Maharashtra Board Class 9 Science Chapter 1 Laws of Motion Solutions

Std 9 Science Chapter 1 Laws Of Motion Question Answer Maharashtra Board

Class 9 Science Chapter 1 Laws Of Motion Question Answer Maharashtra Board

 

Question 1. Match the first column with appropriate entries in the second and third columns and remake the table.
Answer:

S. No.	Column 1	Column 2	Column 3
1	Negative acceleration	The velocity of the object decreases	A vehicle moving with the velocity of 10 m/s, stops after 5 seconds.
2	Positive acceleration	The velocity of the object increases	A car, initially at rest reaches a velocity of 50 km/hr in 10 seconds
3	Zero acceleration	The velocity of the object remains constant	A vehicle is moving with a velocity of 25 m/s

In simple words: This table correctly matches each type of acceleration with its definition concerning velocity change and provides a corresponding real-world example.

🎯 Exam Tip: Pay close attention to the direction of velocity change when identifying types of acceleration; positive means increasing speed, negative means decreasing speed (retardation), and zero means constant velocity.

2. Clarify The Differences

 

Question. A. Distance and displacement
Answer:

Distance	Displacement
(i) Distance is the length of the actual path travelled by an object.	(i) Displacement is the minimum distance between the starting and finishing points.
(ii) It is a scalar quantity.	(ii) It is a vector quantity.
(iii) It is either equal to or greater than displacement.	(iii) It is either equal to or less than distance.
(iv) Distance travelled is always positive.	(iv) Displacement may be positive or negative or zero.

In simple words: Distance measures the total path covered, while displacement measures the shortest straight-line path from start to end, making distance a scalar and displacement a vector.

🎯 Exam Tip: Remember that displacement can be zero even if distance is non-zero (e.g., a round trip), and its direction matters, unlike distance.

 

Question. B. Uniform and non-uniform motion.
Answer:

Uniform motion	Non-uniform motion
(i) If an object covers equal distances in equal intervals of time it is said to be in uniform motion.	(i) If an object moves unequal distances in equal intervals of time, its motion is said to be nonuniform.
(ii) Distance - time graph for uniform motion is a straight line.	(ii) Distance - time graph for non-uniform motion is not a straight line.
(iii) In uniform motion, acceleration is zero.	(iii) In non-uniform motion acceleration is non-zero.

In simple words: Uniform motion involves equal distances in equal time, resulting in constant velocity and zero acceleration, whereas non-uniform motion has unequal distances in equal time, leading to changing velocity and non-zero acceleration.

🎯 Exam Tip: The key distinction lies in the equality of distance covered over equal time intervals; this directly impacts velocity and acceleration. Pay attention to graph shapes: straight for uniform, curved for non-uniform motion.

 

Question 3. Complete the following table.
Answer:

u (m/s)	a (m/s²)	t (sec)	v = u + at (m/s)
2	4	3	14
10	5	2	20

u (m/s)	a (m/s²)	t (sec)	s = ut + \( \frac{1}{2} \)at² (m)
5	12	3	69
7	8	4	92

u (m/s)	a (m/s²)	s (m)	v² = u² + 2as (m/s)²
4	3	-4/3	8
No real solution	5	8.4	10

In simple words: This table is completed using the three fundamental equations of motion: \( v = u + at \), \( s = ut + \frac{1}{2}at^2 \), and \( v^2 = u^2 + 2as \), where 'u' is initial velocity, 'a' is acceleration, 't' is time, 'v' is final velocity, and 's' is displacement.

🎯 Exam Tip: Accurately identifying the knowns and unknowns is crucial before applying the correct equation of motion. Be mindful of units and signs for direction. If a calculation yields an impossible physical result (e.g., negative square of velocity), it's important to note it, though such cases are rare in well-posed problems.

 

Question 4. Complete the sentences and explain them.
Answer:
a. The minimum distance between the start and finish points of the motion of an object is called the displacement of the object.
b. Deceleration is negative acceleration
c. When an object is in uniform circular motion, its direction changes at every point.
d. During collision total momentum remains constant.
e. The working of a rocket depends on Newton's third law of motion.

In simple words: This question tests fundamental definitions and principles of motion: displacement as the shortest path, deceleration as negative acceleration, changing direction in circular motion, conservation of momentum in collisions, and Newton's third law for rocket propulsion.

🎯 Exam Tip: Memorize key definitions and understand the underlying principles of each concept. For examples like rocket propulsion, link them directly to the relevant Newton's law.

 

Question 5. Give scientific reasons.
Answer:
a. When an object falls freely to the ground, its acceleration is uniform.

• When the body falls freely to the ground, there are equal changes in velocity of the body in equal intervals of time.
• Thus the acceleration of the body is constant, and it possesses uniform acceleration.

In simple words: An object in free fall experiences uniform acceleration because the gravitational force causes its velocity to change by equal amounts over equal time intervals.

🎯 Exam Tip: The constant acceleration due to gravity (\(g\)) is a key concept; remember its value and that it causes uniform acceleration for freely falling objects, neglecting air resistance.

 

Question 5.b. Even though the magnitudes of action force and reaction force are equal and their directions are opposite, their effects do not get cancelled.
Answer:

• Action and reaction forces act on different bodies.
• They don't act on the same body, hence they cannot cancel each other's effect.
• Hence, even though the magnitudes of action force and reaction force are equal, they do not cancel each other.

In simple words: Action and reaction forces don't cancel because they are applied to two *different* objects, not on the same object.

🎯 Exam Tip: A common misconception is that action-reaction pairs cancel out; emphasizing that they act on *different* bodies is crucial for understanding Newton's Third Law.

 

Question 5.c. It is easier to stop a tennis ball as compared to a cricket ball, when both are traveling with the same velocity.
Answer:

• Momentum of an object depends on its mass as well as its velocity.
• Cricket ball is heavier than a tennis ball. Although they are thrown with the same velocity, cricket ball has more momentum than a tennis ball.
• The force required to stop a cricket ball is more than a tennis ball.
• Hence it is easier to stop a tennis ball than a cricket ball moving with same velocity.

In simple words: A cricket ball has more mass than a tennis ball, so even with the same velocity, its momentum is greater, requiring a larger force to bring it to a stop.

🎯 Exam Tip: This question highlights the concept of momentum (mass x velocity); understanding that greater mass leads to greater momentum (and thus more force needed to stop it) is a key learning point.

 

Question 5.d. The velocity of an object at rest is considered to be uniform.
Answer:

• When a body is at rest there is no change in velocity.
• A body with constant velocity is said to be in uniform motion.
• Hence, the state of rest is an example of uniform motion.

In simple words: An object at rest has zero velocity, which is a constant velocity, and thus is considered to be in a state of uniform motion (specifically, uniform motion with zero speed).

🎯 Exam Tip: Recognize that "rest" is a special case of uniform motion where the constant velocity is zero. This demonstrates a deep understanding of uniform motion's definition.

 

Question 6. Take 5 examples from your surroundings and give an explanation based on Newton's laws of motion.
Answer:
Here are 5 examples from surroundings explained using Newton's laws of motion:

1. **Newton's First Law (Law of Inertia):** When a bus suddenly starts, passengers sitting inside tend to fall backward. This happens because their bodies tend to remain in a state of rest (inertia of rest) while the bus moves forward.

2. **Newton's First Law (Law of Inertia):** When you kick a football, it travels some distance and then stops. This is due to friction and air resistance acting as unbalanced forces, opposing the ball's motion and eventually bringing it to rest.

3. **Newton's Second Law (Force and Acceleration):** Pushing a shopping cart. The harder you push (more force), the faster it accelerates. If the cart is full (more mass), you need to apply more force to achieve the same acceleration.

4. **Newton's Second Law (Force and Acceleration):** A cricket player pulling their hands back while catching a fast-moving ball. By increasing the time over which the ball's momentum changes, the player reduces the force exerted on their hands, minimizing injury.

5. **Newton's Third Law (Action-Reaction):** A rocket taking off. The rocket expels hot gases downwards (action), and in response, the gases push the rocket upwards with an equal and opposite force (reaction), propelling it into space.

In simple words: Newton's laws explain everyday phenomena like why you lurch forward in a sudden stop (inertia), why heavier objects are harder to push (force = mass x acceleration), and how rockets fly (action-reaction).

🎯 Exam Tip: For such questions, clearly state the law and then provide a concise, relevant example. Explaining the cause-and-effect relationship for each example demonstrates a strong grasp of the concepts.

 

Question 7. Solve the following examples.
Answer:
a. An object moves 18 m in the first 3 s, 22 m in the next 3 s and 14 m in the last 3 s. What is its average speed? (Ans: 6 m/s)
Given:
Total distance (d) = 18 + 22 + 14 = 54 m
Total time taken (t) = 3 + 3 + 3 = 9 sec
To find:
Average speed = ?
Formula:
Average speed = \( \frac{\text{Total distance covered}}{\text{Total time taken}} \)
Solution:
Average speed = \( \frac{\text{Total distance covered}}{\text{Total time taken}} \)
= \( \frac{54}{9} \)
= 6 m/s
The object moves with an average speed of 6 m/s.

In simple words: Average speed is calculated by dividing the total distance an object travels by the total time it takes to travel that distance.

🎯 Exam Tip: Remember to sum all individual distances for total distance and all individual time intervals for total time when calculating average speed. Units must be consistent (e.g., meters and seconds).

 

Question 7.b) An object of mass 16 kg is moving with an acceleration of 3 m/s². Calculate the applied force. If the same force is applied on an object of mass 24 kg, how much will be the acceleration? (Ans: 48 N, 2 m/s²)
Answer:
Given:
Mass of 1^st body (m₁) = 16 kg
Acceleration of 1^st body (a₁) = 3 m/s²
Mass of 2^nd body (m₂) = 24 kg
To find:
Force on 1^st body (F₁) =?
Acceleration of 2^nd body (a₂) = ?
Formula:
F = m x a
Solution:
F₁ = m₁ x a₁
F₁ = 16 × 3
F₁ = 48 N

F₂ = m₂ x a₂

\( \implies \) a₂ = \( \frac{F_2}{m_2} \)
a₂ = \( \frac{48}{24} \)
a₂ = 2 m/s²
The force acting on the 1^st body is 48 N and the acceleration of the 2^nd body is 2 m/s².

In simple words: This problem uses Newton's Second Law, \( F = ma \), to first calculate the force from a given mass and acceleration, then uses that same force to find the new acceleration for a different mass.

🎯 Exam Tip: When dealing with multiple scenarios involving force, mass, and acceleration, ensure you correctly identify which values belong to which object or situation, and remember that force is a vector, but in one-dimensional problems, its magnitude is primarily considered.

 

Question 7.c) A bullet having a mass of 10 g and moving with a speed of 1.5 m/s, penetrates a thick wooden plank of mass 90 g. The plank was initially at rest. The bullet gets embedded in the plank and both move together. Determine their velocity. (Ans: 0.15 m/s)
Answer:
Given:
Mass of bullet (m₁) = 10g = \( \frac{10}{1000} \) kg
Mass of plank (m₂) = 90g = \( \frac{90}{1000} \) kg
Initial velocity of bullet (u₁) = 1.5 m/s
Initial velocity of plank (u₂) = 0 m/s
To find: Common velocity
Formula: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Solution: v₁ = v₂ = v

\( \implies \) m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

\( \implies \) \( (\frac{10}{1000} \times 1.5) + (\frac{90}{1000} \times 0) = (\frac{10}{1000} \times v) + (\frac{90}{1000} \times v) \)

\( \implies \) \( \frac{15}{1000} + 0 = v (\frac{10}{1000} + \frac{90}{1000}) \)

\( \implies \) \( \frac{15}{1000} = v (\frac{100}{1000}) \)

\( \implies \) 15 = v x 100

\( \implies \) v = \( \frac{15}{100} \)
v = 0.15 m/s
The plank embedded with the bullet moves with a velocity of 0.15 m/s.

In simple words: This problem demonstrates the principle of conservation of momentum in an inelastic collision where two objects stick together, calculating their combined final velocity.

🎯 Exam Tip: For problems involving collisions and objects sticking together, remember to use the principle of conservation of momentum (\( m_1u_1 + m_2u_2 = (m_1+m_2)v \)) and ensure all masses are in consistent units (kg).

 

Question 7.d) A person swims 100 m in the first 40 s, 80 m in the next 40 s and 45 m in the last 20 s. What is the average speed? (Ans: 2.25 m/s)
Answer:
Given:
Total distance (d) = 100 + 80 + 45 = 225 m
Total time taken (t) = 40 + 40 + 20 = 100 sec
To find:
Average speed =?
Formula:
Average speed = \( \frac{\text{Total distance covered}}{\text{Total time taken}} \)
Solution:
Average speed = \( \frac{\text{Total distance covered}}{\text{Total time taken}} \)
= \( \frac{225}{100} \)
= 2.25 m/s
The person swims with an average speed of 2.25 m/s.

In simple words: Average speed is found by dividing the total distance covered by the total time taken for the entire journey, regardless of variations in speed during different segments.

🎯 Exam Tip: Always sum up all distances and all time intervals to calculate average speed. Be careful with units; average speed is in m/s, not m/s².

Class 9 Science Chapter 1 Laws Of Motion Intext Questions And Answers

 

Question. (i) Who will take less time to reach the school and why?
Answer:
Prashant will take less time as the path followed by him is the shortest.

In simple words: The person who takes the shortest path between two points will generally reach faster, assuming they maintain a similar speed.

🎯 Exam Tip: In real-world scenarios, the shortest path (displacement) usually corresponds to the least travel time if speed is maintained, but other factors like obstacles or traffic can influence this.

 

Question. (2) (a)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्ताकार पथ को दर्शाता है जिस पर एक बिंदु 'A' अंकित है। यह चित्र दूरी और विस्थापन के बीच के अंतर को समझाने के लिए दिया गया है, जहाँ एक वस्तु वृत्ताकार पथ पर गति कर सकती है।

 

Question. a) Every morning, Swaralee walks round the edge of a circular field having a radius of 100 m. As shown in figure (a), if she starts from the point A and takes one round, how much distance has she walked and what is her displacement?
Answer:
Radius (r) = 100 m
Distance covered = Circumference of the circle
= 2 \( \pi \)r
= 2 x 3.14 x 100
= 628 m
Displacement = 0 m (Shortest distance between initial and final position is zero)

In simple words: For one full round on a circular path, the distance covered is the circumference of the circle, but the displacement is zero because the starting and ending points are the same.

🎯 Exam Tip: Clearly distinguish between distance (total path length) and displacement (straight-line distance from start to end). For a complete circular path, displacement is always zero.

 

Question. b) If a car, starting from point P, goes to point Q (see figure 1.9) and then returns to point P, how much distance has it travelled and what is its displacement?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सीधी सड़क पर स्थित दो बिंदुओं, P और Q को दर्शाता है, जिनके बीच की दूरी 360 मीटर है। यह चित्र एक कार की गति को दर्शाता है जो P से Q तक जाती है और फिर P पर वापस आती है, जिससे छात्रों को दूरी और विस्थापन के बीच का अंतर समझ में आए।
Answer:
Distance covered = PQ + QP
= 360 + 360
= 720 m
Displacement = 0 m (The shortest distance between initial and final position is zero)

In simple words: When an object travels from one point to another and then returns to its starting point, the total distance covered is twice the one-way path, but the net displacement is zero.

🎯 Exam Tip: Always remember that displacement considers only the initial and final positions, making it zero for any round trip, while distance accumulates the length of the entire path traversed.

Class 9 Science Chapter 1 Laws Of Motion Additional Important Questions And Answers

(A) Choose And Write The Correct Option:

 

Question 1. The displacement that occurs in unit time is called ............

(a) displacement
(b) distance
(c) velocity
(d) acceleration
Answer: (c) velocity

In simple words: Velocity is defined as the rate of change of displacement, meaning how much an object's position changes per unit time in a specific direction.

🎯 Exam Tip: This question tests a core definition in kinematics. Remember that velocity includes direction, unlike speed.

 

Question 2. The unit of velocity in the SI system is ............

(a) cm/s
(b) m/s²
(c) µm/s²
(d) m/s
Answer: (d) m/s

In simple words: The standard international (SI) unit for velocity, which measures displacement over time, is meters per second.

🎯 Exam Tip: Always use SI units (meters, seconds, kilograms) for calculations unless specified otherwise, as it prevents errors in multi-step problems.

 

Question 3. v² = u² + 2as is the relation between and ............

(a) speed and velocity
(b) distance and acceleration
(c) displacement and velocity
(d) speed and distance
Answer: (c) displacement and velocity

In simple words: The equation \( v^2 = u^2 + 2as \) links the final velocity, initial velocity, acceleration, and displacement of an object in uniformly accelerated motion.

🎯 Exam Tip: This is one of the three kinematic equations; know when to apply it, particularly when time is not a given or required variable.

 

Question 4. ............ is the relation between displacement and time.

(a) v = u + at
(b) v² = u² + 2as
(c) s = ut + 1/2 at²
(d) v = u + 2as
Answer: (c) s = ut + 1/2 at²

In simple words: The equation \( s = ut + \frac{1}{2}at^2 \) describes how displacement changes over time for an object moving with constant acceleration.

🎯 Exam Tip: This equation is fundamental for problems where you need to find displacement given initial velocity, acceleration, and time, or vice-versa.

 

Question 5. The force necessary to cause an acceleration of 1 m/s² in an object of mass 1 kg is called ............

(a) 1 dyne
(b) 1 m/s
(c) 1 Newton
(d) 1 cm/s
Answer: (c) 1 Newton

In simple words: A Newton is the SI unit of force, defined as the amount of force required to accelerate a 1-kilogram mass at a rate of 1 meter per second squared.

🎯 Exam Tip: This is the definition of the Newton, derived from \( F=ma \). Understanding this definition is essential for SI unit conversions and force calculations.

 

Question 6. Even if the displacement of an object is zero, the actual distance traversed by it ............

(a) may not be zero.
(b) will be zero
(c) will be constant
(d) will be infinity
Answer: (a) may not be zero

In simple words: An object can travel a significant distance and still have zero displacement if it returns to its starting point, like completing a lap on a track.

🎯 Exam Tip: This question highlights the critical difference between distance (scalar, total path) and displacement (vector, net change in position). A common mistake is equating the two.

 

Question 7. If the velocity changes by equal amounts in equal time intervals, the object is said to be in ............

(a) uniform acceleration
(b) uniform velocity
(c) non-uniform acceleration
(d) non-uniform motion
Answer: (a) uniform acceleration

In simple words: Uniform acceleration means that the object's velocity is changing at a constant rate, increasing or decreasing by the same amount in every equal time interval.

🎯 Exam Tip: Distinguish carefully: uniform velocity means no acceleration; uniform acceleration means velocity changes uniformly.

 

Question 8. If an object is moving with a uniform velocity ............

(a) its speed remains the same, but direction of motion changes
(b) its speed changes but direction of motion is same
(c) its speed and direction both change
(d) its speed and direction both remain the same
Answer: (d) its speed and direction both remain the same

In simple words: Uniform velocity implies that both the object's speed and its direction of motion are constant, meaning there is no acceleration.

🎯 Exam Tip: Velocity is a vector quantity, so "uniform" means both its magnitude (speed) and direction are unchanging. If either changes, the velocity is not uniform.

 

Question 9. ............ is an example of positive acceleration.

(a) A stone is thrown vertically upwards
(b) A stone falls freely towards the earth
(c) Brakes are applied by the truck driver
(d) The train arriving at the station
Answer: (b) a stone falls freely towards the earth

In simple words: Positive acceleration occurs when an object's velocity increases in the direction of motion, such as a stone speeding up as it falls due to gravity.

🎯 Exam Tip: Positive acceleration corresponds to an increase in speed (or velocity in the positive direction), while negative acceleration (deceleration) corresponds to a decrease in speed or acceleration in the opposite direction.

 

Question 10. An object continues to remain at rest or in a state of uniform motion along a straight line unless an ............ acts on it.

(a) internal imbalanced force
(b) external unbalanced force
(c) internal balanced force
(d) external balanced force
Answer: (b) external unbalanced force

In simple words: According to Newton's First Law, an object's state of motion (at rest or constant velocity) will only change if a net external force acts upon it.

🎯 Exam Tip: This question is a direct test of Newton's First Law of Motion, often called the law of inertia. Ensure you understand the role of *unbalanced* and *external* forces.

 

Question 11. The ............ is proportional to the applied force and it occurs in the direction of the force.

(a) change of momentum
(b) rate of change of velocity
(c) change of velocity
(d) rate of change of momentum
Answer: (d) rate of change of momentum

In simple words: Newton's Second Law states that the applied force is directly proportional to the rate at which an object's momentum changes, and this change happens in the direction of the force.

🎯 Exam Tip: This is the definition of Newton's Second Law in terms of momentum. Remember that force not only causes a change in momentum but also dictates the direction of that change.

 

Question 12. ............ is a relative concept.

(a) Motion
(b) Direction
(c) Power
(d) Acceleration
Answer: (a) Motion

In simple words: Whether an object is in motion or at rest depends entirely on the observer's frame of reference, making motion a relative concept.

🎯 Exam Tip: Understand that motion is always described relative to a chosen reference point. What appears to be at rest from one frame may be in motion from another.

 

Question 13. A body is said to be in motion if it changes its ............ with respect to its surroundings.

(a) position
(b) direction
(c) speed
(d) force
Answer: (a) position

In simple words: Motion is fundamentally defined by a change in an object's location relative to its environment over time.

🎯 Exam Tip: The simplest definition of motion involves a change in position relative to a reference point. Other options describe aspects of motion but not its fundamental nature.

 

Question 14. A body is said to be at ............ if it does not change its position with respect to its surroundings.

(a) Motion
(b) Rest
(c) Gravity
(d) Force
Answer: (b) Rest

In simple words: An object is considered to be at rest if its position remains constant when observed from its surroundings.

🎯 Exam Tip: Just like motion, rest is a relative concept and is defined by the absence of a change in position relative to a chosen frame of reference.

 

Question 15. ............ is the length of the actual path travelled by an object in motion while going from one point to another.

(a) Distance
(b) Displacement
(c) Speed
(d) Velocity
Answer: (a) Distance

In simple words: Distance refers to the total path length covered by a moving object, irrespective of its starting or ending points.

🎯 Exam Tip: Distance is a scalar quantity, always positive, and measures the entire route taken, a fundamental concept in kinematics.

 

Question 16. The distance covered by a body in unit time is called its ............

(a) velocity
(b) speed
(c) displacement
(d) rest
Answer: (b) speed

In simple words: Speed is a scalar quantity that measures how quickly an object covers distance, defined as distance travelled per unit of time.

🎯 Exam Tip: Differentiate speed from velocity: speed is scalar (magnitude only), while velocity is vector (magnitude and direction). This definition is key to understanding basic motion.

 

Question 17. S.I. unit of speed is ............ and in C.G.S unit it is ............

(a) m/s and cm/s
(b) km/s and cm/s
(c) m/s and mm/s
(d) m/s and nm/s
Answer: (a) m/s and cm/s

In simple words: The SI unit for speed is meters per second (m/s), while in the CGS system, it is centimeters per second (cm/s).

🎯 Exam Tip: Be proficient in converting between different unit systems (SI and CGS) for speed and other physical quantities, as problems often mix them.

 

Question 18. The distance travelled in a particular direction by an object in unit time is called its ............

(a) velocity
(b) speed
(c) displacement
(d) rest
Answer: (a) velocity

In simple words: Velocity is a vector quantity that describes both the speed of an object and its direction of motion.

🎯 Exam Tip: The phrase "in a particular direction" is key here, explicitly pointing to velocity rather than just speed.

 

Question 19. Units of speed and velocity are the ............

(a) Same
(b) Different
(c) Greater than each other
(d) Unequal
Answer: (a) Same

In simple words: While speed and velocity are different concepts (scalar vs. vector), their units, representing distance/displacement over time, are identical.

🎯 Exam Tip: Do not confuse the conceptual difference between speed and velocity with their units; both are measured in units like m/s or km/h.

 

Question 20. ............ is related to distance, while ............ is related to displacement.

(a) Gravity and magnetism
(b) Speed and force
(c) Speed and velocity
(d) Motion and rest
Answer: (c) Speed, velocity

In simple words: Speed is derived from the total distance covered over time, whereas velocity is derived from the displacement (change in position) over time.

🎯 Exam Tip: This question reinforces the scalar-vector distinction: distance is scalar and relates to speed, while displacement is vector and relates to velocity.

 

Question 21. If an object covers equal distances in equal time intervals, it is said to be moving with ............ speed.

(a) Uniform
(b) Non uniform
(c) Changing
(d) Random
Answer: (a) Uniform

In simple words: Uniform speed means an object travels the same amount of distance in every equal interval of time.

🎯 Exam Tip: This is the definition of uniform speed. Understanding this concept is crucial for grasping constant motion in physics.

 

Question 22. If an object covers unequal distances in equal time Intervals, it is said to be moving with ............ speed.

(a) Uniform
(b) Non uniform
(c) Changing
(d) Random
Answer: (b) Non uniform

In simple words: Non-uniform speed means an object's speed is changing, covering different distances in the same amount of time.

🎯 Exam Tip: Non-uniform speed implies acceleration (or deceleration). This is a common type of motion in everyday life, where speeds rarely remain perfectly constant.

 

Question 23.
The rate of change of velocity is called
(a) Speed
(b) Acceleration
(c) Velocity
(d) Rest
Answer: (b) Acceleration
In simple words: Acceleration is how quickly an object's velocity changes, either in speed or direction.

🎯 Exam Tip: Remember that acceleration measures the rate of change of velocity, not just speed. Pay attention to both magnitude and direction.

 

Question 24.
Speed of light in dry air is ............. m/s.
(a) 3 x 10⁷
(b) 3 x 10⁸
(c) 3 x 10⁹
(d) 3 x 10³
Answer: (b) 3 x 10⁸
In simple words: The speed of light in a vacuum, and approximately in dry air, is an extremely high constant value, about 300,000,000 meters per second.

🎯 Exam Tip: Know fundamental physical constants like the speed of light. It's often tested as a basic knowledge point.

 

Question 25.
When velocity of a body increases, its acceleration is .................
(a) Negative
(b) Zero
(c) Positive
(d) Equal
Answer: (c) positive
In simple words: If an object's velocity is increasing, it means it is speeding up in the direction of its motion, which corresponds to positive acceleration.

🎯 Exam Tip: Positive acceleration means increasing velocity, while negative acceleration (or deceleration/retardation) means decreasing velocity.

 

Question 26.
When velocity of a body decreases, its acceleration is .................
(a) Negative
(b) Zero
(c) Positive
(d) Equal
Answer: (a) negative
In simple words: When an object slows down, its acceleration is in the opposite direction to its velocity, resulting in negative acceleration or deceleration.

🎯 Exam Tip: Understand the difference between positive, negative, and zero acceleration based on how velocity changes over time.

 

Question 27.
Negative acceleration is also called or
(a) Deceleration or retardation
(b) Deceleration or acceleration
(c) acceleration or retardation
(d) Zero
Answer: (a) deceleration or retardation
In simple words: Negative acceleration simply means that an object is slowing down, and this phenomenon is also known as deceleration or retardation.

🎯 Exam Tip: These terms are often used interchangeably in physics to describe a decrease in speed or velocity.

 

Question 28.
In case of motion, object travels equal ................. in equal intervals of time.
(a) Uniform, distance
(b) Non-Uniform, distance
(c) Uniform, displacement
(d) Uniform, displacement
Answer: (a) uniform, distances
In simple words: For uniform motion, an object covers the same amount of distance in every equal time interval.

🎯 Exam Tip: Uniform motion is characterized by constant speed, meaning equal distances are covered in equal time intervals.

 

Question 29.
Motion of an object was studied by .................
(a) Sir Albert Einstein
(b) Sir Thomas Edison
(c) Sir Isaac Newton
(d) Sir Ravindranath Tagore
Answer: (c) Sir Issac Newton
In simple words: Sir Isaac Newton laid the foundational laws that describe how objects move and interact with forces.

🎯 Exam Tip: Newton's Laws of Motion are fundamental to classical mechanics and explain most everyday movements.

 

Question 30.
When an object moves in a circular path with uniform speed, its motion is ................. motion.
(a) Non uniform circular
(b) Random circular
(c) Uniform circular
(d) Uniform linear
Answer: (c) uniform circular
In simple words: Motion in a circle at a constant speed is called uniform circular motion.

🎯 Exam Tip: In uniform circular motion, speed is constant, but velocity changes due to continuous change in direction, leading to acceleration.

 

Question 31.
When a coin moves along a circular path, the direction of its motion at every point is .................
(a) Circular
(b) Concave
(c) Tangential
(d) Convex
Answer: (c) tangential
In simple words: At any point on a circular path, an object's instantaneous direction of motion is along the tangent to the circle at that point.

🎯 Exam Tip: The tangential direction is crucial for understanding the velocity and forces involved in circular motion.

 

Question 32.
For all uniformly accelerated motions, the velocity-time graph is a .................
(a) Curved line
(b) Straight line
(c) Negative line
(d) Positive line
Answer: (b) straight line
In simple words: When an object's acceleration is constant, its velocity changes at a steady rate, making its velocity-time graph a straight line.

🎯 Exam Tip: A straight line on a velocity-time graph indicates uniform acceleration, where the slope represents the acceleration.

 

Question 33.
In the distance-time graph, the slope of the straight line indicates .................
(a) Acceleration
(b) Velocity
(c) Speed
(d) Rest
Answer: (b) velocity
In simple words: The steepness of a straight line on a distance-time graph tells us how fast an object is moving and in what direction, which is its velocity.

🎯 Exam Tip: For distance-time graphs, slope (change in distance/change in time) directly gives speed or velocity, depending on whether distance or displacement is plotted.

 

Question 34.
The first equation of motion gives relation between ................. and time.
(a) Acceleration
(b) Velocity
(c) Speed
(d) Rest
Answer: (b) velocity
In simple words: The first equation of motion, \( v = u + at \), connects an object's initial velocity, final velocity, acceleration, and the time taken.

🎯 Exam Tip: Master the equations of motion (\( v = u + at \), \( s = ut + \frac{1}{2}at^2 \), \( v^2 = u^2 + 2as \)) and know which variables each relates.

 

Question 35.
Newton's first law explains the phenomenon of
(a) Rest
(b) Inertia
(c) Speed
(d) Velocity
Answer: (b) inertia
In simple words: Newton's first law describes inertia, which is an object's natural tendency to resist changes in its state of motion (stay at rest or keep moving at a constant velocity).

🎯 Exam Tip: Inertia is a key concept in physics; remember it's related to mass – more mass means more inertia.

 

Question 36.
....................... cause a change in the state of an object at rest or in uniform motion.
(a) Balanced forces
(b) Zero forces
(c) Unbalanced forces
(d) None of them
Answer: (c) Unbalanced forces
In simple words: Only an unbalanced force, where the net force is not zero, can make an object start moving, stop moving, or change its direction.

🎯 Exam Tip: Balanced forces result in no change in motion (object stays at rest or continues uniform motion), while unbalanced forces cause acceleration.

 

Question 37.
To describe an object's momentum, we must specify its ................. and .................
(a) Mass and displacement
(b) Mass and direction
(c) Mass and velocity
(d) Mass and acceleration
Answer: (c) mass and velocity
In simple words: Momentum is a measure of an object's motion, determined by both its mass and its velocity.

🎯 Exam Tip: Momentum is a vector quantity, meaning it has both magnitude (from mass and speed) and direction (from velocity).

 

Question 38.
....................... is the product of mass and velocity of an object.
(a) Speed
(b) Acceleration
(c) Momentum
(d) Force
Answer: (c) Momentum
In simple words: Momentum is calculated by multiplying an object's mass by its velocity, indicating the 'quantity of motion' it possesses.

🎯 Exam Tip: Understand the formula for momentum: \( p = mv \), where p is momentum, m is mass, and v is velocity.

 

Question 39.
The rate of change of momentum is proportional to the applied .................
(a) Balanced force
(b) Unbalanced force
(c) Mass
(d) Velocity
Answer: (b) unbalanced force
In simple words: According to Newton's second law, the harder an unbalanced force pushes on an object, the faster its momentum will change.

🎯 Exam Tip: Newton's second law (\( F = ma \)) can also be expressed in terms of momentum: force equals the rate of change of momentum.

 

Question 40.
S.I. unit of momentum is
(a) kg cm/s
(b) kg m/s
(c) gm/s
(d) m/s
Answer: (b) kg m/s
In simple words: The standard international unit for momentum is kilograms times meters per second, reflecting its definition as mass times velocity.

🎯 Exam Tip: Always use SI units (kilograms, meters, seconds) for calculations unless specified otherwise, to avoid errors.

 

Question 41.
....................... is always conserved in a collision.
(a) Force
(b) Power
(c) Speed
(d) Total momentum
Answer: (d) Total momentum
In simple words: In any collision, the total momentum of the interacting objects before the collision is equal to their total momentum after the collision, assuming no external forces.

🎯 Exam Tip: The law of conservation of momentum is a fundamental principle in physics, especially important for analyzing collisions and explosions.

 

Question 42.
When a bullet is fired from the gun, the gun moves in backward direction. This motion is called as .................
(a) Momentum
(b) Velocity
(c) Acceleration
(d) Recoil
Answer: (d) Recoil
In simple words: The backward movement of a gun after firing a bullet is called recoil, a direct consequence of Newton's third law of motion.

🎯 Exam Tip: Recoil is an excellent example of Newton's third law (action-reaction) and the conservation of momentum in action.

 

Question 43.
In CGS system, the unit of force is .................
(a) Newton
(b) Watt
(c) Horse power
(d) Dyne<
Answer: (d) Dyne.
In simple words: While Newton is the SI unit of force, the CGS system uses the dyne as its unit.

🎯 Exam Tip: Differentiate between SI (System International) and CGS (Centimeter-Gram-Second) units for various physical quantities, especially for force and momentum.

 

(B) 1. Find The Odd Man Out:

 

Question 1.
Displacement, Force, Momentum, Mass
Answer: Mass
In simple words: Mass is the odd one out because displacement, force, and momentum are all vector quantities (having direction), whereas mass is a scalar quantity (only magnitude).

🎯 Exam Tip: Understand the fundamental difference between scalar quantities (like mass, speed, distance) and vector quantities (like displacement, velocity, force, momentum).

 

Question 2.
Speed, Power, Energy, Acceleration
Answer: Acceleration
In simple words: Acceleration is a vector quantity, while speed, power, and energy are all scalar quantities.

🎯 Exam Tip: Carefully categorize physical quantities as either scalar or vector; this distinction is crucial in solving problems and understanding concepts.

 

Question 3.
Newton's 1st law, Newton's 2nd law, Newton's 3rd law, Kepler's laws of motion
Answer: Newton's 3rd law
In simple words: Kepler's laws describe planetary motion, while Newton's laws describe general motion and forces, so Kepler's laws are the odd one out.

🎯 Exam Tip: While all relate to motion, Kepler's laws are specific to celestial mechanics, whereas Newton's laws are universal principles of classical mechanics.

 

(B) 2. Find Out The Correlation

 

Question 1.
Speed zero: Body at rest :: Negative acceleration : Retardation
Answer: Retardation
In simple words: Just as zero speed means a body is at rest, negative acceleration is another term for retardation, meaning the body is slowing down.

🎯 Exam Tip: Recognize synonyms and related concepts in physics (e.g., negative acceleration, deceleration, retardation all mean slowing down).

 

Question 2.
Displacement : Vector quantity :: Distance : Scalar quantity
Answer: Scalar quantity
In simple words: Displacement is a vector because it has both magnitude and direction, whereas distance is a scalar as it only has magnitude.

🎯 Exam Tip: The core difference between displacement and distance lies in direction: displacement considers it, distance does not.

 

Question 3.
When body comes to rest at the end of the motion : Final velocity is zero :: When body is at rest at the starting of motion : Initial velocity is zero
Answer: Initial velocity is zero
In simple words: If an object starts from a resting position, its initial velocity is zero, mirroring how its final velocity is zero if it comes to a stop.

🎯 Exam Tip: Clearly identify initial and final conditions in motion problems, as "at rest" always implies zero velocity.

 

Question 4.
Uniform circular motion: Displacement is zero :: Uniform velocity : Acceleration is zero
Answer: Acceleration is zero
In simple words: In uniform velocity, since there's no change in velocity, the acceleration is zero, just as displacement is zero after a full uniform circular path.

🎯 Exam Tip: Constant velocity implies zero acceleration. For uniform circular motion, displacement is zero only after completing a full circle.

 

Question 5.
Inertia : Newton's 1st law :: Rate of change of momentum : Newton's 2nd law
Answer: Newton's 2nd law
In simple words: Newton's first law defines inertia, and his second law relates the rate of change of momentum to the applied force.

🎯 Exam Tip: Each of Newton's laws has a core concept; link inertia to the first law and force/momentum change to the second law.

 

Question 6.
Balanced force : Body at rest :: Force equation : Mass x acceleration
Answer: Body at rest
In simple words: When forces are balanced, an object remains at rest or in uniform motion, similar to how the force equation is defined as mass times acceleration.

🎯 Exam Tip: Recall that balanced forces mean no net force, leading to either a state of rest or constant velocity.

 

(B) 3. Distinguish Between:

 

Question 1.
Positive acceleration and Negative acceleration
Answer:

Positive acceleration	Negative acceleration
(i) When the velocity of a body increases, acceleration is said to be positive acceleration.	(i) When the velocity of a body decreases, acceleration is said to be negative acceleration.

In simple words: Positive acceleration means an object is speeding up, while negative acceleration means it is slowing down.

🎯 Exam Tip: Clearly state the effect on velocity for each type of acceleration and ensure precise terminology.

 

Question 2.
Scalar quantity and Vector quantity
Answer:

Scalar quantity	Vector quantity
(i) Scalar quantities are physical quantities having magnitude only.	(i) Vector quantities are physical quantities having both magnitude and direction.

In simple words: Scalar quantities only describe "how much," while vector quantities describe "how much" and "in which direction."

🎯 Exam Tip: Provide clear definitions and examples for both scalar and vector quantities to demonstrate full understanding.

 

Question 3.
Balanced force and Unbalanced force
Answer:

Balanced force	Unbalanced force
(i) Two equal forces applied on a body in the opposite direction.	(i) Two unequal forces applied on a body.
(ii) This force does not change the state of rest or the state of uniform motion	(ii) This force can change the state of rest or the state of uniform motion of a body in a straight line.

In simple words: Balanced forces cancel each other out, causing no change in motion, while unbalanced forces result in a net force that causes acceleration.

🎯 Exam Tip: Focus on the net effect of the forces: balanced forces lead to zero net force and no acceleration, while unbalanced forces lead to a non-zero net force and acceleration.

 

(B) 4. State Whether The Following Statements Are True Or False:

 

Question 1.
The velocity of a body is given by the distance covered by it in unit time in a given direction.
Answer: True
In simple words: Velocity measures how fast an object is moving and in what specific direction, which is precisely the distance covered per unit time in a particular direction.

🎯 Exam Tip: Understanding that velocity includes direction is key; if direction isn't specified, it's typically speed.

 

Question 2.
Displacement is a scalar quantity.
Answer: False
In simple words: Displacement is a vector quantity because it indicates both the distance and the direction from the starting point to the ending point.

🎯 Exam Tip: Remember that displacement is the shortest path between two points, and it always includes direction, making it a vector.

 

Question 3.
Uniform acceleration means that the body is moving with a uniform velocity.
Answer: False
In simple words: Uniform acceleration means the velocity changes at a constant rate, but the velocity itself is not uniform; it's constantly increasing or decreasing.

🎯 Exam Tip: Uniform velocity means zero acceleration. Uniform acceleration means the rate of change of velocity is constant, not that velocity itself is constant.

 

Question 4.
The direction of acceleration can be opposite to that of velocity.
Answer: True
In simple words: Yes, when an object is slowing down, its acceleration acts in the opposite direction to its velocity.

🎯 Exam Tip: This situation occurs during deceleration or retardation, where the object is braking or moving against a resistive force.

 

Question 5.
Work is a vector quantity.
Answer: False
In simple words: Work is a scalar quantity because it only has magnitude, representing the amount of energy transferred, and not a specific direction.

🎯 Exam Tip: Distinguish between work, which is scalar, and force and displacement, which are vectors, even though work is calculated from them.

 

Question 6.
Displacement is always greater than distance.
Answer: False
In simple words: Displacement is either less than or equal to the distance travelled, but never greater than it.

🎯 Exam Tip: The magnitude of displacement is the shortest path, while distance is the actual path, so distance can be equal to or greater than displacement.

 

Question 7.
The distance and displacement are equal only if, motion is along a straight path.
Answer: True
In simple words: For distance and displacement to be the same, an object must move in a single, unchanging direction along a straight line.

🎯 Exam Tip: Any deviation from a straight line or change in direction will make the distance greater than the magnitude of displacement.

 

Question 8.
If an object experiences acceleration, a force is acting on it.
Answer: True
In simple words: According to Newton's second law, acceleration is directly caused by a net force acting on an object.

🎯 Exam Tip: This is a direct application of Newton's second law (\( F = ma \)), where acceleration is the effect of an applied force.

 

Question 9.
A train pulling out from a station is in uniform motion.
Answer: False
In simple words: A train starting from rest and gaining speed is undergoing accelerated motion, not uniform motion.

🎯 Exam Tip: Uniform motion means constant velocity (zero acceleration). Starting or stopping always involves non-uniform motion.

 

Question 10.
If a bus in motion is suddenly stopped, the passengers fall backwards.
Answer: False
In simple words: When a bus suddenly stops, passengers continue to move forward due to inertia, so they fall forwards, not backwards.

🎯 Exam Tip: This is an example of inertia (Newton's first law): passengers resist the change in motion and tend to maintain their original forward velocity.

 

Question 11.
If a single force is acting on an object, it will always accelerate.
Answer: True
In simple words: A single, unopposed force will always cause an object to accelerate, changing its velocity.

🎯 Exam Tip: This is a direct consequence of Newton's second law; a non-zero net force (which a single force implies) results in acceleration.

 

Question 12.
In circular motion, direction of motion is tangential.
Answer: True
In simple words: At any point on a circular path, the object's velocity vector points along the tangent to the circle at that specific point.

🎯 Exam Tip: The tangential direction of velocity is crucial for understanding how objects would fly off if the centripetal force ceased.

 

Question 13.
The inertia of a body is measured in terms of its mass.
Answer: True
In simple words: Mass is the quantitative measure of an object's inertia; the more massive an object, the greater its resistance to changes in motion.

🎯 Exam Tip: Remember that mass is the intrinsic property that determines an object's inertia, meaning its reluctance to accelerate.

 

(B) 5. Name The Following:

 

Question 1.
The scientist who summarized motion in a set of equations of motion.
Answer: Isaac Newton
In simple words: Isaac Newton is credited with formulating the three fundamental laws and equations that describe motion.

🎯 Exam Tip: Newton's equations (laws) of motion are the backbone of classical mechanics and are applied universally.

 

Question 2.
Motion of an object along a circular path with uniform speed.
Answer: Uniform circular motion
In simple words: When an object travels in a perfect circle at a steady speed, it's called uniform circular motion.

🎯 Exam Tip: Even with uniform speed, uniform circular motion involves continuous acceleration due to the changing direction of velocity.

 

Question 3.
What is the backward motion of the gun called?
Answer: Recoil
In simple words: The backward jerk experienced by a gun when a bullet is fired is known as recoil.

🎯 Exam Tip: Recoil illustrates the principle of conservation of momentum and Newton's third law of motion.

 

Question 4.
The motion in which the object covers equal distance in equal intervals of time.
Answer: Uniform motion
In simple words: Uniform motion describes movement where an object maintains a constant speed and direction, covering equal distances in equal time periods.

🎯 Exam Tip: For uniform motion, both speed and direction must be constant, resulting in zero acceleration.

 

Question 5.
S. I. unit of acceleration.
Answer: m/s²
In simple words: The standard unit for acceleration is meters per second squared, indicating the change in velocity (meters per second) over time (second).

🎯 Exam Tip: Always remember the SI units for basic physical quantities to avoid calculation errors in problems.

 

Question 6.
CGS unit of momentum
Answer: g cm/s
In simple words: In the CGS system, momentum is measured in grams times centimeters per second.

🎯 Exam Tip: Be mindful of the different unit systems (SI vs. CGS) when solving problems, as using the wrong units can lead to incorrect answers.

 

(B) 6. Answer The Following In One Sentence:

 

Question 1.
When is acceleration said to be positive?
Answer: When the velocity of a body increases, acceleration is said to be positive acceleration.
In simple words: Acceleration is positive when an object is speeding up in the direction of its motion.

🎯 Exam Tip: Positive acceleration means that the acceleration vector points in the same direction as the velocity vector.

 

Question 2.
What is negative acceleration?
Answer: When the velocity of a body decreases, acceleration is said to be negative acceleration.
In simple words: Negative acceleration means an object is slowing down, or its acceleration is in the opposite direction to its velocity.

🎯 Exam Tip: This is also known as deceleration or retardation; the acceleration vector points opposite to the velocity vector.

 

Question 3.
What is the direction of velocity of an object performing uniform circular motion?
Answer: The direction of velocity is along the tangential direction to its position.
In simple words: In uniform circular motion, the object's velocity is always tangent to the circular path at its current location.

🎯 Exam Tip: This tangential velocity is constantly changing direction, which is why there is always a centripetal acceleration towards the center of the circle.

 

Question 4.
Give the mathematical expression used to determine velocity of an object moving with uniform circular motion.
Answer: \( \text{Velocity} = \frac{\text{Circumference}}{\text{Time}} = \frac{2\pi r}{t} \) is the expression used to determine velocity of a body moving with uniform circular motion.
In simple words: For uniform circular motion, velocity can be calculated by dividing the total distance around the circle (circumference) by the time it takes to complete one round.

🎯 Exam Tip: This formula calculates the magnitude of the velocity (speed) in uniform circular motion, where 'r' is the radius and 't' is the time for one revolution.

 

Question 5.
What kind of force keeps the body at rest?
Answer: Balanced force keeps the body at rest.
In simple words: A balanced force (or zero net force) is what allows an object to remain motionless.

🎯 Exam Tip: According to Newton's first law, an object at rest will stay at rest if the net force acting on it is zero (i.e., forces are balanced).

 

Question 6.
Which law of motion gives the measure of force?
Answer: Newton's second law of motion gives the measure of force.
In simple words: Newton's second law quantitatively defines force as the product of mass and acceleration.

🎯 Exam Tip: The formula \( F = ma \) is the mathematical representation of Newton's second law, directly relating force to mass and acceleration.

 

Question 7.
What are vectors and scalars?
Answer: Scalars are physical quantities having magnitude only whereas, vectors are physical quantities having both magnitude and direction.
In simple words: Scalars are quantities that only tell you "how much," while vectors tell you "how much" and "in which direction."

🎯 Exam Tip: Always remember that the key distinguishing factor is the presence or absence of direction. Examples are essential for clarity.

 

Question 8.
Which of the quantities distance, speed, velocity, time and displacement are scalars and which are vectors?
Answer: Distance, speed and time are scalars displacement and velocity are vectors.
In simple words: Distance, speed, and time are scalars because they only have magnitude; displacement and velocity are vectors because they also include direction.

🎯 Exam Tip: Practice classifying various physical quantities to solidify your understanding of scalars and vectors.

 

Give Formula:

 

Question 1.
Force =
Answer: Mass x Acceleration = \( ma \)
In simple words: Force is calculated by multiplying an object's mass by the acceleration it undergoes.

🎯 Exam Tip: This fundamental formula (\( F=ma \)) is Newton's second law and is used extensively in dynamics problems.

 

Question 2.
Final velocity (v) =
Answer: Initial Velocity + (Acceleration x Time) = \( u + at \)
In simple words: To find the final velocity, you add the change in velocity (acceleration times time) to the initial velocity.

🎯 Exam Tip: This is the first equation of motion, useful when time, initial velocity, and acceleration are known.

 

Question 3.
Displacement (s) =
Answer: (Initial Velocity x Time) + \( \frac{1}{2} \) x Acceleration x Time² = \( (ut) + \frac{1}{2} at^2 \)
In simple words: Displacement is found by combining the distance covered at initial velocity with the additional distance due to acceleration over time.

🎯 Exam Tip: This is the second equation of motion, particularly useful when initial velocity, acceleration, and time are given to find displacement.

 

Question 4.
Final velocity² (v²) =
Answer: Initial Velocity² + 2 x Acceleration x Displacement = \( u^2 + 2as \)
In simple words: The square of the final velocity is equal to the square of the initial velocity plus twice the product of acceleration and displacement.

🎯 Exam Tip: This is the third equation of motion, very useful when time is not given or not required in a problem involving velocity, acceleration, and displacement.

 

Question 5.
velocity of an object moving with uniform circular motion =
Answer: \( \text{Velocity} = \frac{\text{Circumference}}{\text{Time}} = \frac{2\pi r}{t} \)
In simple words: The magnitude of velocity (speed) for an object in uniform circular motion is found by dividing the distance of one full circle by the time taken to complete it.

🎯 Exam Tip: Remember that while the speed is constant in uniform circular motion, the velocity changes due to continuous change in direction.

 

Give Scientific Reasons:

 

Question 5.
Motion is relative.
Answer:
• The motion of an object depends on the observer, hence a body may appear moving for one person and at the same time at rest for another one.
• Hence, motion is relative.
In simple words: Whether something is considered moving or not depends entirely on the observer's own frame of reference.

🎯 Exam Tip: Always consider the frame of reference when discussing motion. A common example is a passenger in a moving train: they are at rest relative to the train but moving relative to the ground.

 

Question 6.
Newton's first law of motion is called as law of inertia.
OR
Heavier objects offer more inertia.
Answer:
• Inertia is related to the mass of the object.
• As mass is the quantity of matter in a body, we need to exert more force to push a heavier body.
• Hence heavier objects offer more inertia.
• As the same property is described by Newton's first law of motion, it is called as Law of Inertia.
In simple words: Newton's first law is called the law of inertia because it states that objects resist changes to their motion, a property quantified by their mass; heavier objects have more inertia and resist changes more.

🎯 Exam Tip: Connect the concepts of mass and inertia directly. A clear explanation of why heavier objects have more inertia shows a deeper understanding.

 

Question 7.
The launching of a rocket is based on Newton's third law of motion.
Answer:
• Newton's third law of motion states that 'Every action force has an equal and opposite reaction force which acts simultaneously.'
• When the fuel in a rocket is ignited, it burns as a result of chemical reaction.
• The exhaust gases escape with a great force in the backward direction.
• It exerts an equal and opposite reaction force on the rocket, due to which the rocket moves in the forward direction.
• Thus, the principle of launching of rocket is based on Newton's third law of motion.
In simple words: A rocket launches because the powerful downward expulsion of exhaust gases creates an equal and opposite upward force, propelling the rocket, as explained by Newton's third law.

🎯 Exam Tip: This is a classic example of Newton's third law; ensure you clearly describe the action (gas expulsion) and the reaction (rocket propulsion).

 

Solve The Following Numerical:

 

Question 1.
An athlete is running on a circular track. He runs a distance of 400 m in 25 s before returning to his original position. What is his average speed and velocity?
Answer:
Given:
Total distance travelled = 400 m
Total displacement = 0, as he returns to his original position.
Total time = 25 seconds.
To find:
Average speed = ?
Average velocity = ?
Formula:
Average speed = \( \frac{\text{Total distance covered}}{\text{Total time taken}} \)
Average velocity = \( \frac{\text{Total displacement}}{\text{Total time taken}} \)
Solution:
Average speed = \( \frac{400}{25} \)
Average speed = 16 m/s
Average velocity = \( \frac{0}{25} \)
Average velocity = 0 m/s
The athlete runs at an average speed of 16 m/s and velocity 0 m/s.
In simple words: The athlete's average speed is calculated from the total distance covered over the time taken, while the average velocity is zero because he returns to his starting point, making his total displacement zero.

🎯 Exam Tip: Remember that average speed considers the total path length, while average velocity considers only the net change in position (displacement).

 

Question 2. An aeroplane taxies on the runway for \(30\) s with an acceleration of \(3.2\) \(m/s^2\) before taking off. How much distance would it have covered on the runway? (Ans: \(1440\) \(m\))

Answer:Given: \(a = 3.2 \ m/s^2\), \(t = 30 \ s\), \(u = 0\) To find: \(s = ?\) Formula: \(s = ut + \frac{1}{2} at^2\) Solution: \(s = ut + \frac{1}{2} at^2\)
\( = 0 \times 30 + \frac{1}{2} \times 3.2 \times 30^2 \)
\( = 1440 \ m \) The distance covered on the runway is \(1440 \ m\).In simple words: We used the second equation of motion, starting from rest, to calculate the total distance covered by the aeroplane using its acceleration and the time it taxied.

🎯 Exam Tip: Numerical problems often require correctly identifying given values and selecting the appropriate formula. Show all steps clearly for full marks.

 

Question 3. A kangaroo can jump \(2.5\) m vertically. What must be the initial velocity of the kangaroo? (Ans: \(7 \ m/s\))

Answer:Given: \(a = 9.8 \ m/s^2\), \(s = 2.5 \ m\), \(v = 0\) To find: \(u = ?\) Formula: \(v^2 = u^2 + 2as\) Solution: \(v^2 = u^2 + 2as\) \(0^2 = u^2 + 2 \times (-9.8) (2.5)\) : (Negative sign is used as the acceleration is in the direction opposite to that of velocity.) \(0 = u^2 - 49\) \(u^2 = 49\) \(u = 7 \ m/s\) The initial velocity of the kangaroo must be \(7 \ m/s\).In simple words: To find the initial velocity required for the kangaroo to jump \(2.5\) meters high, we use the third equation of motion, considering the final velocity at the peak of the jump to be zero and acceleration due to gravity acting downwards.

🎯 Exam Tip: Pay attention to the direction of acceleration (e.g., gravity) and use appropriate signs (positive/negative) in equations for projectile motion.

 

Question 4. A motorboat starts from rest and moves with uniform acceleration, if it attains the velocity of \(15 \ m/s\) in \(5s\), calculate the acceleration and the distance travelled in that time.

Answer:Given: Initial velocity, \(u = 0\) Final velocity, \(v = 15 \ m/s\) Time, \(t = 5 \ s\) To find: Acceleration (\(a\)) = ? Distance (\(s\)) = ? Solution: From the first equation of motion \(a = \frac{v-u}{t}\)
\( = \frac{15-0}{5}\) \(a = 3 \ m/s^2\) From the second equation of motion, the distance covered will be \(s = ut + \frac{1}{2} at^2\) \(s = 0 \times 5 + \frac{1}{2} \times 3 \times 25\)
\( = \frac{0+75}{2}\)
\( = 37.5 \ m\) The acceleration is \(3 \ m/s^2\) and distance travelled is \(37.5 \ m\).In simple words: We first found the acceleration using the change in velocity over time, and then used this acceleration with the initial velocity and time to calculate the total distance covered by the motorboat.

🎯 Exam Tip: For problems requiring multiple calculations (like acceleration and distance), always clearly state intermediate results and ensure units are consistent throughout the problem.

 

Question 5. The mass of a cannon is \(500 \ kg\) and it recoils with a speed of \(0.25 \ m/s\). What is the momentum of the cannon?

Answer:Given: mass of the cannon = \(500 \ kg\) recoil speed = \(0.25 \ m/s\) To find: Momentum = ? Formula: Momentum = \(m \times v\) Solution: Momentum = \(m \times v\)
\( = 500 \times 0.25 \)
\( = 125 \ kg \ m/s \) The momentum of cannon is \(125 \ kg \ m/s\).In simple words: Momentum is calculated by multiplying an object's mass by its velocity. Here, we directly apply this definition to find the cannon's momentum after recoil.

🎯 Exam Tip: Remember the units for momentum (\(kg \ m/s\)). Clearly listing given values, the formula, and the solution steps helps avoid errors.

Answer The Following In Short:

 

Question 1. Explain the three different ways to change the velocity.

Answer:As velocity is related to speed and direction, it can be changed by :
• Changing the speed while keeping the direction constant.
• Changing direction while keeping speed constant.
• Changing both speed as well as direction of motion.In simple words: Velocity can be changed by altering how fast an object moves (speed), by changing its path (direction), or by changing both.

🎯 Exam Tip: Understanding the vector nature of velocity (magnitude and direction) is crucial for explaining how it can be altered.

 

Question 2. Explain what is positive, negative and zero acceleration.

Answer:
• Positive Acceleration: When the velocity of an object increases, the acceleration is positive. In this case, the acceleration is in the direction of velocity.
• Negative Acceleration: When the velocity of an object decreases with time, it has negative acceleration. Negative acceleration is also called deceleration. Its direction is opposite to the direction of velocity.
• Zero Acceleration: If the velocity of the object does not change with time, it has zero acceleration.In simple words: Positive acceleration means an object is speeding up, negative acceleration (deceleration) means it's slowing down, and zero acceleration means its velocity isn't changing at all.

🎯 Exam Tip: Clearly defining each type of acceleration with respect to changes in velocity and direction is important.

 

Question 3. What inference do we draw from the velocity-time graph for a uniformly accelerated motion?

Answer:
• From velocity-time graph we can infer whether velocity changes by equal amounts in equal intervals of time or not.
• Thus, for all uniformly accelerated motion, the velocity - time graph is a straight line and slope of the line gives the acceleration.
• For non-uniformly accelerated motion, velocity-time graph can have any shape according to variation in velocity with respect to time.In simple words: A velocity-time graph for uniformly accelerated motion will be a straight line, and its slope tells us the constant acceleration. For non-uniform motion, the graph will be a curved line.

🎯 Exam Tip: Being able to interpret different types of motion from velocity-time graphs (e.g., slope for acceleration, area for displacement) is a key skill.

 

Question 4. State the three equations of motion and give the relationship explained by them.

Answer:
• \(v = u + at\): This is the relation between velocity and time.
• \(s = ut + \frac{1}{2} at^2\) : This is the relation between displacement and time
• \(v^2 = u^2 + 2as\) : This is the relation between displacement and velocity.In simple words: These three equations connect initial velocity (\(u\)), final velocity (\(v\)), acceleration (\(a\)), time (\(t\)), and displacement (\(s\)) for objects moving with constant acceleration.

🎯 Exam Tip: Memorize these three equations and understand when to apply each one based on the known and unknown variables in a problem.

 

Question 5. What are the implications of Newton's Third Law of motion?

Answer:
• Action and reaction are terms that express force.
• These forces act in pairs. One force cannot exist by itself.
• Action and reaction forces act simultaneously.
• Action and reaction forces act on different objects. They do not act on the same object and hence cannot cancel each other's effect.In simple words: Newton's Third Law states that forces always come in pairs (action-reaction), are equal and opposite, happen at the same time, and act on different objects, meaning they don't cancel each other out.

🎯 Exam Tip: Emphasize that action-reaction forces always act on *different* bodies, which is a common point of confusion for students.

 

Question 6. Explain recoil and recoil velocity. Derive its expression.

Answer:
• Let us consider the example of a bullet fired from a gun. When a bullet of mass \(m_1\) is fired from a gun of mass \(m_2\), its velocity becomes \(v_1\), and its momentum becomes \(m_1v_1\). Before firing the bullet, both the gun and the bullet are at rest. Hence, total initial momentum is zero.
• According to Newton's third law of motion, the total final momentum also has to be zero. Thus, the forward-moving bullet causes the gun to move backward after firing.
• This backward motion of the gun is called its recoil. The velocity of recoil, \(v_2\) is such that, \(m_1v_1 + m_2v_2 = 0\)
\( \implies v_2 = -\frac{m_1}{m_2} \times v_1 \)In simple words: Recoil is the backward movement of a gun when a bullet is fired, caused by the equal and opposite reaction force. The recoil velocity is calculated using the conservation of momentum, showing it's in the opposite direction and dependent on the masses and bullet's velocity.

🎯 Exam Tip: When deriving expressions involving conservation of momentum, clearly define all variables (masses and velocities before and after interaction) and state the principle applied.

Complete The Flow Chart:

 

Question 1. (1) Types of force and their effects

Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह प्रवाह चार्ट 'बल' (Force) के विभिन्न प्रकारों और उनके प्रभावों को दर्शाता है। बल को 'संतुलित बल' (Balanced force) और 'असंतुलित बल' (Unbalanced force) में विभाजित किया गया है। असंतुलित बल के तहत, यह गति में वस्तु, चलती वस्तु को रोकना, और विराम में वस्तु जैसे प्रभाव दिखाता है।

🎯 Exam Tip: Flowcharts are effective for visually organizing concepts. Ensure the categories and their relationships are logical and accurate.

 

Question 2. (2) Newton's laws

Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह प्रवाह चार्ट न्यूटन के गति के नियमों को उनकी मुख्य अवधारणाओं के साथ दिखाता है। न्यूटन के नियम को पहले नियम (जड़ता), दूसरे नियम (संवेग परिवर्तन की दर), और तीसरे नियम (क्रिया और प्रतिक्रिया बल) में विभाजित किया गया है।

🎯 Exam Tip: Understanding the core concept behind each of Newton's laws is fundamental. Use clear, concise explanations for each law.

Distinguish Between:

 

Question 1. Speed and velocity.

Answer:

Speed	Velocity
(i) Speed is the distance covered by a body in unit time.	(i) The displacement that occurs in unit time is called velocity.
(ii) It is a scalar quantity.	(ii) It is a vector quantity.
(iii) Speed is either equal to or greater than displacement.	(iii) Velocity is equal to or less than speed.
(iv) Speed = distance/time	(iv) Velocity = displacement/time
(v) It is always positive or zero but never negative.	(v) It may be positive, Zero or negative.

In simple words: Speed is how fast an object is moving, while velocity is how fast it's moving in a specific direction. Speed is a scalar quantity, only having magnitude, whereas velocity is a vector quantity, having both magnitude and direction.

🎯 Exam Tip: When distinguishing between concepts, use clear, contrasting points, ideally presented in a table format for clarity and easy comparison.

 

Question 2. Balanced force and unbalanced force.

Answer:

Balanced force	Unbalanced force
(i) Two equal forces applied on a body in the opposite direction.	(i) Two unequal forces applied on a body.
(ii) This force does not change the state of rest or the state of uniform motion	(ii) This force can change the state of rest or the state of uniform motion of a body in a straight line.

In simple words: Balanced forces result in no change in an object's motion (it stays at rest or continues uniform motion), whereas unbalanced forces cause a change in motion, like acceleration or deceleration.

🎯 Exam Tip: Highlight the key difference: balanced forces result in zero net force, while unbalanced forces result in a non-zero net force, leading to acceleration.

Give Examples:

 

Question 1. Scalar quantities

Answer: Time, Volume, Speed, Mass, Temperature, Distance, Entropy, Energy, and WorkIn simple words: Scalar quantities are physical measurements that only have a size or amount (magnitude), without any specific direction.

🎯 Exam Tip: Provide a diverse range of examples to demonstrate a comprehensive understanding of scalar quantities.

 

Question 2. Vector quantities

Answer: Acceleration, Velocity, Momentum, Force, and WeightIn simple words: Vector quantities are physical measurements that have both a size (magnitude) and a specific direction associated with them.

🎯 Exam Tip: Remember that for vector quantities, specifying both magnitude and direction is essential for a complete description.

Answer The Following Questions:

Observe The Figure And Answer The Questions:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो घरों (शीतल और प्रशांत का घर, और संगीता का घर) तथा एक स्कूल के बीच की विभिन्न रास्तों को दर्शाता है। इसमें बिंदु A और B के बीच दो अलग-अलग पथ (पथ A और पथ B) दिखाए गए हैं, जिनकी दूरियां (500 मी. और 1300 मी.) और एक बिंदीदार सीधी रेखा भी इंगित की गई है जो सबसे कम दूरी को दर्शाती है।

🎯 Exam Tip: Diagrams are often used to test understanding of distance and displacement. Always analyze the paths shown and identify initial and final positions.

 

Question (a). Measure the distance between points A and B in different ways as shown in figure (I).

Answer: Distances measured may be of different lengths depending on the path taken.In simple words: The total length traveled from point A to point B varies depending on the specific path chosen, as each path could have a different route.

🎯 Exam Tip: Recognize that "distance" refers to the actual path length, which can differ between two points.

 

Question (b). Now measure the distance along the dotted line. Which distance is correct according to you and why?

Answer: Dotted line shows the shortest way of reaching from A to B.In simple words: The dotted line represents the shortest path between points A and B, which corresponds to the concept of displacement.

🎯 Exam Tip: The shortest distance between two points is always a straight line, representing displacement, not necessarily the actual distance traveled.

 

Question (c). Observe the following figures. If you increase the number of sides of the polygon and make it infinite, how many times will you have to change the direction? What will be the shape of the path?

Answer: If we increase the number of sides of the polygon and make it infinite, then we will have to change the direction an infinite number of times. The shape of the path thus obtained will be a circle.In simple words: As a polygon gains more and more sides, eventually an infinite number, its shape approaches a perfect circle, requiring continuous, infinite changes in direction.

🎯 Exam Tip: This question relates to the concept of uniform circular motion, where direction continuously changes even if speed is constant.

Observe The Figure And Answer The Questions

ℹ️ चित्र व्याख्या (Diagram Explanation): ये चित्र एक मोटरसाइकिल चालक को विभिन्न सड़क स्थितियों में दिखाते हैं: एक सीधी सड़क पर, एक घुमावदार पहाड़ी सड़क पर, और एक रेसिंग ट्रैक पर। यह चित्र गति और दिशा में परिवर्तन के साथ मोटरसाइकिल की वेग पर पड़ने वाले प्रभावों को दर्शाते हैं।

🎯 Exam Tip: Visualizing real-world scenarios like vehicle motion helps understand abstract concepts like velocity changes due to speed and/or direction changes.

 

Question 1. What will be the effect on the velocity of the motorcycle if its speed increases or decreases, but its direction remains unchanged?

Answer: If the speed is increased the velocity of the motorcycle will increase and if the speed is decreased the velocity of the motorcycle will decreaseIn simple words: If the direction of motion stays the same, changing the motorcycle's speed will directly change its velocity—increasing speed increases velocity, and decreasing speed decreases velocity.

🎯 Exam Tip: For motion in a straight line, velocity's magnitude changes directly with speed, as direction is constant.

 

Question 2. In case of a turning on the road, will the velocity and speed be same?

Answer: As speed is scalar quantity while velocity is vector quantity so by turning velocity will change while speed remains sameIn simple words: No, on a turning road, speed and velocity will not be the same because even if the speed (magnitude) is constant, the velocity's direction changes, making the velocity different.

🎯 Exam Tip: Distinguish clearly between scalar (speed) and vector (velocity) quantities. A change in direction alone is sufficient to change velocity.

 

Question 3. If, on a turning, we change the direction as well as the speed of the motorcycle, what will be the effect on its velocity?

Answer: Its velocity will change because velocity depends on speed as well as direction and here both speed and direction are changed.In simple words: If both the speed and direction of the motorcycle change on a turn, its velocity will definitely change, as velocity is dependent on both of these factors.

🎯 Exam Tip: Reinforce that any change in either the magnitude (speed) or direction of motion, or both, will result in a change in velocity.

Numerical:

 

Question 1. A person travels a distance of \(72 \ km\) in \(4\) hours. Calculate average speed in \(m/s\).

Answer:Given: Total distance (\(d\)) \( = 72 \ km \) \( = 72 \times 1000 \) \( = 72000 \ m \) Total time taken (\(t\)) \( = 4 \ hours \) \( = 4 \times 3600 \) (\(1 \ hour = 3600 \ sec\)) \( = 14400 \ s \) To find: Average speed = ? Formula: Average speed \( = \frac{\text{Total distance covered}}{\text{Total time taken}} \) Solution: Average speed \( = \frac{\text{Total distance covered}}{\text{Total time taken}} \)
\( = \frac{72000}{14400} \)
\( = \frac{720}{144} \ m/s \)
\( = \frac{10}{2} \ m/s \)
\( = 5 \ m/s \) The person travels with average speed of \(5 \ m/s\).In simple words: To find the average speed in meters per second, we convert the total distance traveled from kilometers to meters and the total time from hours to seconds, then divide the total distance by the total time.

🎯 Exam Tip: Always pay attention to units specified in the question (e.g., \(m/s\)) and convert all given values to the appropriate units before calculation.

 

Question 2. balls have masses of \(50 \ gm\) and \(100 \ gm\) and they are moving along the same line in the same direction with velocities of \(3 \ m/s\) and \(1.5 \ m/s\) respectively. They collide with each other and after the collision, the first ball moves with a velocity of \(2.5 \ m/s\). Calculate the velocity of the other ball after collision.

Answer:Given: Mass of 1st ball (\(m_1\)) \( = 50 \ g = \frac{50}{1000} \ kg \). Mass of 2nd ball (\(m_2\)) \( = 100 \ g = \frac{100}{1000} \ kg \). Initial velocity of 1st ball (\(u_1\)) \( = 3 \ m/s \) Initial velocity of 2nd ball (\(u_2\)) \( = 1.5 \ m/s \) Final velocity of 1st ball (\(v_1\)) \( = 2.5 \ m/s \) To find: Final velocity of 2nd ball (\(v_2\)) = ? Formula: \(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\) Solution: \(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\) \((0.05 \times 3) + (0.1 \times 1.5) = (0.05 \times 2.5) + (0.1 \times v_2)\)
\( \frac{150}{1000} + \frac{150}{1000} = \frac{125}{1000} + \frac{100v_2}{1000} \)
\( \frac{300}{1000} = \frac{125+100v_2}{1000} \) \(300 = 125 + 100v_2\) \(100v_2 = 175\)
\( v_2 = \frac{175}{100} \) \(v_2 = 1.75 \ m/s\) Final velocity of second ball after collision is \(1.75 \ m/s\).In simple words: By applying the law of conservation of momentum, we equate the total momentum before the collision to the total momentum after the collision to find the unknown velocity of the second ball.

🎯 Exam Tip: For collision problems, ensure all masses are in kilograms and velocities are in meters per second. The principle of conservation of momentum (\(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\)) is key.

Write Laws And Explain Write Implications:

 

Question 1. Newton's third law of motion

Answer: 'Every action force has an equal and opposite reaction force which acts simultaneously'.
• Action and reaction are terms that express force.
• These forces act in pairs. One force cannot exist by itself.
• Action and reaction forces act simultaneously.
• Action and reaction forces act on different objects. They do not act on the same object and hence cannot cancel each other's effect.In simple words: Newton's Third Law states that for every force (action), there's an equal and opposite force (reaction) that happens at the same time but acts on a different object.

🎯 Exam Tip: Always emphasize the simultaneous and distinct nature of action-reaction forces when explaining Newton's Third Law.

 

Question 2. Explain Newton's second law of motion and derive the formula.

Answer: Newton's second law explains about change in momentum. It states that 'The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force.' Suppose an object of mass 'm' has an initial velocity 'u'. When a force 'F' is applied in direction of its velocity for time 't', its velocity becomes 'v'. Then, the total initial momentum of the body = 'mu'. Its final momentum after time t = 'mv'. So, the rate of change of momentum \( = \frac{\text{Change in momentum}}{\text{time}} \)
\( = \frac{mv-mu}{t} \)
\( = \frac{m(v-u)}{t} \)
\( \implies = ma \) \( \left( \because a = \frac{v-u}{t} \right) \) Hence by Newtons second law of motion, the rate of change of momentum is proportional to the applied force.
\( \therefore ma \propto F \)
\( \therefore F \propto ma \)
\( \therefore F = kma \) (\(k = \text{Constant of proportionality and value is 1}\)).
\( \therefore F = ma \)In simple words: Newton's Second Law says that the force applied to an object is directly proportional to how quickly its momentum changes, and this change happens in the direction of the force, leading to the formula \(F = ma\).

🎯 Exam Tip: Clearly define momentum and show each step of the derivation of \(F=ma\), explaining how the rate of change of momentum leads to acceleration.

 

Question 3. State the law of conservation of momentum and derive the formula.

Answer:Let mass of object A and B be \(m_1\) and \(m_2\) respectively Let their initial velocity be \(u_1\) and \(u_2\) Let their final velocity be \(v_1\) and \(v_2\) We know, \(P = mv\) Let their initial momentum be \(m_1u_1\) and \(m_2u_2\) Let their final momentum be \(m_1v_1\) and \(m_2v_2\) Total initial momentum \( = (m_1u_1 + m_2u_2) \) Total final momentum \( = (m_1v_1 + m_2v_2) \) If \(F_2\) is the force that acts on object B, \(F_2 = -F_1\) \(m_2a_2 = -m_1a_1 \) [... \(F = ma\)]
\(m_2 \times \frac{v_2-u_2}{t} = -m_1 \times \frac{v_1-u_1}{t} \) \( \left( \because a = \frac{v-u}{t} \right) \)
\( \therefore m_2(v_2-u_2) = -m_1(v_1-u_1) \)
\( \therefore m_2v_2 - m_2u_2 = -m_1v_1 + m_1u_1 \)
\( \therefore (m_2v_2 + m_1v_1) = (m_2u_2 + m_1u_1) \) i.e. The magnitude of total final momentum = the magnitude of total initial momentumIn simple words: The law of conservation of momentum states that in an isolated system, the total momentum before a collision or interaction is equal to the total momentum after the interaction, meaning momentum is neither lost nor gained.

🎯 Exam Tip: The derivation of the conservation of momentum relies heavily on Newton's Third Law and the definition of acceleration. Clearly state the "before" and "after" scenarios.

Complete The Paragraph:

 

Question 1. Moving Objects

Answer: 'Distance' is the length of the actual path travelled by an object in motion while going from one point to another, whereas displacement is the minimum distance between the starting and finishing points. The displacement that occurs in unit time is called velocity. The units of speed and velocity are the same. In the SI system, the unit is \(m/s\) while in the CGS system; it is \(cm/s\). Speed is related to distance while velocity is related to the displacement. If the motion is along a straight line, the values of speed and velocity are the same, otherwise they can be different. The first scientist to measure speed as the distance /time was Galileo. The speed of sound in dry air is \(343.2 \ m/s\) while the speed of light is about \(3 \times 10^8 \ m/s\). The speed of revolution of the earth around the sun is about \(29770 \ m/s\).In simple words: This paragraph defines key concepts in motion like distance, displacement, speed, and velocity, highlighting their units and how they differ or relate, especially in straight-line versus other types of motion.

🎯 Exam Tip: Be precise with definitions of fundamental concepts like distance, displacement, speed, and velocity, and remember their standard units in both SI and CGS systems.

 

Question 2. Types of motion

Answer: If an object covers unequal distances in equal time intervals, it is said to be moving with non-uniform speed. For example the motion of a vehicle being driven through heavy traffic. If an object covers equal distances in equal time intervals, it is said to be moving with uniform speed. The rate of change of velocity is called acceleration. If the velocity changes by equal amounts in equal time intervals, the object is said to be in uniform acceleration. If the velocity changes by unequal amounts in equal time intervals, the object are said to be non-uniform acceleration. The speed of the tip is constant, but the direction of its displacement and therefore, its velocity is constantly changing. As the tip is moving along a circular path, its motion is called uniform circular motion.In simple words: This paragraph explains different types of motion—uniform speed, non-uniform speed, uniform acceleration, non-uniform acceleration, and uniform circular motion—based on how distance and velocity change over time.

🎯 Exam Tip: Differentiate between uniform and non-uniform motion for both speed and acceleration. Understand that uniform circular motion involves constant speed but changing velocity due to continuous change in direction.

 

Question 3. Newton's laws and conservation of momentum

Answer: An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it. The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force. Momentum has magnitude as well as direction. Its direction is the same as that of velocity. In SI system, the unit of momentum is \(kg \ m/s\), while in CGS system, it is \(g \ cm/s\). If an unbalanced force applied on an object causes a change in the velocity of the object, then it also causes a change in its momentum. The force necessary to cause a change in the momentum of an object depends upon the rate of change of momentum. Every action force has an equal and opposite reaction force which acts simultaneously. As the mass of the gun is much higher than the mass of the bullet, the velocity of the gun is much smaller than the velocity of the bullet. The magnitude of the momentum of the bullet and that of the gun are equal and their directions are opposite. Thus, the total momentum is constant. Total momentum is also constant during the launch of a rocket.In simple words: This paragraph summarizes Newton's laws of motion, covering inertia, the relationship between force and momentum change, and action-reaction pairs, all of which lead to the principle of conservation of momentum in various scenarios.

🎯 Exam Tip: This comprehensive paragraph touches on all three of Newton's laws and the conservation of momentum. Ensure you can explain each concept and its relevance in real-world examples.

Answer The Following In Detail:

 

Question 1. What is speed? State its units and types. Explain instantaneous speed and average speed.

Answer: The speed of a body is the distance travelled in unit time. The units of speed in CGS system is \(cm/s\) and in SI system is \(m/s\). There are two types of speed :
• Uniform speed : When a body covers equal distance in equal intervals of time throughout its motion, it is said to have uniform speed.
• Non-uniform or variable speed: When a body covers unequal distance in equal intervals of time it is said to have non-uniform speed. The speed of the body at any instant is called instantaneous speed. Average speed is the ratio of total distance covered to total time taken.In simple words: Speed is the rate at which an object covers distance. It can be uniform (equal distance in equal time) or non-uniform (unequal distance in equal time). Instantaneous speed is speed at a specific moment, while average speed is total distance divided by total time.

🎯 Exam Tip: Clearly differentiate between instantaneous and average speed. Remember that speed is a scalar quantity and its units in SI and CGS systems.

 

Question 2. What is velocity? State its units and types.

Answer: The velocity of a body is the distance travelled by a body in a particular direction in unit time. Thus, rate of change of displacement is called velocity. \(v = s/t\) where: \(s = \text{displacement}\); \(t = \text{time taken}\); \(v = \text{velocity}\) (MKS unit: \(m/s\) CGS unit: \(cm/s\)) There are two types of velocities :In simple words: Velocity measures how fast an object moves in a specific direction (rate of change of displacement). It can be uniform, meaning constant speed in a constant direction, or non-uniform, meaning either speed or direction, or both, are changing.

🎯 Exam Tip: Emphasize that velocity is a vector quantity. Differentiate between uniform and non-uniform velocities by considering both magnitude (speed) and direction of motion.

 

Question 3. What is acceleration? State its units and types.
Answer:
(i) Acceleration is a rate of change in velocity. It is a vector quantity, \( a = \frac{v-u}{t} \) where : v = final velocity; u = initial velocity; a = acceleration
Units of acceleration in SI system is m/s² and CGS system is cm/s².
(ii) Types of acceleration: .
(a) Uniform acceleration : If the change in velocity is equal in equal intervals of time, the acceleration is uniform acceleration.
(b) Non-uniform acceleration : If the change in velocity is unequal in equal intervals of time, the acceleration is a non-uniform acceleration.
(iii) Kinds of acceleration:
Positive acceleration : When the velocity of an object goes on increasing, it is said to have Positive acceleration.
Negative acceleration : When the velocity of an object goes on decreasing, it is said to have negative acceleration or retardation or deceleration.
Zero acceleration : If the velocity of the object does not change with time, it has zero acceleration.
In simple words: Acceleration is how fast an object's velocity changes over time, including both speed and direction. It can be uniform (constant change), non-uniform (varying change), positive (speeding up), negative (slowing down), or zero (constant velocity).

🎯 Exam Tip: Understanding the different types of acceleration and their definitions is crucial for conceptual clarity and scoring in descriptive questions.

 

Question 4. Explain Newton's second law of motion and derive the formula.
Answer:
Newton's second law explains about change in momentum. It states that 'The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force.'
Suppose an object of mass 'm' has an initial velocity 'u'. When a force 'F' is applied in the direction of its velocity for time 't', its velocity becomes 'v'. Then, the total initial momentum of the body = 'mu'. Its final momentum after time t = 'mv'.
So, the rate of change of momentum
\[ = \frac{\text{Change in momentum}}{\text{time}} \]
\[ = \frac{mv-mu}{t} \]
\[ = \frac{m(v-u)}{t} \]
\[ = ma \]
\[ \left( a = \frac{v-u}{t} \right) \]
Hence by Newton's second law of motion, the rate of change of momentum is proportional to the applied force.
\[ \therefore ma \propto F \]
\[ \therefore F \propto ma \]
\[ \therefore F = kma \text{ (k = Constant of proportionality and value is 1).} \]
\[ \therefore F = ma \]
In simple words: Newton's second law states that the force applied to an object is directly proportional to the rate at which its momentum changes and acts in the direction of the force. This relationship can be expressed by the formula F=ma, where F is force, m is mass, and a is acceleration.

🎯 Exam Tip: Be precise when stating Newton's laws and ensure correct derivation steps for the formula F=ma.

 

Question 5. State the law of conservation of momentum and derive the formula.
Answer:
(i) Let mass of objects A and B be m₁ and m₂ respectively
Let their initial velocity be u₁ and u₂
Let their final velocity be v₁ and v₂
(ii) We know,
P = mv
Let their initial momentum be m₁u₁ and m₂u₂
Let their final momentum be m₁v₁ and m₂v₂
(iii) Total initial momentum = (m₁u₁ + m₂u₂)
Total final momentum = (m₁v₁ + m₂v₂)
(iv) If F₂ is the force that acts on object B,
\[ F_2 = -F_1 \]
\[ \therefore m_2a_2 = -m_1a_1 \]
\[ \therefore m_2 \times \frac{v_2-u_2}{t} = -m_1 \times \frac{v_1-u_1}{t} \]
\[ (\therefore F = ma) \]
\[ \left( a = \frac{v-u}{t} \right) \]
\[ \therefore m_2(v_2-u_2) = -m_1(v_1-u_1) \]
\[ \therefore m_2v_2 - m_2u_2 = -m_1v_1 + m_1u_1 \]
\[ \therefore (m_2v_2 + m_1v_1) = (m_2u_2 + m_1u_1) \]
i.e. The magnitude of total final momentum = the magnitude of total initial momentum.
In simple words: The law of conservation of momentum states that in a closed system, the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction. This means momentum is neither lost nor gained, but transferred.

🎯 Exam Tip: When deriving the conservation of momentum, clearly state the initial and final momentum of each object and the system, and apply Newton's third law.

 

Question 6. Obtain the equations of motion by graphical method:
(a) Equation for velocity-time relation.
Answer:
Velocity-time graph: shows the change in velocity with time of a uniformly accelerated object. The object starts from the point D in the graph with velocity v. Its velocity keeps increasing and after time t it reaches the point B on the graph.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वेग-समय ग्राफ है जो एक समान रूप से त्वरित वस्तु की गति को दर्शाता है। क्षैतिज अक्ष पर समय (t) और ऊर्ध्वाधर अक्ष पर वेग (v) है। ग्राफ एक सीधी रेखा AB है, जिसमें प्रारंभिक वेग OD (u) और अंतिम वेग OC (v) है, और समय अंतराल OE (t) है।
The initial velocity of the object = u = OD
The final velocity of the object = v = OC
Time = t = OE
\[ \text{Acceleration (a)} = \frac{\text{Change in velocity}}{\text{Time}} \]
\[ = \frac{\text{(Final velocity - Initial velocity)}}{\text{Time}} \]
\[ = \frac{\text{(OC-OD)}}{t} \]
\[ \therefore \text{ at = OC-OD} \]
From the graph,
OC = v and OD = u
\[ \therefore \text{ at = v-u ... (from i)} \]
\[ \therefore \text{ v = u + at} \]
This is the first equation of motion.
In simple words: The first equation of motion, v = u + at, shows how the final velocity of an object is related to its initial velocity, acceleration, and the time taken, assuming constant acceleration. It's derived from the definition of acceleration from a velocity-time graph.

🎯 Exam Tip: When deriving graphically, clearly label the axes and points on your velocity-time graph, and explain each step in relating the graph's features to the variables (v, u, a, t).

 

(b) Equation for displacement-time relation.
Answer:
Suppose that an object is in uniform acceleration 'a' and it has covered the distance 's' within time 't'. From the graph the distance covered by the object during time 't' is given by the area of quadrangle DOEB.
\[ \therefore \text{ s = area of quadrangle DOEB} \]
\[ = \text{area (rectangle DOEA) + area of triangle (DAB)} \]
\[ = \text{(AE}\times\text{OE)} + \left(\frac{1}{2} \times \text{[AB} \times \text{DA]}\right) \]
But, AE = u, OE = t and (OE = DA = t)
AB = at ... (AB = CD) ..... from (CD = OC-OD = v - u = at from (i))
\[ \therefore s = u \times t + \frac{1}{2} \times a t \times t \]
Newton's second equation of motion is
\[ s = ut + \frac{1}{2}at^2 \].
In simple words: The second equation of motion, s = ut + (1/2)at², calculates the total displacement of an object based on its initial velocity, acceleration, and the time taken. It's derived by finding the area under the velocity-time graph.

🎯 Exam Tip: For graphical derivation of displacement-time, clearly indicate that displacement is the area under the velocity-time graph, and correctly calculate the area of the rectangle and triangle.

 

(c) Equation for displacement-velocity relation.
Answer:
We can determine the distance covered by the object in time t from the area of the quadrangle DOEB. DOEB is a trapezium. So we can use the formula for its area.
\[ \therefore \text{ s = area of trapezium DOEB} \]
\[ \therefore s = \frac{1}{2} \times \text{sum of lengths of parallel sides} \]
\[ \times \text{ distance between the parallel sides} \]
\[ \therefore s = \frac{1}{2} \times \text{(OD + BE)} \times \text{OE} \]
But, OD = u, BE = v, and OE = t
\[ \therefore s = \frac{1}{2} \times \text{(u + v)} \times t \] ...(ii)
But, \[ a = \frac{\text{(v - u)}}{t} \]
\[ \therefore t = \frac{\text{(v - u)}}{a} \] ...(iii)
\[ \therefore s = \frac{1}{2} \times \text{(u+v)} \times \frac{\text{(v-u)}}{a} \]
\[ \therefore s = \frac{\text{(u + v) (v - u)}}{2a} \]
\[ \therefore 2as = \text{(u + v) (v - u)} = v^2 - u^2 \]
\[ \therefore v^2 = u^2 + 2as \]
This is Newton's third equation of motion.
In simple words: The third equation of motion, v² = u² + 2as, relates the final velocity, initial velocity, acceleration, and displacement without needing the time variable. It's derived using the area of a trapezium on the velocity-time graph.

🎯 Exam Tip: When deriving the displacement-velocity relation, remember to treat the area under the graph as a trapezium and substitute 't' from the first equation of motion.

Read The Paragraph And Answer The Questions:

(a) When a bat strikes a ball, the ball exerts an equal and opposite force on the bat. The force : acting on the ball projects it with high velocity, Due to the large mass of bat compared to ball, reaction force on the bat slows down the bat's motion.
(b) When a bullet is fired from a gun, the gun exerts a force on the bullet in the forward direction. This is the action force. The bullet also exerts an equal force on the gun in the backward direction. This is the reaction force. Due to the large mass of the gun, it j moves only a little distance backward. This backward movement of the gun is called the recoil of the gun.
(c) In a rocket, burning fuel creates a push on the front of the rocket pushing it forward. This creates an equal and opposite push on the exhaust gas backwards.

 

Question 1. (i) Which of Newton's law examples are given here?
Answer:
Newton's 3rd law is stated by the above example
In simple words: The examples provided illustrate Newton's Third Law of Motion, which describes how forces always occur in action-reaction pairs.

🎯 Exam Tip: Clearly identify the action-reaction pairs in each example to demonstrate understanding of Newton's third law.

 

Question 1. (ii) When a rifle is fired it is pushed back this movement is called what?
Answer:
Movement of rifle getting pushed back after firing is called recoil.
In simple words: When a rifle is fired, it moves backward due to a reaction force, and this backward movement is known as recoil.

🎯 Exam Tip: Define recoil accurately as the backward movement of a gun or similar device caused by the action-reaction force of firing a projectile.

 

Question 1. (iii) What does the ball acquire after it gets velocity?
Answer:
The ball acquires momentum after it gets velocity.
In simple words: Once the ball gains velocity, it also gains momentum, which is a measure of its mass in motion.

🎯 Exam Tip: Remember that momentum is a vector quantity dependent on both mass and velocity, and it is acquired whenever an object starts moving.

 

Question 1. (iv) State newton's 1st law of motion
Answer:
An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it.
In simple words: Newton's First Law, also known as the law of inertia, states that an object will keep its current state of motion (either at rest or moving at a constant velocity) unless an external force pushes or pulls it.

🎯 Exam Tip: State Newton's First Law precisely, emphasizing both the state of rest and uniform motion, and the role of external unbalanced force.

 

Question 1. (v) Which force is required to produce motion in an object?
Answer:
Unbalanced force is required to produce motion in an object.
In simple words: To make an object move or change its existing motion, an unbalanced force must act upon it, meaning the net force is not zero.

🎯 Exam Tip: Distinguish between balanced and unbalanced forces; only unbalanced forces cause acceleration or changes in motion.

 

Question 2. Paragraph 2
constant velocity. If that velocity is zero, then the object remains at rest. If an external force is applied, the velocity will change because of the force. The second law explains how the velocity of an object changes when it is subjected to an external force. The law defines a force to be equal to changes in momentum (mass times velocity) per change in time. Newton also developed the calculus of mathematics, and the "changes" "expressed in j the second law are most accurately defined in differential forms. (Calculus can also be used to determine the velocity and location variations experienced by an object subjected to an external force.)
For an object with a constant mass the second law states that the force F is the product of an objects mass and its acceleration a:
F = ma
For an external applied force the change in velocity depends on the mass of the object.
A force will cause in velocity; and likewise, a change in velocity will generate a force. The equation works both ways.
The third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal force on object: A. Notice that the forces are exerted on different objects. The third law can be used to explain the generation of lift by a wing and the production of thrust by a jet engine

 

Question 2. (i) A chalk kept on the table remains in the position of rest until moved by the teacher. Which law of motion is followed in this situation?
Answer:
Newton's 1st law is followed in this situation
In simple words: The phenomenon of the chalk remaining at rest until an external force (the teacher moving it) acts upon it demonstrates Newton's First Law of Motion, also known as the law of inertia.

🎯 Exam Tip: Provide examples to illustrate Newton's First Law, linking the concept of inertia to objects at rest or in uniform motion.

 

Question 2. (ii) What will happen to momentum if the mass and acceleration both are doubled?
Answer:
If the mass and acceleration both are doubled then the momentum will be increased four times
In simple words: Momentum (mass × velocity) is directly affected by both mass and velocity. If acceleration doubles, velocity will also increase, potentially doubling over time. If both mass and acceleration (leading to increased velocity) are doubled, the momentum will be quadrupled.

🎯 Exam Tip: Remember the formula for momentum (p = mv) and acceleration (a = Δv/t). If 'a' doubles, 'v' changes more rapidly, leading to a larger 'v' for the same time 't'. If mass is also doubled, the effect on momentum is multiplicative.

 

Question 2. (iii) What will happen to momentum if the mass and acceleration both are halved?
Answer:
If the mass and acceleration both are halved, then the momentum will be decreased to one-fourth.
In simple words: If both the mass and the acceleration (which influences the velocity) of an object are halved, its momentum will be reduced to one-fourth of its original value.

🎯 Exam Tip: Understanding the direct relationship between momentum, mass, and velocity is key. Halving both mass and the rate of velocity change (acceleration) results in a proportional reduction in momentum by a factor of four.

 

Question 2. (iv) A moving stone filled truck collides with a moving car coming from opposite direction. Why is it observed that only the car is pushed backward?
Answer:
A stone filled truck has more mass than a car hence it has more momentum thus it is observed that only the car is pushed backward
In simple words: Due to its much larger mass, a stone-filled truck has significantly more momentum than a car. When they collide, the truck's greater momentum allows it to continue moving forward, while the car is pushed backward due to the conservation of momentum.

🎯 Exam Tip: Emphasize that in a collision, total momentum is conserved. The truck's larger mass (and thus momentum) allows it to dominate the interaction, causing a more significant change in the car's motion.

 

Question 2. (v) What will happen to the force if the jet engines do not produce enough thrust to push the aeroplane in mid-air?
Answer:
If the jet engines do not produce enough thrust to push the aeroplane in mid-air the force of aeroplane and resistance of air will become balanced thus putting the aeroplane at rest which will result in a crash.
In simple words: If jet engines fail to generate sufficient thrust, the forward force becomes balanced by air resistance and the airplane's weight, causing it to lose speed, descend, and ultimately crash as it can no longer counteract gravity and drag.

🎯 Exam Tip: Relate this scenario to Newton's First Law and the concept of unbalanced forces. Insufficient thrust means the forces are either balanced or gravity/drag are dominant, preventing flight.