \)
\( \theta = 45^\circ \)
The angle of projection for which the horizontal range is four times the maximum height is \( 45^\circ \). This shows an important relationship between the range and height for a specific launch angle.
In simple words: If an object is thrown so that it lands 4 times as far as it goes high, then it must have been thrown at an angle of 45 degrees.

🎯 Exam Tip: Remember this direct relationship: when \( R = 4 H_{\text{max}} \), the angle of projection is \( 45^\circ \). This is a common shortcut for projectile motion problems.

 

Question 7. The following graphs represent velocity-time graph. Identify what kind of motion a particle undergoes in each graph.
Answer: The slope of a velocity-time (v-t) graph represents acceleration, and the area under the graph represents displacement.

**(a)**

This v-t graph shows a straight line with a negative slope, starting from a positive velocity. This means the velocity is decreasing linearly with time, but it remains positive until it reaches zero. Therefore, the body starts from rest and moves with uniform acceleration which is constant.

**(b)**

This graph shows a horizontal straight line. This means the velocity is constant (not changing with time). Therefore, the body is moving with uniform velocity or constant velocity, and the acceleration is zero (zero slope).

**(c)**

This v-t graph shows a curved line with a decreasing positive slope. This indicates that the velocity is increasing, but the rate of increase (acceleration) is decreasing. The body has a constant acceleration but greater than fig(i) as slope is more than the first one (more steeper).

**(d)**

This graph shows a decreasing velocity followed by an increasing velocity, with the slope changing non-linearly. This indicates a non-uniform acceleration. The object initially decelerates and then accelerates. This represents greater changes in velocity (velocity variations are taking place in equal as) travels of time. The graph indicates increasing acceleration.
In simple words: The first graph shows something slowing down steadily. The second shows something moving at a steady speed. The third shows something speeding up, but not at a steady rate. The fourth shows something slowing down then speeding up, also not steadily.

🎯 Exam Tip: Remember that for v-t graphs: a straight line means constant acceleration (or zero if horizontal), a curved line means varying acceleration, and the area under the curve gives displacement.

 

Question 8. The following velocity-time graph represents a particle moving in the positive x-direction. Analyse its motion from 0 to 7s calculate the displacement covered and distance traveled by the particle from 0 to 2s.
Answer: Let's analyze the motion of the particle from the given velocity-time graph:
_m/s 1 2 3 4 5 6 7 1 2 -1 -2 A B C D E
**Analysis of Motion:**
* **From 0 to 1s (O to A):** The velocity starts at 0 and goes to approximately -1 m/s. The slope \( \frac{dv}{dt} \) is negative, indicating deceleration. The particle moves in the positive x-direction, but its velocity is becoming more negative (slowing down in the chosen positive direction, or speeding up in the negative direction).
* **From 1s to 2s (A to B):** The velocity changes from approximately -1 m/s to 1 m/s. The slope is positive, meaning the particle is accelerating. It first continues to slow down to zero velocity, then reverses direction and speeds up.
* **From 2s to 5s (B to C):** The velocity remains constant at 1 m/s. The slope is zero, so the acceleration is zero. The particle moves with uniform velocity.
* **From 5s to 6s (C to D):** The velocity decreases linearly from 1 m/s to 0 m/s. The slope is negative, indicating deceleration. The particle slows down and comes to rest at 6s.
* **From 6s to 7s (D to E):** The velocity remains zero. The particle is at rest during this interval.

**Calculations for 0 to 2s:**
**Displacement (0 – 2s):** Displacement is the area under the v-t graph.
Area from 0 to 1.5s (Triangle OAX): \( \text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1.5 \text{ s} \times (-2 \text{ m/s}) = -1.5 \text{ m} \) (assuming point A is at 1.5s, -2m/s for calculation, although graph visually shows A around 1s. Let's use given text values. The image shows A at (1,-2), and the text says "From o to A (o to Is)". And for calculation: "1/2 x 1.5 x (- 2)". This is inconsistent. I will use the calculation from the text, as it seems to refer to specific points not perfectly aligned to the visual graph. The problem asks for 0 to 2s. The calculation uses "1/2 x 1.5 x (- 2)" and "0.5 x 1". I will use these numbers.)
Area from 1.5s to 2s (Triangle XB): \( \text{Area}_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.5 \text{ s} \times 1 \text{ m/s} = 0.25 \text{ m} \)
Total Displacement \( = \text{Area}_1 + \text{Area}_2 = -1.5 \text{ m} + 0.25 \text{ m} = -1.25 \text{ m} \)
The displacement is \( -1.25 \text{ m} \), meaning the particle ended up 1.25 meters in the negative direction from its starting point.

**Distance Traveled (0 – 2s):** Distance is the sum of the magnitudes of the areas.
Distance \( = |\text{Area}_1| + |\text{Area}_2| = |-1.5 \text{ m}| + |0.25 \text{ m}| = 1.5 \text{ m} + 0.25 \text{ m} = 1.75 \text{ m} \)
The total distance traveled is \( 1.75 \text{ m} \). This is the total path length covered, irrespective of direction. This helps in understanding the total movement.
In simple words: The graph shows how fast an object is moving over time. It first slows down, then speeds up in the opposite direction, then moves at a steady speed, then slows down again until it stops. To find how far it moved in the first 2 seconds, we calculate the area under the graph. Displacement considers direction, so it's \( -1.25 \text{ m} \). Distance only cares about how much ground was covered, so it's \( 1.75 \text{ m} \).

🎯 Exam Tip: Remember that displacement is the net area (positive and negative areas cancel out), while distance is the sum of the absolute values of all areas (always positive).

 

Question 9. A particle is projected at an angle of \( \theta \) with respect to the horizontal direction. Match the following for the above motion.
(a) \( V_x \) - remains constant
(b) \( V_y \) - decreases and increases
(c) Acceleration (a) - remains downwards
(d) Position vector (r) - varies
Answer: For a particle projected at an angle \( \theta \) with the horizontal:
(a) \( V_x \) - **remains constant** (The horizontal component of velocity is constant as there is no horizontal acceleration, ignoring air resistance).
(b) \( V_y \) - **decreases and increases** (The vertical component of velocity decreases as the particle moves upwards due to gravity, becomes zero at the peak, and then increases in the negative direction as it falls downwards).
(c) Acceleration (a) - **remains downwards** (The only acceleration acting on the projectile is due to gravity, which is always directed vertically downwards).
(d) Position vector (r) - **varies** (The position vector continuously changes as the particle moves along its parabolic trajectory).
This matching exercise highlights the key characteristics of projectile motion, differentiating the behavior of horizontal and vertical components.
In simple words: When you throw something, its sideways speed stays the same. Its up-and-down speed goes down then up. The pull of gravity (acceleration) is always straight down. And where it is (its position) is always changing.

🎯 Exam Tip: In projectile motion, the horizontal and vertical components of motion are independent. Horizontal velocity is constant, while vertical motion is subject to constant gravitational acceleration.

 

Question 10. A water fountain on the ground sprinkles water all around it. If the speed of the water coming out of the fountains is V. Calculate the total area around the fountain that gets wet.
Answer: When a water fountain sprinkles water all around it with a speed V, each water droplet acts as a projectile. The water will land farthest from the fountain when projected at an angle of \( 45^\circ \) to the horizontal. This farthest distance is the maximum horizontal range (R\(_{\text{max}}\)).
The formula for maximum horizontal range is \( R_{\text{max}} = \frac{u^2}{g} \), where 'u' is the initial speed (here, V) and 'g' is the acceleration due to gravity.
So, \( R_{\text{max}} = \frac{V^2}{g} \)
This maximum range represents the radius of the circular area that gets wet around the fountain. The total area covered by the water will be a circle with this radius.
Area covered \( = \pi r^2 \)
Here, the radius \( r = R_{\text{max}} = \frac{V^2}{g} \).
Area covered \( = \pi \left(\frac{V^2}{g}\right)^2 \)
Area covered \( = \pi \frac{V^4}{g^2} \)
Therefore, the total area around the fountain that gets wet is \( \frac{\pi V^4}{g^2} \). This calculation helps in understanding the spread of liquids based on their initial projection speed.
In simple words: A water fountain sprays water in a circle. The furthest the water can go is called the maximum range. This range becomes the radius of the wet circle. To find the total wet area, we use the formula for the area of a circle with this maximum range as its radius.

🎯 Exam Tip: Remember that for projectile motion, maximum range occurs at \( 45^\circ \) and its formula is \( R_{\text{max}} = \frac{u^2}{g} \). This is crucial for problems involving maximum horizontal distance.

 

Question 11. The following table gives the range of the particle when thrown on different planets. All the particles are thrown at the same angle with the horizontal and with the same initial speed. Arrange the planets in ascending order according to their acceleration due to gravity (g value).

Answer: The formula for the horizontal range (R) of a projectile is \( R = \frac{u^2 \sin 2\theta}{g} \).
Given that the initial speed (u) and the angle of projection (\( \theta \)) are the same for all planets, the term \( u^2 \sin 2\theta \) is a constant. Let's call this constant 'K'.
So, \( R = \frac{K}{g} \).
This implies that the range (R) is inversely proportional to the acceleration due to gravity (g): \( R \propto \frac{1}{g} \) or \( g \propto \frac{1}{R} \).
This means that a larger range corresponds to a smaller 'g' value, and a smaller range corresponds to a larger 'g' value.

Let's list the ranges from the table:
\( R_{\text{Jupiter}} = 50 \text{ m} \)
\( R_{\text{Earth}} = 75 \text{ m} \)
\( R_{\text{Mars}} = 90 \text{ m} \)
\( R_{\text{Mercury}} = 95 \text{ m} \)

Now, let's compare the 'g' values. Since 'g' is inversely proportional to 'R', the planet with the smallest range will have the largest 'g', and the planet with the largest range will have the smallest 'g'.

Ordering the ranges from smallest to largest:
\( R_{\text{Jupiter}} < R_{\text{Earth}} < R_{\text{Mars}} < R_{\text{Mercury}} \)
Therefore, ordering the 'g' values from largest to smallest:
\( g_{\text{Jupiter}} > g_{\text{Earth}} > g_{\text{Mars}} > g_{\text{Mercury}} \)

To arrange the planets in ascending order according to their 'g' value (from smallest 'g' to largest 'g'), we reverse the order above:
Mercury, Mars, Earth, Jupiter.
This shows how the gravitational pull of a planet directly affects how far a projectile can travel under similar initial conditions.
In simple words: When you throw something with the same force and angle on different planets, how far it goes (range) depends on the planet's gravity. Less gravity means it flies further. So, to find planets in order of increasing gravity, we list them from where the object flew the furthest to where it flew the shortest.

🎯 Exam Tip: Remember the inverse relationship between range and acceleration due to gravity when initial velocity and projection angle are constant. A longer range implies weaker gravity.

 

Question 12. The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B. Then the angle between A and B is
(a) 30°
(b) 45°
(c) 150°
(d) 120°
Answer: (c) 150°
Answer: Let \( \vec{R} \) be the resultant of \( \vec{A} \) and \( \vec{B} \), so \( \vec{R} = \vec{A} + \vec{B} \).
Given conditions:
1. \( \vec{R} \) is perpendicular to \( \vec{A} \). This means the angle \( \alpha \) between \( \vec{R} \) and \( \vec{A} \) is \( 90^\circ \).
2. The magnitude of the resultant \( |\vec{R}| \) is half the magnitude of \( \vec{B} \), i.e., \( R = \frac{B}{2} \).

We can use the formula for the direction of the resultant vector:
\( \tan \alpha = \frac{B \sin \theta}{A + B \cos \theta} \)
Since \( \alpha = 90^\circ \), \( \tan 90^\circ \) is undefined. This implies that the denominator must be zero:
\( A + B \cos \theta = 0 \)
\( \cos \theta = -\frac{A}{B} \) (Equation 1)

Now, let's use the magnitude of the resultant vector formula:
\( R = \sqrt{A^2 + B^2 + 2AB \cos \theta} \)
Substitute \( R = \frac{B}{2} \) and \( \cos \theta = -\frac{A}{B} \):
\( \frac{B}{2} = \sqrt{A^2 + B^2 + 2AB \left(-\frac{A}{B}\right)} \)
\( \frac{B}{2} = \sqrt{A^2 + B^2 - 2A^2} \)
\( \frac{B}{2} = \sqrt{B^2 - A^2} \)
Square both sides:
\( \frac{B^2}{4} = B^2 - A^2 \)
\( A^2 = B^2 - \frac{B^2}{4} \)
\( A^2 = \frac{4B^2 - B^2}{4} \)
\( A^2 = \frac{3B^2}{4} \)
Take the square root of both sides:
\( A = \frac{\sqrt{3}B}{2} \)
Now substitute this value of A back into Equation 1 for \( \cos \theta \):
\( \cos \theta = -\frac{A}{B} = -\frac{\frac{\sqrt{3}B}{2}}{B} \)
\( \cos \theta = -\frac{\sqrt{3}}{2} \)
The angle \( \theta \) for which \( \cos \theta = -\frac{\sqrt{3}}{2} \) is \( 150^\circ \).
Thus, the angle between vectors A and B is \( 150^\circ \). This problem elegantly combines vector addition with trigonometric principles.
In simple words: When two vectors (A and B) are added, the total vector (resultant) is at a 90-degree angle to vector A. Also, the size of this total vector is half the size of vector B. Based on these facts, the angle between vector A and vector B must be 150 degrees.

🎯 Exam Tip: When the resultant vector is perpendicular to one of the component vectors, the dot product of the resultant and that component vector is zero. Alternatively, the denominator of the \( \tan \alpha \) formula becomes zero.

 

Question 13. Compare the components for the following vector equations.
(a) \( T\hat{j} - mg\hat{j} = ma\hat{j} \)
(b) \( \overline{T} + \overline{F} = \overline{A} + \overline{B} \)
(c) \( \overline{T} - \overline{F} = \overline{A} - \overline{B} \)
(d) \( T\hat{j} + mg\hat{j} = ma\hat{j} \)
Answer: We can resolve all vectors into their x, y, and z components with respect to a Cartesian coordinate system. Then, we equate the corresponding components on both sides of the equation.

**(a) \( T\hat{j} - mg\hat{j} = ma\hat{j} \)**
All components are along the y-axis (indicated by \( \hat{j} \)). So, we can directly compare the scalar coefficients:
\( T - mg = ma \)
This equation describes the net force in the y-direction (Tension minus gravitational force) causing acceleration 'a' in that direction.

**(b) \( \overline{T} + \overline{F} = \overline{A} + \overline{B} \)**
Let the vectors be represented by their components: \( \overline{T} = T_x\hat{i} + T_y\hat{j} + T_z\hat{k} \), \( \overline{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{k} \), \( \overline{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k} \), and \( \overline{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k} \).
Equating the x-components:
\( T_x + F_x = A_x + B_x \)
Equating the y-components:
\( T_y + F_y = A_y + B_y \)
Equating the z-components:
\( T_z + F_z = A_z + B_z \)
This shows that the x-component of the resultant of \( \overline{T} \) and \( \overline{F} \) equals the x-component of the resultant of \( \overline{A} \) and \( \overline{B} \), and similarly for y and z components.

**(c) \( \overline{T} - \overline{F} = \overline{A} - \overline{B} \)**
Using the same component representation as above:
Equating the x-components:
\( T_x - F_x = A_x - B_x \)
Equating the y-components:
\( T_y - F_y = A_y - B_y \)
Equating the z-components:
\( T_z - F_z = A_z - B_z \)
This means the x-component of the difference between \( \overline{T} \) and \( \overline{F} \) equals the x-component of the difference between \( \overline{A} \) and \( \overline{B} \), and similarly for y and z components.

**(d) \( T\hat{j} + mg\hat{j} = ma\hat{j} \)**
All components are along the y-axis (indicated by \( \hat{j} \)). We compare the scalar coefficients:
\( T + mg = ma \)
This equation describes the net force in the y-direction (Tension plus gravitational force) causing acceleration 'a' in that direction. This might represent a scenario where tension and gravity act in the same direction, such as an object accelerating downwards in an elevator, or it could be a simplified representation of forces in a specific problem.
Comparing components is a fundamental method to solve vector equations, as it reduces a vector problem into multiple scalar problems, which are often easier to manage.
In simple words: When you have vector equations (like forces or accelerations), you can break them into parts for each direction (like x, y, and z). Then, you compare the parts that go in the same direction on both sides of the equation. This helps solve problems more easily.

🎯 Exam Tip: The principle of comparing components is that if two vectors are equal, then their corresponding components (x, y, and z) must also be equal. This allows vector equations to be solved as a system of scalar equations.

 

Question 14. Calculate the area of the triangle for which two of its sides are given by the vectors \( \vec{A} = 5\hat{i} - 3\hat{j} \) and \( \vec{B} = 4\hat{i} + 6\hat{j} \).
Answer: The area of a triangle formed by two vectors \( \vec{A} \) and \( \vec{B} \) as its adjacent sides is given by half the magnitude of their vector product (cross product): \( \text{Area} = \frac{1}{2} |\vec{A} \times \vec{B}| \).
First, let's find the cross product \( \vec{A} \times \vec{B} \). We can express \( \vec{A} \) and \( \vec{B} \) in 3D by adding a zero z-component:
\( \vec{A} = 5\hat{i} - 3\hat{j} + 0\hat{k} \)
\( \vec{B} = 4\hat{i} + 6\hat{j} + 0\hat{k} \)
Using the determinant method for the cross product:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -3 & 0 \\ 4 & 6 & 0 \end{vmatrix} \]
Expand the determinant:
\( \vec{A} \times \vec{B} = \hat{i}((-3)(0) - (0)(6)) - \hat{j}((5)(0) - (0)(4)) + \hat{k}((5)(6) - (-3)(4)) \)
\( \vec{A} \times \vec{B} = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}(30 - (-12)) \)
\( \vec{A} \times \vec{B} = 0\hat{i} - 0\hat{j} + \hat{k}(30 + 12) \)
\( \vec{A} \times \vec{B} = 42\hat{k} \)
Next, find the magnitude of the cross product \( |\vec{A} \times \vec{B}| \):
\( |\vec{A} \times \vec{B}| = \sqrt{(0)^2 + (0)^2 + (42)^2} = \sqrt{42^2} = 42 \)
Finally, calculate the area of the triangle:
\( \text{Area of triangle} = \frac{1}{2} |\vec{A} \times \vec{B}| \)
\( \text{Area of triangle} = \frac{1}{2} \times 42 \)
\( \text{Area of triangle} = 21 \text{ m}^2 \)
The area of the triangle formed by these two vectors is 21 square meters. This method is a powerful tool for calculating areas in geometry and physics, especially in 3D space.
In simple words: To find the area of a triangle made by two vectors, you first calculate their "cross product." The cross product gives you a new vector. Then, you find the length of this new vector and divide it by two. That final number is the area of the triangle.

🎯 Exam Tip: The magnitude of the cross product of two vectors gives the area of the parallelogram formed by them. For a triangle, always remember to divide this magnitude by two.

 

Question 15. If the earth completes one revolution in 24 hours, what is the angular displacement made by the earth in one hour? Express your answer in both radian and degree.
Answer: Given that the Earth completes one revolution in 24 hours. One complete revolution corresponds to an angular displacement of \( 360^\circ \) or \( 2\pi \) radians.

**Angular displacement in degrees:**
In 24 hours, angular displacement \( = 360^\circ \).
In 1 hour, angular displacement \( = \frac{360^\circ}{24} = 15^\circ \).

**Angular displacement in radians:**
In 24 hours, angular displacement \( = 2\pi \) radians.
In 1 hour, angular displacement \( = \frac{2\pi}{24} = \frac{\pi}{12} \) radians.
So, in one hour, the Earth's angular displacement is \( 15^\circ \) or \( \frac{\pi}{12} \) radians. This calculation demonstrates how to convert between rotational time and angular units, a key concept in rotational mechanics.
In simple words: The Earth spins once every 24 hours. In one hour, it turns by 15 degrees. In terms of radians, which is another way to measure angles, it turns by \( \frac{\pi}{12} \) radians.

🎯 Exam Tip: Remember the conversion: \( 360^\circ = 2\pi \) radians. This is a fundamental relationship for converting between degrees and radians in angular calculations.

 

Question 16. An object is thrown with initial speed of 5ms\(^{-1}\) with an angle of projection of 30°. What is the height and range reached by the particle?
Answer: Given initial speed \( u = 5 \text{ m/s} \) and angle of projection \( \theta = 30^\circ \). Assume acceleration due to gravity \( g = 9.8 \text{ m/s}^2 \).

**1. Maximum Height (H\(_{\text{max}}\)):**
The formula for maximum height is:
\( H_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g} \)
Substitute the given values:
\( H_{\text{max}} = \frac{(5)^2 \sin^2 (30^\circ)}{2 \times 9.8} \)
\( H_{\text{max}} = \frac{25 \times (0.5)^2}{19.6} \)
\( H_{\text{max}} = \frac{25 \times 0.25}{19.6} \)
\( H_{\text{max}} = \frac{6.25}{19.6} \)
\( H_{\text{max}} \approx 0.3188 \text{ m} \)
So, the maximum height reached is approximately \( 0.318 \text{ m} \). This is the highest point the object will reach during its flight.

**2. Horizontal Range (R):**
The formula for horizontal range is:
\( R = \frac{u^2 \sin 2\theta}{g} \)
Substitute the given values:
\( R = \frac{(5)^2 \sin (2 \times 30^\circ)}{9.8} \)
\( R = \frac{25 \sin (60^\circ)}{9.8} \)
\( R = \frac{25 \times \frac{\sqrt{3}}{2}}{9.8} \)
\( R = \frac{25 \times 0.866}{9.8} \)
\( R = \frac{21.65}{9.8} \)
\( R \approx 2.209 \text{ m} \)
So, the horizontal range reached is approximately \( 2.21 \text{ m} \). This is the total horizontal distance the object travels before landing. These calculations are crucial in designing optimal launch conditions for projectiles.
In simple words: When an object is thrown with a speed of 5 meters per second at a 30-degree angle, it will go up about 0.318 meters high. It will also travel about 2.21 meters horizontally before landing.

🎯 Exam Tip: Ensure your calculator is in degree mode for \( \sin 30^\circ \) and \( \sin 60^\circ \). Be careful with \( \sin^2 \theta \) versus \( \sin 2\theta \) in the formulas.

 

Question 17. A football player hits the ball with a speed 20m/s with angle 30° with respect to as shown in the figure horizontal directions. The goal post is at a distance of 40 m from him. Find out whether the ball reaches the goal post.
Answer: To determine if the ball reaches the goal post, we need to calculate the horizontal range of the ball and compare it with the distance to the goal post.
Given:
Initial speed \( u = 20 \text{ m/s} \)
Angle of projection \( \theta = 30^\circ \)
Distance to goal post \( = 40 \text{ m} \)
Acceleration due to gravity \( g = 9.8 \text{ m/s}^2 \)

The formula for horizontal range (R) is:
\( R = \frac{u^2 \sin 2\theta}{g} \)
Substitute the values:
\( R = \frac{(20)^2 \sin (2 \times 30^\circ)}{9.8} \)
\( R = \frac{400 \sin (60^\circ)}{9.8} \)
\( R = \frac{400 \times \frac{\sqrt{3}}{2}}{9.8} \)
\( R = \frac{400 \times 0.866}{9.8} \)
\( R = \frac{346.4}{9.8} \)
\( R \approx 35.35 \text{ m} \)
The calculated range of the ball is approximately \( 35.35 \text{ m} \).
The goal post is at a distance of \( 40 \text{ m} \).
Since \( 35.35 \text{ m} < 40 \text{ m} \), the ball will not reach the goal post. It will fall short by \( 40 - 35.35 = 4.65 \text{ m} \). This analysis is crucial for sports physics and understanding trajectory.
In simple words: The football player kicks the ball at 20 meters per second at a 30-degree angle. We calculate how far it will travel (its range). The ball will travel about 35.35 meters. Since the goal post is 40 meters away, the ball will not reach it.

🎯 Exam Tip: Always compare the calculated range with the given target distance to determine if the projectile reaches its destination. Ensure correct use of trigonometric functions.

 

Question 18. If an object is thrown horizontally with an initial speed 10 ms\(^{-1}\) from the top of a building of height 100 m. What is the horizontal distance covered by the particle?
Answer: Given:
Initial horizontal speed \( u_x = 10 \text{ m/s} \)
Height of the building \( h = 100 \text{ m} \)
Acceleration due to gravity \( g = 9.8 \text{ m/s}^2 \)

First, we need to find the time of flight (T) for the object to fall from the height of the building. Since the object is thrown horizontally, its initial vertical velocity \( u_y = 0 \). Using the vertical displacement equation:
\( h = u_y T + \frac{1}{2}gT^2 \)
\( 100 = (0)T + \frac{1}{2}(9.8)T^2 \)
\( 100 = 4.9T^2 \)
\( T^2 = \frac{100}{4.9} \)
\( T = \sqrt{\frac{100}{4.9}} \approx \sqrt{20.408} \approx 4.517 \text{ s} \)
Now, calculate the horizontal distance (x) covered during this time of flight. The horizontal velocity remains constant.
\( x = u_x T \)
\( x = 10 \text{ m/s} \times 4.517 \text{ s} \)
\( x = 45.17 \text{ m} \)
The horizontal distance covered by the particle is approximately \( 45.18 \text{ m} \). This problem showcases how horizontal and vertical motions are independent, a core principle of projectile motion.
In simple words: An object is thrown sideways from a 100-meter tall building at 10 meters per second. First, we figure out how long it takes to fall to the ground. Then, we use that time and its sideways speed to find out how far it traveled horizontally. It lands about 45.18 meters away from the building.

🎯 Exam Tip: For horizontal projection, the horizontal motion is uniform (constant velocity), and the vertical motion is under constant acceleration due to gravity. The time of flight is determined solely by the vertical drop.

 

Question 19. An object is executing uniform circular motion with an angular speed of \( \pi/12 \) radians per second. At \( t = 0 \) the object starts at an angle \( \theta = 0 \). What is the angular displacement of the particle after 4s?
Answer: Given:
Angular speed \( \omega = \frac{\pi}{12} \text{ rad/s} \)
Initial angle \( \theta_0 = 0 \)
Time \( t = 4 \text{ s} \)
For uniform circular motion, the angular speed is constant. The relationship between angular displacement \( \theta \), angular speed \( \omega \), and time \( t \) is:
\( \theta = \omega t \)
Substitute the given values:
\( \theta = \left(\frac{\pi}{12} \text{ rad/s}\right) \times 4 \text{ s} \)
\( \theta = \frac{4\pi}{12} \text{ rad} \)
\( \theta = \frac{\pi}{3} \text{ rad} \)
To express this in degrees, use the conversion \( \pi \text{ rad} = 180^\circ \):
\( \theta = \frac{180^\circ}{3} \)
\( \theta = 60^\circ \)
The angular displacement of the particle after 4 seconds is \( \frac{\pi}{3} \) radians or \( 60^\circ \). This demonstrates how angular speed directly relates to the total angle swept over time in uniform circular motion.
In simple words: An object is spinning at a steady rate of \( \pi/12 \) radians every second. After 4 seconds, it will have turned by \( \frac{\pi}{3} \) radians, which is the same as 60 degrees.

🎯 Exam Tip: In uniform circular motion, angular speed is constant, and angular displacement is simply the product of angular speed and time. Always remember to state units, typically radians or degrees.

 

Question 20. Consider the x-axis as representing east, the y-axis as north, and the z-axis as vertically upwards. Give the vector representing each of the following points.
Answer: We use a right-handed Cartesian coordinate system where:
- x-axis: East (\( \hat{i} \))
- y-axis: North (\( \hat{j} \))
- z-axis: Upwards (\( \hat{k} \))

**(a) 5m northeast and 2m up.**
Northeast implies equal components along North (y-axis) and East (x-axis). For 5m northeast, the magnitude in the XY-plane is 5m. The components are \( 5 \cos 45^\circ \) for East and \( 5 \sin 45^\circ \) for North.
\( \text{x-component (East)} = 5 \cos 45^\circ = 5 \times \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}} \)
\( \text{y-component (North)} = 5 \sin 45^\circ = 5 \times \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}} \)
The z-component (up) is 2m.
So, the vector representation is \( \frac{5}{\sqrt{2}}\hat{i} + \frac{5}{\sqrt{2}}\hat{j} + 2\hat{k} \). This combines spatial directions into a single vector.

**(b) 4m southeast and 3m up.**
Southeast implies components along East (x-axis) and South (negative y-axis). For 4m southeast, the magnitude in the XY-plane is 4m. The components are \( 4 \cos 45^\circ \) for East and \( -4 \sin 45^\circ \) for South.
\( \text{x-component (East)} = 4 \cos 45^\circ = 4 \times \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} \)
\( \text{y-component (South)} = -4 \sin 45^\circ = -4 \times \frac{1}{\sqrt{2}} = -\frac{4}{\sqrt{2}} \)
The z-component (up) is 3m.
So, the vector representation is \( \frac{4}{\sqrt{2}}\hat{i} - \frac{4}{\sqrt{2}}\hat{j} + 3\hat{k} \). This demonstrates handling negative directions.

**(c) 2m northwest and 4m up.**
Northwest implies components along North (y-axis) and West (negative x-axis). For 2m northwest, the magnitude in the XY-plane is 2m. The components are \( -2 \cos 45^\circ \) for West and \( 2 \sin 45^\circ \) for North.
\( \text{x-component (West)} = -2 \cos 45^\circ = -2 \times \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} \)
\( \text{y-component (North)} = 2 \sin 45^\circ = 2 \times \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} \)
The z-component (up) is 4m.
So, the vector representation is \( -\frac{2}{\sqrt{2}}\hat{i} + \frac{2}{\sqrt{2}}\hat{j} + 4\hat{k} \). This provides a complete description of position in 3D space relative to a given origin.
In simple words: We imagine a map where East is positive x, North is positive y, and Up is positive z. Then, we break down each movement into how much it goes East/West, North/South, and Up. For example, "5m northeast and 2m up" means it goes \( \frac{5}{\sqrt{2}} \) meters East, \( \frac{5}{\sqrt{2}} \) meters North, and 2 meters Up.

🎯 Exam Tip: When converting compass directions to vectors, remember that angles are measured from the positive x-axis (East). Northeast is \( 45^\circ \), Southeast is \( -45^\circ \) or \( 315^\circ \), Northwest is \( 135^\circ \), and Southwest is \( -135^\circ \) or \( 225^\circ \).

 

Question 21. The moon is orbiting the earth approximately once in 27 days. What is the angle transversed by the moon per day?
Answer: Given that the moon orbits the Earth once in 27 days.
One full orbit (revolution) corresponds to an angular displacement of \( 360^\circ \) or \( 2\pi \) radians.

**Angle transversed per day in degrees:**
In 27 days, the angular displacement is \( 360^\circ \).
So, in 1 day, the angle transversed \( = \frac{360^\circ}{27} \)
\( \text{Angle per day} \approx 13.33^\circ \)

**Angle transversed per day in radians:**
In 27 days, the angular displacement is \( 2\pi \) radians.
So, in 1 day, the angle transversed \( = \frac{2\pi}{27} \) radians.
\( \text{Angle per day} \approx \frac{2 \times 3.14159}{27} \approx 0.2327 \) radians.
The angle transversed by the moon per day is approximately \( 13.3^\circ \) or \( \frac{2\pi}{27} \) radians. This calculation is a simple application of angular motion, linking orbital period to daily angular change.
In simple words: The moon goes around the Earth one time in 27 days. So, each day, it moves a small part of that full circle. This daily movement is about 13.3 degrees or \( \frac{2\pi}{27} \) radians.

🎯 Exam Tip: Ensure you understand the relationship between total angular displacement (one revolution \( = 360^\circ \) or \( 2\pi \) radians) and the time taken for that displacement.

 

Question 22. An object of mass m has an angular acceleration \( \alpha = 0.2 \text{ rad/s}^2 \). What is the angular displacement covered by the object after 3 seconds? (Assume that the object started with angle zero with zero angular velocity)
Answer: Given:
Angular acceleration \( \alpha = 0.2 \text{ rad/s}^2 \)
Time \( t = 3 \text{ s} \)
Initial angular velocity \( \omega_0 = 0 \) (started with zero angular velocity)
Initial angle \( \theta_0 = 0 \) (started with angle zero)

We can use the kinematic equation for angular displacement:
\( \theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2 \)
Substitute the given values:
\( \theta = 0 + (0)(3) + \frac{1}{2}(0.2)(3)^2 \)
\( \theta = 0 + 0 + \frac{1}{2}(0.2)(9) \)
\( \theta = \frac{1}{2}(1.8) \)
\( \theta = 0.9 \text{ rad} \)
To express this in degrees, use the conversion \( \pi \text{ rad} = 180^\circ \):
\( 0.9 \text{ rad} = 0.9 \times \frac{180^\circ}{\pi} \)
\( 0.9 \times \frac{180^\circ}{3.14159} \approx 51.57^\circ \)
So, the angular displacement covered by the object after 3 seconds is \( 0.9 \) radians or approximately \( 51.6^\circ \). This calculation helps in understanding how objects rotate when they have a steady increase in their spinning speed.
In simple words: An object starts spinning from rest and speeds up its spin steadily. After 3 seconds, it will have turned by 0.9 radians, which is about 51.6 degrees.

🎯 Exam Tip: Always make sure to use consistent units (radians for \( \alpha \) and \( \theta \) in equations) and ensure you are using the correct angular kinematic equation for the given conditions (e.g., zero initial angular velocity).

 

Question 23. The variation of velocity of a particle with time moving along a straight line is illustrated in the following figure. The distance travelled by the particle in 4s is
Answer: The distance traveled by a particle is the total area under the absolute value of the velocity-time (v-t) graph. We need to calculate the areas of the geometric shapes formed under the graph up to 4 seconds.

We can break the area under the graph into three parts:
**Part 1: From \( t=0 \) to \( t=2 \text{ s} \)** (Triangle)
This is a triangle from (0,10) to (2,30). The initial velocity is 10 and final velocity is 30 at t=2s. The shape is a trapezoid from t=0 to t=2s.
No, looking at the graph carefully, the points are (0,10), (1,20), (2,30). This represents increasing velocity.
From 0 to 1s: This is a trapezoid. Base 1: 10, Base 2: 20, Height: 1s.
Area 1 (0-1s) = \( \frac{1}{2} (10 + 20) \times 1 = \frac{1}{2} \times 30 \times 1 = 15 \text{ m} \)
From 1 to 2s: This is also a trapezoid. Base 1: 20, Base 2: 30, Height: 1s.
Area 2 (1-2s) = \( \frac{1}{2} (20 + 30) \times 1 = \frac{1}{2} \times 50 \times 1 = 25 \text{ m} \)
From 2 to 3s: This is a rectangle. Height: 30, Base: 1s.
Area 3 (2-3s) = \( 30 \times 1 = 30 \text{ m} \)
From 3 to 4s: This is a rectangle. Height: 10, Base: 1s.
Area 4 (3-4s) = \( 10 \times 1 = 10 \text{ m} \)
Wait, the graph I'm looking at and the description from OCR are different. The OCR graph: V goes from 10 to 20 (at t=1) then 20 to 30 (at t=2) then stays 30 (at t=3) then drops to 10 (at t=4). Let's use the provided graph image. Points on graph: (0, 10) (1, 20) (2, 30) (3, 30) (4, 10) Area calculation for distance traveled in 4s: Area 1 (0 to 1s): Trapezoid. \( \text{height}=1 \), parallel sides are \( v(0)=10 \) and \( v(1)=20 \). \( A_1 = \frac{1}{2} (10+20) \times 1 = 15 \text{ m} \) Area 2 (1 to 2s): Trapezoid. \( \text{height}=1 \), parallel sides are \( v(1)=20 \) and \( v(2)=30 \). \( A_2 = \frac{1}{2} (20+30) \times 1 = 25 \text{ m} \) Area 3 (2 to 3s): Rectangle. \( \text{base}=1 \), \( \text{height}=v(2)=v(3)=30 \). \( A_3 = 30 \times 1 = 30 \text{ m} \) Area 4 (3 to 4s): Trapezoid. \( \text{height}=1 \), parallel sides are \( v(3)=30 \) and \( v(4)=10 \). \( A_4 = \frac{1}{2} (30+10) \times 1 = \frac{1}{2} \times 40 \times 1 = 20 \text{ m} \) Total distance traveled \( = A_1 + A_2 + A_3 + A_4 \) \( = 15 + 25 + 30 + 20 \) \( = 90 \text{ m} \) The distance traveled by the particle in 4s is 90m. This calculation demonstrates how the area under a velocity-time graph represents distance, a key concept in kinematics for understanding total path covered.
In simple words: To find the total distance an object travels from its speed-time graph, you calculate the area of all the shapes under the graph. In this case, we add up the areas of three trapezoids and one rectangle for a total of 90 meters.

🎯 Exam Tip: For distance, always sum the absolute values of the areas under the v-t graph. For displacement, consider the signs of the areas (areas above the t-axis are positive, below are negative).

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