Samacheer Kalvi Class 11 Physics Solutions Chapter 3 Laws of Motion

Part - I:

I. Multiple choice questions:

 

Question 1. Sudden left turn in the curved road, passengers are pushed towards the right due to
(a) inertia of direction
(b) inertia of motion
(c) inertia of rest
(d) absence of inertia
Answer: (a) inertia of direction
In simple words: When a car takes a sudden left turn, your body tries to keep moving in its original direction. This resistance to changing direction makes you feel pushed to the right.

🎯 Exam Tip: Remember that inertia is the tendency of an object to resist changes in its state of motion or rest. In this case, it's about resisting a change in direction.

 

Question 2. An object of mass m held against a vertical wall by applying horizontal force F as shown in the figure. The minimum value of the force F is __________ (IIT JEE 1994)
(a) Less than mg
(b) Equal to mg
(c) Greater than mg
(d) Cannot determine
Answer: (c) Greater than mg
In simple words: To hold an object against a vertical wall, the horizontal force F must be strong enough to create a friction force that is greater than or equal to the object's weight (mg). This ensures the object does not slide down.

🎯 Exam Tip: For an object to remain stationary against a vertical wall, the static friction force must balance the gravitational force. This friction force depends on the normal force (F in this case) and the coefficient of static friction.

 

Question 3. A vehicle is moving along the positive x-direction, if a sudden brake is applied, then __________ .
(a) frictional force acting on the vehicle is along negative x-direction
(b) frictional force acting on the vehicle is along the positive x-direction
(c) no frictional force acts on the vehicle
(d) frictional force acts in a downward direction
Answer: (a) frictional force acting on the vehicle is along negative x-direction
In simple words: When you hit the brakes, the wheels try to stop, but the vehicle wants to keep moving forward. So, the friction from the road pushes against the direction of motion, helping the vehicle slow down.

🎯 Exam Tip: Frictional force always acts in the opposite direction to the relative motion or the tendency of motion between surfaces in contact. When braking, the tendency of motion is forward, so friction acts backward.

 

Question 4. A book is at rest on the table which exerts a normal force on the book. If this force is considered as a reaction force, what is the action force according to Newton's third law?
(a) Gravitational force exerted by Earth on the book
(b) Gravitational force exerted by the book on Earth
(c) Normal force exerted by the book on the table
(d) None of the above
Answer: (c) Normal force exerted by the book on the table
In simple words: Newton's third law states that for every action, there is an equal and opposite reaction. If the table pushes up on the book (reaction), then the book must push down on the table (action).

🎯 Exam Tip: Action-reaction pairs always act on different objects. The normal force exerted *by* the table *on* the book has its reaction as the normal force exerted *by* the book *on* the table.

 

Question 5. Two masses \( m_1 \) and \( m_2 \) are experiencing the same force where \( m_1 < m_2 \). The ratio of their acceleration \( a_1/a_2 \) is
(a) 1
(b) less than 1
(c) greater than 1
(d) all the three cases
Answer: (c) greater than 1
In simple words: If two objects feel the same push, the lighter one will speed up more than the heavier one. Since \( m_1 \) is lighter than \( m_2 \), \( a_1 \) will be bigger than \( a_2 \), making the ratio greater than one.

🎯 Exam Tip: Remember Newton's second law, \( F = ma \), which means \( a = F/m \). For a constant force \( F \), acceleration \( a \) is inversely proportional to mass \( m \). Thus, a smaller mass will have a larger acceleration.

 

Question 6. Choose an appropriate free body diagram for the particle experiencing net acceleration along the negative y-direction. (Each arrow mark represents the system).
(a) Diagram with two forces pulling up and left, and one pulling down and right.
(b) Diagram with two forces pulling up and right, and one pulling down and left.
(c) Diagram with one force pulling straight up and one pulling straight down, but the downward force is larger.
(d) Diagram with two forces pulling down and right, and one pulling up and left.
Answer: (c) Diagram with one force pulling straight up and one pulling straight down, but the downward force is larger.
In simple words: If something is speeding up downwards, it means the total push or pull downwards is stronger than any upward push or pull. The diagram showing a larger downward force correctly represents this.

🎯 Exam Tip: Net acceleration in a specific direction means the sum of all forces (vector sum) has a component in that direction. For negative y-direction, the net force must point downwards.

 

Question 7. A particle of mass m sliding on the smooth double inclined plane (shown in the figure) will experience
(a) greater acceleration along the path AB
(b) greater acceleration along the path AC
(c) same acceleration in both the paths
(d) no acceleration in both the paths
Answer: (b) greater acceleration along the path AC
In simple words: When an object slides down a smooth slope, its acceleration depends on the steepness of the slope. A steeper slope (like path AC with 45 degrees) will cause faster acceleration than a less steep one (like path AB with 30 degrees).

🎯 Exam Tip: For an object sliding down a smooth inclined plane, the acceleration is given by \( a = g \sin \theta \). A larger angle \( \theta \) results in a larger acceleration.

 

Question 8. Two blocks of masses m and 2m are placed on a smooth horizontal surface as shown. In the first case, only a force \( F_1 \) is applied from the left. Later only a force \( F_2 \) is applied from the right. If the force acting at the interface of the two blocks in the two cases is the same, then \( F_1 : F_2 \) is
(a) 1:1
(b) 1:2
(c) 2:1
(d) 1:3
Answer: (c) 2:1
In simple words: To get the same push between two blocks, you need different total forces depending on which side you push from. When pushing the 'm' block from the left towards '2m', the force \( F_1 \) has to move both blocks. When pushing the '2m' block from the right towards 'm', the force \( F_2 \) also moves both blocks, but the internal force changes the ratio. The math shows the ratio is 2:1.

🎯 Exam Tip: Solve such problems by drawing free body diagrams for each block in both scenarios and applying Newton's second law, \( F=ma \), to relate the applied forces to the acceleration and the contact forces.

 

Question 9. Force acting on the particle moving with constant speed is
(a) always zero
(b) need not be zero
(c) always non zero
(d) cannot be concluded
Answer: (b) need not be zero
In simple words: If an object moves at a constant speed, it means its acceleration is zero. According to Newton's second law, zero acceleration implies zero net force. However, there might be forces acting on the object that cancel each other out, like friction and an applied force.

🎯 Exam Tip: Constant speed does not always mean no forces are acting. It means the *net* force is zero. For example, a car moving at a constant speed on a highway still has engine force and air resistance acting on it, but they are balanced.

 

Question 10. An object of mass m begins to move on the plane inclined at an angle \( \theta \). The coefficient of static friction of inclined surface is \( \mu_s \). The maximum static friction experienced by the mass is
(b) \( \mu_s \) mg
(c) \( \mu_s \) mg sin \( \theta \)
(d) \( \mu_s \) mg cos \( \theta \)
Answer: (d) \( \mu_s \) mg cos \( \theta \)
In simple words: The friction that stops an object from sliding on a slope depends on how hard the slope pushes back on the object, which is called the normal force. On a slope, the normal force is not the full weight (mg) but a part of it, specifically \( mg \cos \theta \). The maximum static friction is then this normal force multiplied by the coefficient of static friction.

🎯 Exam Tip: Always resolve forces into components parallel and perpendicular to the inclined plane. The normal force is perpendicular to the surface and is typically balanced by the component of gravity perpendicular to the surface, \( mg \cos \theta \).

 

Question 11. When the object is moving at a constant velocity on the rough surface
(a) net force on the object is zero
(b) no force acts on the object
(c) only external force acts on the object
(d) only kinetic friction acts on the object
Answer: (a) net force on the object is zero
In simple words: If something is moving at a steady speed without getting faster or slower, it means all the forces pushing and pulling on it are perfectly balanced. The total force adds up to nothing.

🎯 Exam Tip: Constant velocity (both speed and direction) is a key indicator of zero net acceleration. By Newton's second law (\( F = ma \)), if \( a=0 \), then the net force \( F \) must also be zero.

 

Question 12. When an object is at rest on the inclined rough surface
(a) static and kinetic frictions acting on the object is zero
(b) static friction is zero but kinetic friction is not zero.
(c) static friction is not zero and kinetic friction is zero.
(d) static and kinetic frictions are not zero.
Answer: (c) static friction is not zero and kinetic friction is zero.
In simple words: When an object is sitting still on a rough slope, there's a static friction force holding it in place, stopping it from sliding. Since it's not moving, there's no kinetic friction acting on it. Static friction works to prevent movement.

🎯 Exam Tip: Static friction acts when an object is at rest but there's a force trying to make it move. Kinetic friction acts only when an object is in motion. An object at rest cannot experience kinetic friction.

 

Question 13. The centrifugal force appears to exist
(a) only in inertial frames
(b) only in rotating frames
(c) in an accelerated frame
(d) non-inertial frames
Answer: (b) only in rotating frames
In simple words: Centrifugal force is a 'fake' force that you feel when you are in something that is spinning or turning, like in a rotating car or amusement park ride. It makes you feel pushed outwards, but it's not a real force in the same way gravity is.

🎯 Exam Tip: Centrifugal force is a pseudo force or inertial force, meaning it appears only in non-inertial (accelerating) reference frames, specifically rotating ones, to explain observed phenomena from that frame's perspective.

 

Question 14. Choose the correct statement from the following.
(a) Centrifugal and centripetal forces are action-reaction pairs.
(b) Centripetal forces is a natural force.
(c) Centrifugal force arises from the gravitational force.
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion.
Answer: (d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion.
In simple words: Centripetal force is a real force that pulls an object towards the center to make it move in a circle. Centrifugal force is what you feel pushing you outwards when you're in a spinning motion; it's an apparent force, not a real one, and it helps explain motion from within that spinning system.

🎯 Exam Tip: Understand that centripetal force is a *real* force (e.g., tension, gravity, friction) causing circular motion, while centrifugal force is a *pseudo* (fictitious) force observed *only* in the rotating reference frame. They are not an action-reaction pair as they don't act on different bodies as required by Newton's third law.

 

Question 15. If a person moving from pole to equator, the centrifugal force acting on him
(a) increase
(b) decreases
(c) remains the same
(d) increase and then decreases
Answer: (a) increase
In simple words: The Earth spins, causing a centrifugal force that pushes objects away from its axis. This outward push is strongest at the equator, where you are furthest from the Earth's axis of rotation, and weakest at the poles. So, moving towards the equator means the outward force increases.

🎯 Exam Tip: Centrifugal force depends on the distance from the axis of rotation. At the poles, this distance is zero, so centrifugal force is zero. At the equator, the distance is maximum (Earth's radius), so centrifugal force is maximum.

 

II. Short Answer Questions:

 

Question 1. Explain the concept of Inertia. Write two examples each for Inertia of motion, inertia of rest and inertia of direction.
Answer: Inertia is the property of an object that makes it resist any changes to its state of motion or rest. This means an object will stay still if it's still, or keep moving at the same speed and direction if it's moving, unless an outside force acts on it. There are three kinds of inertia:
1. Inertia of rest: This is when an object cannot change its state of rest by itself. It wants to stay still.
• Example: When a bus that was stopped suddenly moves, passengers feel a push backward because their bodies try to stay in their resting position.
• Example: A book on a table will not move unless someone pushes or pulls it.
2. Inertia of motion: This is when an object cannot change its state of moving at a constant speed by itself. It wants to keep moving.
• Example: When a moving bus suddenly brakes, passengers are thrown forward because their bodies want to continue moving.
• Example: A runner continues to move forward for a short distance even after crossing the finish line.
3. Inertia of direction: This is when an object cannot change its direction of motion by itself. It wants to keep going straight.
• Example: If you swing a stone on a string and the string breaks, the stone flies off in a straight line, not in a circle.
• Example: When a bus turns left, passengers are thrown to the right as their bodies try to continue moving in a straight line.
In simple words: Inertia is an object's natural tendency to keep doing what it's already doing—staying still or moving steadily. We see it when we get pushed back in a starting bus (inertia of rest), thrown forward in a braking bus (inertia of motion), or pushed sideways in a turning bus (inertia of direction).

🎯 Exam Tip: Clearly define inertia first. Then, for each type, give a concise definition and two distinct, easy-to-understand examples that illustrate the specific type of inertia.

 

Question 2. State Newton's second law:
Answer: Newton's second law states that the force acting on an object is directly related to how fast its momentum changes. We can write this as:
\( F = \frac { dp }{ dt } \)
If an object has mass \( m \) and its velocity is \( v \), its momentum \( p \) is \( mv \). So, the equation can also be written as:
\( F = \frac { d }{ dt } (mv) \)
If the mass \( m \) of the object does not change, then:
\( F = m \frac { dv }{ dt } \)
Since \( \frac { dv }{ dt } \) is acceleration \( a \), this simplifies to the well-known formula:
\( F = ma \). This law shows that a greater force causes a greater change in an object's motion.
In simple words: Newton's second law says that when you push something, how much it speeds up depends on how hard you push and how heavy it is. A harder push on a lighter object makes it speed up a lot.

🎯 Exam Tip: Start with the momentum definition of force (\( F = dp/dt \)) as it's more fundamental, then derive \( F=ma \) assuming constant mass. This shows a deeper understanding.

 

Question 3. Define one newton:
Answer: One newton is defined as the amount of force needed to make an object with a mass of 1 kilogram speed up by 1 meter per second squared in the direction the force is applied. It is the standard unit of force in the International System of Units.
In simple words: One newton is the force that makes a 1 kg object go faster by 1 meter per second, every second.

🎯 Exam Tip: When defining units, always include the magnitude (1), the quantity (kilogram and meter per second squared), and the direction (in the direction of force) for a complete definition.

 

Question 4. Show that impulse is the change of momentum.
Answer: When a very strong force acts on an object for a very short time, this force is called an impulsive force, or simply impulse. According to Newton's second law, the change in momentum \( dp \) over a tiny time \( dt \) is equal to the force \( F \) applied:
\( dp = F \cdot dt \)
To find the total change in momentum over a longer time interval (from \( t_1 \) to \( t_2 \)), we can integrate this equation:
\[ \int_{p_i}^{p_f} dp = \int_{t_1}^{t_2} F dt \] The left side gives the total change in momentum, \( P_f - P_i \), where \( P_i \) is the initial momentum and \( P_f \) is the final momentum. The right side is defined as impulse, \( J \).
So, we get:
\( P_f - P_i = J \)
\( \Delta P = J \)
This shows that the impulse applied to an object is equal to the change in its momentum. This is a crucial concept in understanding collisions and impacts.
In simple words: When you hit something with a quick, strong push (that's impulse), you change how much it's moving (that's change in momentum). The bigger the push, the bigger the change in its motion.

🎯 Exam Tip: Clearly state Newton's second law in terms of momentum. Use the integral form to show the relationship between total impulse and total change in momentum over a time interval.

 

Question 5. Using free body diagram, show that it is easy to pull an object than to push it.
Answer: It is generally easier to pull an object than to push it, especially on a rough surface. This can be understood by looking at the forces involved, particularly the normal force and friction.

Case 1: Pushing an object at an angle \( \theta \)
When you push an object with a force \( F \) at an angle \( \theta \) downwards (or into the surface), this force can be split into two parts: one horizontal and one vertical. The horizontal part, \( F \sin \theta \), helps move the object forward. The vertical part, \( F \cos \theta \), pushes the object further into the ground. This increases the total downward force on the ground.
The total downward force on the ground becomes \( mg + F \cos \theta \). This is equal to the normal force \( N_{\text{push}} \).
\( N_{\text{push}} = mg + F \cos \theta \)
The maximum static friction, \( F_{s(\text{max})} \), which opposes motion, is proportional to the normal force:
\( F_{s(\text{max})} = \mu_s N_{\text{push}} = \mu_s (mg + F \cos \theta) \).

Case 2: Pulling an object at an angle \( \theta \)
When you pull an object with a force \( F \) at an angle \( \theta \) upwards (or away from the surface), this force also splits into two parts. The horizontal part, \( F \sin \theta \), helps move the object forward. The vertical part, \( F \cos \theta \), lifts the object slightly, reducing the total downward force on the ground.
The total downward force on the ground becomes \( mg - F \cos \theta \). This is equal to the normal force \( N_{\text{pull}} \).
\( N_{\text{pull}} = mg - F \cos \theta \)
The maximum static friction, \( F_{s(\text{max})} \), is then:
\( F_{s(\text{max})} = \mu_s N_{\text{pull}} = \mu_s (mg - F \cos \theta) \).

Conclusion:
Comparing the two cases, we see that the normal force \( N_{\text{pull}} \) (when pulling) is less than the normal force \( N_{\text{push}} \) (when pushing). Since the maximum static friction is directly proportional to the normal force, the friction to overcome is smaller when pulling. This means you need less force to start moving an object by pulling it at an angle than by pushing it at the same angle. This principle helps in designing luggage with handles.
In simple words: When you push something, you also push it down into the floor, making it harder to slide. When you pull something upwards, you slightly lift it off the floor, which makes it easier for it to slide. So, pulling needs less effort.

🎯 Exam Tip: Clearly draw and label free body diagrams for both pushing and pulling scenarios. Focus on how the vertical component of the applied force affects the normal force, and consequently, the friction force.

 

Question 6. Explain the various types of friction suggest a few methods to reduce friction.
Answer: There are mainly two types of friction:
(1) Static Friction: This is the force that stops an object from starting to move when a force is applied. It acts when an object is at rest. The static frictional force, \( f_s \), can change and goes from zero up to a maximum value. This maximum static friction is given by \( f_{s(\text{max})} = \mu_s N \), where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. For instance, it's harder to get a heavy box moving than to keep it moving.
(2) Kinetic Friction: This is the force that resists the motion of an object once it is already sliding. It's also known as sliding friction or dynamic friction. Kinetic friction is usually constant for a given pair of surfaces and is expressed as \( f_k = \mu_k N \), where \( \mu_k \) is the coefficient of kinetic friction. Kinetic friction is always less than or equal to maximum static friction.

Methods to reduce friction:
Friction can be reduced in several ways to make things move more easily or to prevent wear and tear:
• By using lubricants: Applying oil or grease between surfaces creates a thin layer that reduces direct contact, thereby reducing friction. For example, oiling bicycle chains makes them run smoother.
• By using Ball bearings: These are small metal balls or rollers placed between moving parts to replace sliding friction with rolling friction, which is much lower. This is why car wheels and skateboards have bearings.
• By polishing surfaces: Making surfaces smoother reduces the interlocking of tiny bumps and valleys, leading to less friction.
• By streamlining: Shaping objects to reduce air or water resistance (a type of fluid friction) helps them move more efficiently. For example, the design of sports cars and airplanes. This minimizes the energy lost to friction.
In simple words: Friction is the force that makes things hard to move. There's 'static friction' when something is still, and 'kinetic friction' when it's sliding. We can make things move easier by oiling them, using tiny rolling balls (bearings), making surfaces smooth, or shaping objects to cut through air or water better.

🎯 Exam Tip: Define each type of friction clearly, distinguishing when they act. For reduction methods, list at least three and briefly explain how each works, providing a real-world example if possible.

 

Question 7. What is the meaning of "Pseudo force"?
Answer: A "pseudo force," also known as an inertial force or fictitious force, is a force that appears to act on objects when motion is observed from a non-inertial (accelerating) reference frame. These forces do not come from actual interactions with other objects (like gravity or electromagnetism) but arise because the observer's frame of reference is accelerating. For example, centrifugal force is a pseudo force that appears in rotating frames of reference, pushing objects outwards. Pseudo forces are introduced to make Newton's laws of motion seem to work in accelerating frames. For example, if you are in a car that suddenly stops, you feel a forward jerk, which can be thought of as a pseudo force.
In simple words: A pseudo force is a pretend force that you feel when you are inside something that is speeding up, slowing down, or turning. It's not a real push or pull, but it helps explain why things move the way they do from your point of view inside that moving object.

🎯 Exam Tip: Emphasize that pseudo forces are not "real" interaction forces and only appear in non-inertial (accelerating) reference frames, such as rotating or linearly accelerating frames.

 

Question 8. State of the empirical laws of static and kinetic friction.
Answer: The empirical laws of static and kinetic friction describe how these forces behave. Here is a comparison:

These laws help us predict how objects will behave when they interact with surfaces, which is important for engineering and everyday life. For example, knowing these laws helps in designing braking systems.
In simple words: Static friction tries to stop things from starting to move, and its strength can change up to a limit. Kinetic friction acts when things are already sliding, and its strength is usually steady. Both depend on the surfaces touching but not on how big the contact area is.

🎯 Exam Tip: When asked to state empirical laws, presenting them in a comparative table format is highly effective for clarity and ensures all key differences and similarities are covered for full marks.

 

Question 9. State Newton's third law.
Answer: Newton's third law states that for every action, there is an equal and opposite reaction. This means that whenever one object pushes or pulls on a second object, the second object simultaneously pushes or pulls back on the first object with the same amount of force but in the opposite direction. These action-reaction forces always occur in pairs and act on different objects. This law is fundamental to understanding how objects interact in the universe.
In simple words: Newton's third law means that if you push something, it pushes back on you just as hard, but in the opposite direction. It's like when you jump, your feet push the ground down, and the ground pushes your feet up.

🎯 Exam Tip: The two crucial points to remember for Newton's third law are: 1) the forces are equal in magnitude and opposite in direction, and 2) they always act on *different* objects.

 

Question 10. What are inertial frames?
Answer: An inertial frame of reference is a special kind of reference frame where Newton's first law of motion (the law of inertia) holds true. In simpler terms, in an inertial frame, an object at rest will stay at rest, and an object in motion will continue to move at a constant velocity (constant speed in a straight line), unless it is acted upon by an external force. This also means that in an inertial frame, an object with no net force acting on it will have zero acceleration. A classic example is a laboratory on Earth, which is often considered an inertial frame for many experiments. A frame moving at a constant velocity relative to an inertial frame is also an inertial frame.
In simple words: An inertial frame is like a steady, calm place where things stay still or move in a straight line at a steady speed unless something pushes them. It's a special viewpoint where Newton's laws of motion work simply.

🎯 Exam Tip: The defining characteristic of an inertial frame is that Newton's first law (and thus Newton's second law in its simplest form, \( F=ma \)) is valid in it. It is a non-accelerating frame of reference.

 

Question 11. Under what condition will a car skid on a leveled circular road?
Answer: A car will skid on a leveled circular road if the static friction between its tires and the road is not strong enough to provide the necessary centripetal force to keep it moving in a circle. The static friction provides the inward force needed for the car to turn. If the car's speed becomes too high, or the radius of the turn is too small, the required centripetal force can exceed the maximum available static friction. The condition for skidding is when the coefficient of static friction \( \mu_s \) is less than the ratio of the square of the velocity \( v \) to the product of the radius \( r \) and acceleration due to gravity \( g \):
\( \mu_s < \frac{v^2}{rg} \)
This means that the car will skid if it tries to turn too sharply or too quickly for the given road conditions. For safer turns, roads are often banked, which means they are tilted.
In simple words: A car skids on a circular road when its tires can't grip the road enough to make it turn. This happens if the car goes too fast or tries to make too sharp a turn, and the friction needed to keep it turning is more than the tires can provide.

🎯 Exam Tip: Understand that static friction provides the necessary centripetal force for a car to turn on a level road. Skidding occurs when the demand for centripetal force exceeds the maximum static friction available.

 

III. Long Answer Questions:

 

Question 1. Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it.
Answer:
Law of Conservation of Linear Momentum:
If no external forces act on a system of objects, then the total linear momentum of that system remains constant. In simpler terms, the total momentum before an event (like a collision or an explosion) is equal to the total momentum after the event.

Proof from Newton's Third Law:
Consider two particles, 1 and 2, interacting with each other. According to Newton's third law, the force exerted by particle 1 on particle 2 (let's call it \( \vec{F}_{12} \)) is equal in magnitude and opposite in direction to the force exerted by particle 2 on particle 1 (let's call it \( \vec{F}_{21} \)).
\( \vec{F}_{12} = - \vec{F}_{21} \)
According to Newton's second law, force is the rate of change of momentum. So:
\( \vec{F}_{12} = \frac{d\vec{p}_1}{dt} \)
and
\( \vec{F}_{21} = \frac{d\vec{p}_2}{dt} \)
Substituting these into the third law equation:
\( \frac{d\vec{p}_1}{dt} = - \frac{d\vec{p}_2}{dt} \)

This means:
\( \frac{d\vec{p}_1}{dt} + \frac{d\vec{p}_2}{dt} = 0 \)

We can write this as:
\( \frac{d}{dt} (\vec{p}_1 + \vec{p}_2) = 0 \)

If the rate of change of the total momentum \( (\vec{p}_1 + \vec{p}_2) \) is zero, it means the total momentum is constant:
\( \vec{p}_1 + \vec{p}_2 = \text{constant vector} \)
So, the total linear momentum of the two-particle system is conserved, provided no external forces are acting. This principle is fundamental in many areas of physics, from celestial mechanics to particle physics.

Recoil velocity of a gun:
Let's consider a gun and a bullet as a system. Before firing, both are at rest, so the total initial momentum of the system is zero.
Initial momentum \( P_{\text{initial}} = (M_{\text{gun}} + m_{\text{bullet}}) \cdot 0 = 0 \)
After firing, the bullet moves forward with a velocity \( v_b \), and the gun recoils backward with a velocity \( v_g \).
Final momentum \( P_{\text{final}} = M_{\text{gun}} v_g + m_{\text{bullet}} v_b \)
According to the law of conservation of linear momentum:
\( P_{\text{initial}} = P_{\text{final}} \)
\( 0 = M_{\text{gun}} v_g + m_{\text{bullet}} v_b \)
This can be rearranged to find the recoil velocity of the gun:
\( M_{\text{gun}} v_g = - m_{\text{bullet}} v_b \)

\( v_g = - \frac{m_{\text{bullet}} v_b}{M_{\text{gun}}} \)
The negative sign indicates that the gun recoils in the opposite direction to the bullet's motion. This shows how the gun's mass affects its recoil, with heavier guns having less recoil for the same bullet momentum.
In simple words: If nothing from outside pushes or pulls a group of objects, their total movement strength (momentum) stays the same, even if they hit each other or explode. For a gun, when the bullet shoots forward, the gun must kick backward with the same "push strength" to keep the total movement equal, making the gun recoil.

🎯 Exam Tip: For the proof, clearly state Newton's third law and express forces as rates of change of momentum. For the gun recoil, remember the system's initial momentum is zero and final momentum must also be zero, and pay attention to the direction (negative sign) of recoil.

 

Question 2. What are concurrent forces? State Lamis theorem.
Answer:
Concurrent forces: A group of forces is called concurrent if their lines of action all meet at a single point. If these forces also lie in the same flat plane, they are called coplanar forces. Understanding concurrent forces helps in analyzing the equilibrium of objects where multiple forces act at one spot.
Lami's theorem: This theorem states that if three concurrent and coplanar forces are in balance (equilibrium), then the size of each force is directly proportional to the sine of the angle between the other two forces. This theorem is useful for solving force problems in static systems, where objects are not moving.
In simple words: Concurrent forces meet at one point. Lami's theorem helps us find unknown forces when three forces are balanced and meet at a single point.

🎯 Exam Tip: Remember to clearly define concurrent forces and Lami's theorem. When applying Lami's theorem, ensure the forces are both concurrent and coplanar, and the system is in equilibrium.

 

Question 3. Explain the motion of blocks connected by a string in 1) vertical motion 2) horizontal motion.
Answer: When blocks are connected by strings and a force is applied, tension (T) develops in the string, which influences the acceleration. We will look at both vertical and horizontal motion.
Case 1: Vertical Motion of Connected Bodies
Consider two blocks, mass \( m_1 \) and mass \( m_2 \), where \( m_1 > m_2 \), connected by an unbreakable string over a smooth pulley. When the system is released, \( m_2 \) moves up vertically, and \( m_1 \) moves down vertically, both with the same acceleration \( a \). The gravitational force \( m_1g \) on \( m_1 \) helps lift \( m_2 \).
Applying Newton's Second Law:
For \( m_2 \) (upward motion): \( T - m_2g = m_2a \)
For \( m_1 \) (downward motion): \( m_1g - T = m_1a \)
Adding these two equations gives the acceleration:
\( (m_1 - m_2)g = (m_1 + m_2)a \)
\( \implies a = \frac{m_1 - m_2}{m_1 + m_2} g \)
If \( m_1 = m_2 \), then \( a = 0 \), meaning the system remains at rest with no acceleration. The tension (T) in the string can be found by substituting the value of 'a' into either equation.
\( T = m_2g + m_2a = m_2g + m_2 \left( \frac{m_1 - m_2}{m_1 + m_2} g \right) \)
\( \implies T = m_2g \left( 1 + \frac{m_1 - m_2}{m_1 + m_2} \right) = m_2g \left( \frac{m_1 + m_2 + m_1 - m_2}{m_1 + m_2} \right) \)
\( \implies T = m_2g \left( \frac{2m_1}{m_1 + m_2} \right) = \frac{2m_1m_2}{m_1 + m_2} g \)
This setup is often called an Atwood machine, and it demonstrates how forces and accelerations are distributed in a simple pulley system.
Case 2: Horizontal Motion of Connected Masses
Consider mass \( m_2 \) resting on a horizontal surface (like a table) and connected by a string over a small pulley to mass \( m_1 \), which hangs freely. Assume no friction on the surface. When released, \( m_1 \) moves downward with acceleration \( a \), and \( m_2 \) moves horizontally with the same acceleration \( a \).
The forces acting on \( m_2 \) are:
1. Downward gravitational force (\( m_2g \))
2. Upward normal force (N) from the surface
3. Horizontal tension (T) from the string
Applying Newton's Second Law:
For \( m_1 \) (downward motion): \( m_1g - T = m_1a \) ...(1)
For \( m_2 \) (horizontal motion): \( T = m_2a \) ...(2)
Substitute (2) into (1):
\( m_1g - m_2a = m_1a \)
\( \implies m_1g = (m_1 + m_2)a \)
\( \implies a = \frac{m_1}{m_1 + m_2} g \) ...(4)
The tension can be found by substituting \( a \) back into (2):
\( T = m_2 \left( \frac{m_1}{m_1 + m_2} g \right) = \frac{m_1m_2}{m_1 + m_2} g \) ...(5)
Comparing the two cases, the tension in the string for horizontal motion is half of the tension for vertical motion, assuming the same masses and strings. This highlights how the orientation of forces changes the system's dynamics.
In simple words: When blocks are connected by a string over a pulley, their movement depends on if they are moving up and down (vertical) or one is moving sideways (horizontal). The forces like gravity and string tension make them accelerate, and we can use physics rules to figure out how fast they move and how much the string pulls.

🎯 Exam Tip: When dealing with connected blocks, always draw free-body diagrams for each mass to correctly identify all forces (gravity, tension, normal force) and their directions. This helps apply Newton's laws accurately.

 

Question 4. Briefly explain the origin of friction show that in an inclined plane, angle of friction is equal to angle of repose.
Answer:
Friction is a force that resists the movement of an object over a surface. It always works against the relative motion between two surfaces that are touching. When you try to push an object, friction tries to stop it. This happens because the surfaces are not perfectly smooth; they have tiny bumps and dips that interlock and resist sliding.
Angle of repose equals angle of friction:
The angle of repose is the smallest angle an inclined plane makes with the horizontal when an object placed on it just starts to slide down.
Consider a body of mass \( m \) on an inclined plane with an angle of inclination \( \theta \) to the horizontal.
The gravitational force \( mg \) on the body can be split into two parts:
1. \( mg \sin \theta \) acting parallel to the inclined plane, trying to pull the object down.
2. \( mg \cos \theta \) acting perpendicular to the inclined plane, pressing the object against the surface.
The normal force \( N \) from the surface balances the perpendicular component:
\( N = mg \cos \theta \) ...(1)
When the object is just about to slide, the static friction force \( f_s(\text{max}) \) reaches its maximum value. This maximum static friction opposes the motion down the plane.
At the point of sliding, the downward force equals the maximum static friction:
\( f_s(\text{max}) = mg \sin \theta \) ...(2)
By definition, the coefficient of static friction \( \mu_s \) is \( \frac{f_s(\text{max})}{N} \).
Using equations (1) and (2):
\( \tan \theta = \frac{mg \sin \theta}{mg \cos \theta} = \frac{f_s(\text{max})}{N} \)
Since \( \mu_s = \frac{f_s(\text{max})}{N} \), we have \( \tan \theta = \mu_s \).
This means the angle of repose \( \theta \) is equal to the angle of friction, as \( \tan \theta \) is the coefficient of static friction. So, the angle at which an object starts to slide is directly related to how sticky the surfaces are.
In simple words: Friction is a force that slows things down when they rub together. The "angle of repose" is the steepest angle a slope can be before an object on it starts to slide. We show that this angle is the same as the "angle of friction," which describes how much friction there is.

🎯 Exam Tip: When proving concepts like the angle of repose, clearly define the terms, draw a labeled diagram showing all forces, and apply Newton's second law for equilibrium in both perpendicular and parallel directions to the inclined plane.

 

Question 5. State Newton's three laws and discuss their significance.
Answer: Newton's three laws of motion are fundamental rules that explain how objects move and interact. They are:
1. **First Law (Law of Inertia):** Every object stays in its state of rest or continues to move with steady speed in a straight line, unless an outside force makes it change. This means things won't start or stop moving, or change direction, without something pushing or pulling them. It helps us understand that objects have an inherent "laziness" to change their state of motion.
2. **Second Law (Law of Force and Acceleration):** The force acting on an object is equal to the rate at which its momentum changes. In simpler terms, if you apply a force to an object, it will accelerate (speed up, slow down, or change direction) in the direction of the force. The size of this acceleration depends on how big the force is and how heavy the object is (mass). Mathematically, this is often written as \( \vec{F} = m \vec{a} \), where \( \vec{F} \) is force, \( m \) is mass, and \( \vec{a} \) is acceleration. This law is powerful because it allows us to calculate how an object will move if we know the forces acting on it.
3. **Third Law (Law of Action-Reaction):** For every action, there is an equal and opposite reaction. This means that whenever one object pushes or pulls on another, the second object pushes or pulls back on the first with the same strength but in the opposite direction. For example, when you walk, your feet push backward on the ground, and the ground pushes forward on your feet, moving you ahead. This law explains why forces always come in pairs and cannot exist alone.
**Significance:** These laws are crucial because they form the basis of classical mechanics, allowing us to predict and explain the motion of most everyday objects, from thrown balls to planets orbiting the sun. They help engineers design safe structures and vehicles, and scientists understand the physical world.
In simple words: Newton's three laws tell us how things move. The first says objects keep doing what they're doing unless a force pushes them. The second says force makes things speed up or change direction, and this depends on how heavy they are. The third says for every push, there's an equal and opposite push back. These laws help us understand almost all movement around us.

🎯 Exam Tip: Clearly state each of Newton's laws. For the second law, remember to include both the momentum form (\( \vec{F} = \frac{d\vec{p}}{dt} \)) and the more common \( \vec{F} = m \vec{a} \) form. Emphasize the vector nature of force and acceleration.

 

Question 6. Captain the similarities and differences of between centripetal and centrifugal forces.
Answer: Here are the similarities and differences between centripetal and centrifugal forces:

In simple words: Centripetal force is a real push or pull that makes something move in a circle, always pulling it towards the center. Centrifugal force is an "imaginary" force that you feel pushing you outwards when you are in a spinning system, but it's really just your body's inertia wanting to go straight.

🎯 Exam Tip: Clearly distinguish between centripetal (real, inward, inertial/non-inertial frames) and centrifugal (fictitious, outward, only non-inertial frames). Using a table format helps organize the points effectively.

 

Question 7. Briefly explain centrifugal force with suitable examples.
Answer: Centrifugal force is a "pseudo" or apparent force that seems to push objects outward when they are in a rotating frame of reference. It doesn't come from a direct physical interaction but appears because the observer is in an accelerating (rotating) frame.
To understand it, imagine you are whirling a stone tied to a string in a circle.
* **From an outside (inertial) view:** A person watching from the ground sees the string pulling the stone inward (centripetal force), causing it to move in a circle. If the string breaks, the stone flies off in a straight line because of its inertia.
* **From an inside (rotating) view:** Now imagine you are tiny and sitting on the stone itself, spinning along with it. From your viewpoint, the stone is not moving; it is at rest. However, you would feel a force pushing you outward, away from the center. This outward push is the centrifugal force. To explain why the stone stays "at rest" relative to you, even though the string is pulling it inward, you invent this outward centrifugal force to balance the inward pull.
**Examples:**
1. **Car Turning a Corner:** When a car takes a sharp turn, passengers feel pushed to the outside of the curve. This outward push is the centrifugal force acting on them within the car's rotating reference frame. Their inertia makes them want to continue in a straight line, but the car is turning.
2. **Rotor Ride at an Amusement Park:** In a rotor ride, you stand against a wall, and the cylinder spins. The floor drops, but you stay stuck to the wall. This happens because the wall provides an inward centripetal force, and from your perspective inside the spinning cylinder, you feel a strong centrifugal force pushing you into the wall.
This force is essential for solving problems in rotating systems by making them seem like static (non-moving) problems.
In simple words: Centrifugal force is like an imaginary push you feel outwards when you are spinning around, or riding in a car that turns fast. It's not a real push from another object, but it feels real because your body wants to keep going straight.

🎯 Exam Tip: When explaining centrifugal force, always clarify that it's a "pseudo" or "fictitious" force. Contrast it with centripetal force and provide clear examples from everyday life or physics scenarios.

 

Question 8. Briefly explain 'rolling friction'.
Answer: Rolling friction happens when a round object, like a wheel or ball, rolls over a surface. Unlike sliding friction, where surfaces rub directly against each other, in ideal rolling, the point of contact between the wheel and the surface is always instantly at rest. This means there is no actual sliding between the surfaces at that tiny point.
However, in the real world, surfaces are not perfectly rigid. When a wheel rolls, it slightly deforms the surface it's rolling on, and the wheel itself also deforms a little. This small deformation creates a tiny "dent" or indentation. The wheel has to constantly climb out of this slight dent, which requires a small force. This resistance to rolling is what we call rolling friction. It is much, much smaller than kinetic (sliding) friction, which is why it's easier to roll a heavy object on wheels than to slide it. Rolling friction also occurs because of minor energy losses due to the deformation and recovery of the materials.
In simple words: Rolling friction is the force that makes it hard for a wheel or ball to roll easily. It happens because the wheel and the ground bend a little when they touch, and the wheel always has to push itself slightly uphill out of this small dent. It's much less than sliding friction.

🎯 Exam Tip: When explaining rolling friction, emphasize that it arises from deformation of surfaces, not sliding. Highlighting that it's significantly less than sliding friction is a key point.

 

Question 9. Describe the method of measuring angle of repose.
Answer: The angle of repose is the smallest angle an inclined plane makes with the horizontal when a body placed on it just begins to slide down.
To measure the angle of repose for an object:
1. **Set up an inclined plane:** Place a flat surface, like a wooden board or a tray, on a horizontal table.
2. **Place the object:** Put the object (e.g., a block of wood, a book, or a heap of sand) whose angle of repose you want to measure onto the inclined plane.
3. **Adjust the inclination:** Slowly and carefully raise one end of the inclined plane, increasing its angle with the horizontal. Do this smoothly to avoid sudden jerks.
4. **Observe the start of sliding:** Keep raising the plane until the object just begins to slide down. This is the critical point where the maximum static friction is overcome by the component of gravity pulling the object down the slope.
5. **Measure the angle:** Once the object starts to slide, stop increasing the angle. The angle of inclination of the plane at this moment is the angle of repose (\( \theta \)). You can measure this angle using a protractor or by measuring the height and base of the inclined triangle formed. For example, if 'h' is the height of the raised end and 'L' is the length of the base, then \( \tan \theta = \frac{h}{L} \).
The forces acting on the body at this point are:
(a) Weight \( mg \) of the body acting vertically downwards.
(b) The maximum static friction force \( f_s(\text{max}) \) acting upwards along the inclined plane, resisting the motion.
(c) The normal reaction \( N \) acting perpendicular to the inclined plane.
At the point of impending motion, the component of gravity parallel to the plane equals the maximum static friction:
\( mg \sin \theta = f_s(\text{max}) \) ...(1)
And the normal force balances the component of gravity perpendicular to the plane:
\( N = mg \cos \theta \) ...(2)
Dividing equation (1) by equation (2):
\( \frac{mg \sin \theta}{mg \cos \theta} = \frac{f_s(\text{max})}{N} \)
\( \implies \tan \theta = \frac{f_s(\text{max})}{N} \)
Since the coefficient of static friction \( \mu_s = \frac{f_s(\text{max})}{N} \), it follows that \( \tan \theta = \mu_s \).
Thus, the angle of repose is equal to the angle of friction. This method helps characterize the friction between an object and a surface.
In simple words: To find the angle of repose, slowly lift one end of a flat surface with an object on it. The moment the object starts to slide, measure the angle of the slope. That angle is the angle of repose, which tells you how much friction there is between the object and the surface.

🎯 Exam Tip: When describing the experimental method, ensure to include steps for setting up, adjusting the angle slowly, and precisely identifying the point of impending motion. Clearly explain how the angle of inclination is measured.

 

Question 10. Explain the need for banking of tracks.
Answer: On a flat, circular road, a car can skid easily because the static friction has a limit. This limit depends on how rough the road surface is. To stop cars from skidding, especially outwards, roads are often "banked." This means the outer edge of the road is raised higher than the inner edge. This creates an angle of inclination, known as the banking angle. Banking provides an extra component of the normal force that acts like centripetal force, helping the car turn safely. This design helps cars maintain grip and turn without slipping, especially at higher speeds.
In simple words: Roads that turn are often tilted, with the outer edge higher. This tilt helps cars turn safely without sliding off, especially when driving fast, because it uses the road's push to keep the car on track.

🎯 Exam Tip: When explaining banking, mention both the role of friction (on flat roads) and how the normal force component (on banked roads) provides the necessary centripetal force for safe turns.

 

Question 11. Calculate the centripetal acceleration of moon towards the earth.
Answer: The moon travels around the Earth in a nearly circular path. It takes 27.3 days to complete one orbit.
The distance between the Earth's center and the moon's center (\( R_m \)) is about 60 times the radius of the Earth. If the Earth's radius is \( 6.4 \times 10^6 \text{ m} \), then:
\( R_m = 60 \times 6.4 \times 10^6 \text{ m} \)
\( R_m = 384 \times 10^6 \text{ m} \)
The angular velocity \( \omega \) is given by \( \omega = \frac{2\pi}{T} \), where \( T \) is the time period in seconds.
\( T = 27.3 \text{ days} = 27.3 \times 24 \text{ hours} \times 60 \text{ minutes} \times 60 \text{ seconds} \)
\( \omega = \frac{2\pi}{27.3 \times 24 \times 60 \times 60} \text{ rad/s} \)
The centripetal acceleration \( a_m \) of the moon is calculated as \( a_m = R_m \omega^2 \).
\( a_m = (384 \times 10^6 \text{ m}) \times \left( \frac{2\pi}{27.3 \times 24 \times 60 \times 60} \right)^2 \)
\( a_m = 0.00272 \text{ m/s}^2 \)
\( a_m = 2.72 \times 10^{-3} \text{ m/s}^2 \)
In simple words: To find how fast the moon is accelerating towards Earth, we first find its speed around Earth using how long it takes to orbit and its distance. Then we use that speed and distance to calculate the acceleration that pulls it inwards. The moon's path is not perfectly circular, but this calculation gives a good estimate.

🎯 Exam Tip: Remember to convert the time period from days to seconds when calculating angular velocity for centripetal acceleration problems. Ensure all units are consistent (SI units are preferred).

 

IV. Conceptual Questions:

 

Question 1. Why it is not possible to push a car from inside?
Answer: If a person tries to push a car from inside, the person and the car are considered a single system. According to Newton's third law, any force exerted by the person on the car from the inside is an internal force. Internal forces within a system cannot cause the system to accelerate. The car moves when a person pushes the ground backward, and the ground pushes the person (and thus the car) forward, which is an external force.
In simple words: You can't push a car from the inside because you and the car are one big unit. Your push is an "inside" push, and you need an "outside" push (like pushing the ground) to make something move.

🎯 Exam Tip: Always emphasize the concept of "internal" vs. "external" forces when explaining why a system cannot accelerate itself from within.

 

Question 2. There is a limit beyond which the polishing of a surface increases frictional resistance rather than decreasing it. Why?
Answer: Initially, polishing a surface reduces its roughness and decreases friction. However, if a surface is polished too much, the area of actual contact between the two surfaces increases significantly. When surfaces become extremely smooth, the atoms and molecules on both surfaces come very close to each other. This closeness allows strong attractive forces (like adhesive forces) to act between them, which actually increases the frictional resistance. So, there's an optimal level of smoothness for minimizing friction.
In simple words: Making a surface too smooth can sometimes make friction worse, not better. This happens because if surfaces are too perfect, their tiny particles can stick together, making it harder to slide.

🎯 Exam Tip: Highlight that while roughness usually causes friction, excessive smoothness can lead to strong intermolecular attractive forces, increasing adhesion and thus friction.

 

Question 3. Can a single isolated force exist in nature? Explain your answer?
Answer: No, a single isolated force cannot exist in nature. This is based on Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. Forces always come in pairs. If one object exerts a force on another, the second object simultaneously exerts an equal and opposite force back on the first. Therefore, it's impossible to have a force without its reaction pair.
In simple words: No, forces always come in pairs. If you push something, that something pushes back on you with the same strength. You can't have just one push without the other.

🎯 Exam Tip: Clearly link the non-existence of isolated forces directly to Newton's third law, emphasizing that forces are always interactions between two objects.

 

Question 4. Why does a parachute descend slowly?
Answer: A parachute descends slowly because it experiences a large amount of air resistance. As the parachute falls through the atmosphere, the air pushes up against its large surface area. This upward air resistance force opposes the downward gravitational force acting on the parachute and the person. The larger the surface area of the parachute, the greater the air resistance, which significantly slows down the descent. Eventually, the air resistance becomes equal to the gravitational force, leading to a constant, slow terminal velocity.
In simple words: A parachute has a big shape that catches a lot of air. This air pushes up hard against the parachute, slowing it down so the person floats gently to the ground.

🎯 Exam Tip: The key concepts here are "large surface area" leading to "significant air resistance," which opposes gravity and results in a "slow descent" or "terminal velocity."

 

Question 5. When walking on ice one should take short steps. Why?
Answer: When walking on ice, taking short steps helps prevent slipping because it allows you to maintain a more upright posture. When you take short steps, your feet press down more directly on the ice, increasing the normal force. This increased normal force helps maximize the limited static friction available on the slippery surface. If you take long strides, you exert a force at a larger angle, reducing the vertical component of force and thus the effective normal force. This reduces friction and increases the chance of slipping. Additionally, short steps keep your center of gravity more stable.
In simple words: On ice, take small steps to keep your body straight up and push straight down. This helps your shoes grip the slippery surface better and stops you from falling.

🎯 Exam Tip: Connect short steps to maintaining a larger normal force and better stability, which are crucial for maximizing friction on slippery surfaces like ice.

 

Question 6. When a person walks on a surface the frictional force exerted by the surface on the person is opposite to the direction of motion. True or false.
Answer: False. When a person walks, they push the ground backward. According to Newton's third law, the ground pushes the person forward. This forward push from the ground is the frictional force, and it acts in the *same* direction as the person's motion. So, frictional force helps you move forward.
In simple words: This is false. When you walk, friction pushes you forward, in the direction you are moving, not backward. You push the ground back, and the ground pushes you forward.

🎯 Exam Tip: Clarify that for walking, friction is a *propelling* force, acting in the direction of motion, not an opposing force. This is a common misconception.

 

Question 7. Can the coefficient of friction be more than one?
Answer: Yes, the coefficient of friction can be more than one. While it is often less than one for common surfaces, especially smooth ones, it can exceed one for certain material combinations or very rough surfaces. For example, silicone rubber on a dry, clean surface can have a static coefficient of friction greater than one. This happens when the adhesive forces between the surfaces are very strong, requiring a greater force to initiate motion than the normal force pressing them together.
In simple words: Yes, sometimes the friction number can be bigger than one. This happens with very sticky materials where it takes a lot of force to make them slide.

🎯 Exam Tip: While \( \mu < 1 \) is common, emphasize that it's not a universal rule. Strong adhesive forces or very rough surfaces can lead to \( \mu > 1 \), which is physically possible.

 

Question 8. Can we predict the direction of motion of a body from the direction of force on it?
Answer: We can predict the direction of a body's acceleration from the net force acting on it (Newton's second law: \( \vec{F}_{net} = m\vec{a} \)). However, the direction of *motion* (velocity) is not always the same as the direction of the net force. For example, a projectile thrown upwards initially moves up, but gravity (force) acts downwards. Similarly, in circular motion, the centripetal force acts towards the center, while the velocity is tangential. So, the direction of force tells us about the change in motion (acceleration), but not necessarily the instantaneous direction of motion.
In simple words: No, not always. The direction of the force tells us which way something will *speed up* or *slow down*, but not always the exact direction it is moving at that moment. Like a ball thrown up still moves up even though gravity pulls it down.

🎯 Exam Tip: Differentiate clearly between the direction of force (which dictates acceleration) and the direction of velocity (which dictates instantaneous motion). Provide simple counter-examples like projectile motion to illustrate the difference.

 

Question 9. The momentum of a system of particles is always conserved. True or false.
Answer: False. The momentum of a system of particles is conserved *only if no external forces are acting on it*. If there are external forces, the total momentum of the system will change. This is the law of conservation of linear momentum, which applies to isolated systems where the net external force is zero.
In simple words: This is false. Momentum stays the same only if nothing outside pushes or pulls the system. If outside forces act, the momentum changes.

🎯 Exam Tip: Always state the condition "in the absence of external forces" when discussing the conservation of momentum. Failing to mention this condition makes the statement incorrect.

 

V. Numerical Problems:

 

Question 1. A force of 50N acts on the object of mass 20kg shown in the figure. Calculate the acceleration of the object in x and y-direction.
Answer:Given mass \( m = 20 \text{ kg} \) and force \( F = 50 \text{ N} \). The force acts at an angle of 30 degrees to the x-axis. We need to resolve the force into its x and y components. The x-component of the force is \( F_x = F \cos \theta \). Here, \( F_x = 50 \cos 30^\circ = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \text{ N} \). The y-component of the force is \( F_y = F \sin \theta \). Here, \( F_y = 50 \sin 30^\circ = 50 \times \frac{1}{2} = 25 \text{ N} \). Now, we use Newton's second law, \( F = ma \), to find the acceleration components. Acceleration in the x-direction: \( a_x = \frac{F_x}{m} \) \( a_x = \frac{25\sqrt{3}}{20} = \frac{5\sqrt{3}}{4} \text{ m/s}^2 \). Approximately, \( a_x = 2.165 \text{ m/s}^2 \). Acceleration in the y-direction: \( a_y = \frac{F_y}{m} \) \( a_y = \frac{25}{20} = \frac{5}{4} \text{ m/s}^2 \). Approximately, \( a_y = 1.25 \text{ m/s}^2 \). The force's angle needs to be resolved into horizontal and vertical parts to find how much it pushes in each direction.In simple words: First, break the angled force into two parts: one pushing sideways (x-direction) and one pushing upwards (y-direction). Then, use each part of the force and the object's mass to find how fast it speeds up in each direction.

🎯 Exam Tip: When a force is applied at an angle, always resolve it into its perpendicular components (usually horizontal and vertical) before applying Newton's second law for calculations.

 

Question 2. A spider of mass 50 g is hanging on a string of a cobweb as shown in the figure. What is the tension in the string?
Answer:The spider is hanging still, which means it is in equilibrium. In equilibrium, the upward tension force in the string (T) must balance the downward gravitational force (weight) acting on the spider. Given mass of the spider \( m = 50 \text{ g} = 50 \times 10^{-3} \text{ kg} \). The acceleration due to gravity \( g = 9.8 \text{ m/s}^2 \). The gravitational force (weight) \( F = mg \). So, the tension \( T = mg \). \( T = (50 \times 10^{-3} \text{ kg}) \times (9.8 \text{ m/s}^2) \) \( T = 0.49 \text{ N} \). The tension in the string is equal to the weight of the spider pulling down on it.In simple words: Since the spider is not moving up or down, the pull from the string (tension) must be exactly equal to the spider's weight pulling it down. Calculate the spider's weight to find the tension.

🎯 Exam Tip: For objects in static equilibrium (at rest), the net force is zero. This means all upward forces must balance all downward forces, and all leftward forces must balance all rightward forces.

 

Question 3. What is the reading shown in spring balance?
(i)

Answer: (i) The diagram shows two 4kg masses connected by a string passing over a pulley, with spring balances on both sides. Since both masses are equal (4kg), they will exert equal and opposite forces on the spring balance. When two equal forces pull on a spring balance from opposite sides, the balance reads the magnitude of *one* of those forces. In this setup, the system is in equilibrium, and each spring balance will read the tension in the string due to the hanging mass. Therefore, each spring balance will read 4kg or \( 4 \times 9.8 = 39.2 \text{ N} \). The final answer is zero because the question might be misinterpreted as reading the *net force*, which is zero, but a spring balance reads tension. Given the context of the solution: "Forces on both sides are equal the reading in the spring balance is zero.", this implies a potential question about net force or relative motion, but a standard spring balance measures the tension applied to it. If the question implies a common setup to illustrate a single force measurement, then each balance would read the weight of its hanging mass.
(ii) In the second part, a mass of 2 kg is placed on an inclined plane at 30 degrees, connected to a spring. The spring is pulled by a force along the inclined plane. The force pulling the spring (F) is equal to the component of the mass's weight acting along the inclined plane. \( F = mg \sin \theta \) Given mass \( m = 2 \text{ kg} \), acceleration due to gravity \( g = 9.8 \text{ m/s}^2 \), and angle \( \theta = 30^\circ \). \( F = 2 \times 9.8 \times \sin 30^\circ \) \( F = 2 \times 9.8 \times \frac{1}{2} \) \( F = 9.8 \text{ N} \) The reading on the spring balance is 9.8 N. When a spring balance measures tension, it shows the magnitude of the force acting on it.In simple words: (i) When two equal weights pull on a spring balance from both sides, it reads the weight of just one of them. The "zero" answer suggests a misunderstanding of what a spring balance measures, as it shows tension. (ii) For the mass on the slope, the spring balance shows the part of the mass's weight that pulls it down the slope.

🎯 Exam Tip: A spring balance measures tension, which is a force. If masses are equal and opposite, the tension in the string is equal to the weight of one mass. For inclined planes, remember to resolve the weight component acting along the plane, which is \( mg \sin \theta \).

 

Question 4. The physics books are stacked on each other in the sequence +1 volumes 1 and 2, +2 volumes 1 and 2 on a table, a) identify the forces acting on each book and draw the free body diagram, b) Identify the forces exerted by each books on the other.
Answer: This question asks to identify forces and draw Free Body Diagrams (FBDs) for a stack of books. Let's label the books from top to bottom as A, B, C, D, and the table as T. Assume Book A (2 vol) is on top, then B (1 vol), then C (2 vol), then D (1 vol) resting on the table. Each book has a mass \( m_A, m_B, m_C, m_D \) respectively. **Forces on Book A:** 1. Downward gravitational force exerted by Earth: \( m_A g \) 2. Upward normal force exerted by Book B on A: \( N_{BA} \) (equal to \( m_A g \)) **Forces on Book B:** 1. Downward gravitational force exerted by Earth: \( m_B g \) 2. Downward normal force exerted by Book A on B: \( N_{AB} \) (equal to \( N_{BA} \) and \( m_A g \)) 3. Upward normal force exerted by Book C on B: \( N_{CB} \) (equal to \( m_A g + m_B g \)) **Forces on Book C:** 1. Downward gravitational force exerted by Earth: \( m_C g \) 2. Downward normal force exerted by Book B on C: \( N_{BC} \) (equal to \( N_{CB} \), which is \( m_A g + m_B g \)) 3. Upward normal force exerted by Book D on C: \( N_{DC} \) (equal to \( m_A g + m_B g + m_C g \)) **Forces on Book D:** 1. Downward gravitational force exerted by Earth: \( m_D g \) 2. Downward normal force exerted by Book C on D: \( N_{CD} \) (equal to \( N_{DC} \), which is \( m_A g + m_B g + m_C g \)) 3. Upward normal force exerted by the Table on D: \( N_{TD} \) (equal to \( m_A g + m_B g + m_C g + m_D g \)) The free body diagrams would show these forces for each book. For example, for Book B, there would be \( m_B g \) downwards, \( N_{AB} \) downwards (from Book A), and \( N_{CB} \) upwards (from Book C). All normal forces represent action-reaction pairs between adjacent surfaces. *In simple words:* Imagine each book in the stack separately. For each book, we list all the pushes and pulls acting on it: its own weight pulling down, the push from the book above it (if any), and the push from the book below it (if any). The very bottom book also gets a push from the table. Every push has an equal and opposite push back.

 

Question 9. A foot ball player kicks a 0.8 kg ball and imparts it a velocity \( 12 ms^{-1} \). The contact between foot and ball is only for one sixtieth of a second find the average kicking force.
Answer:
Given:
Mass of ball \( m = 0.8 \text{ kg} \)
Final velocity \( v = 12 \text{ ms}^{-1} \)
Time of contact \( t = \frac{1}{60} \text{ s} \)
Initial velocity \( u = 0 \text{ ms}^{-1} \)

We know that impulse is the change in momentum:
\( \text{Impulse} = \text{Final momentum} - \text{Initial momentum} \)
\( Ft = mv - mu \)
Since the ball starts from rest, \( u = 0 \), so \( mu = 0 \).
\( Ft = mv \)
\( F = \frac{mv}{t} \)
\( F = \frac{0.8 \text{ kg} \times 12 \text{ ms}^{-1}}{\frac{1}{60} \text{ s}} \)
\( F = 0.8 \times 12 \times 60 \)
\( F = 9.6 \times 60 \)
\( F = 576 \text{ N} \)
The average kicking force is \( 576 \text{ N} \). This force is quite large because the time of contact is very short.
In simple words: We calculate the change in the ball's movement and divide it by the very short time the foot touches the ball. This gives us the strong push (force) the player used.

🎯 Exam Tip: Remember that impulse is equal to the change in momentum. For problems involving quick impacts, the force is often large because the time is very small.

 

Question 10. A stone of mass 2kg is attached to a string of length 1m. The string can withstand a maximum tension of 200N, What is the maximum speed that stone can have during the whirling motion.
Answer:
Given:
Mass of stone \( m = 2 \text{ kg} \)
Length of string \( l = 1 \text{ m} \), which is the radius \( r \) of the circular path.
Maximum tension \( T = 200 \text{ N} \)

During whirling motion, the tension in the string provides the necessary centripetal force for the stone to move in a circle. The centripetal force is directed towards the center of the circle.
So, \( T = F_{centripetal} \)
\( T = \frac{mv^2}{r} \)

We need to find the maximum speed \( v \).
\( 200 \text{ N} = \frac{(2 \text{ kg}) \times v^2}{1 \text{ m}} \)
\( 200 = 2v^2 \)
\( v^2 = \frac{200}{2} \)
\( v^2 = 100 \)
\( v = \sqrt{100} \)
\( v = 10 \text{ ms}^{-1} \)
The maximum speed the stone can have is \( 10 \text{ ms}^{-1} \). If it goes faster, the string will break due to excessive tension.
In simple words: The string pulls the stone to keep it moving in a circle. We find the fastest speed the stone can go before this pull (tension) becomes too much for the string to handle.

🎯 Exam Tip: In circular motion problems, the force providing centripetal acceleration (like tension or friction) is crucial. Always identify this force and equate it to \( \frac{mv^2}{r} \).

 

Question 11. Imagine that the gravitational force between Earth and Moon is provided by an invisible string that exists between the Moon and Earth. What is the tension that, exists in this invisible string due to earth's centripetal force?
Answer:
Given:
Mass of the Moon \( m = 7.34 \times 10^{22} \text{ kg} \)
Distance between Earth and Moon \( r = 3.84 \times 10^8 \text{ m} \)
Period of rotation (T) = 27.3 days.

First, convert the period to seconds:
\( T = 27.3 \text{ days} \times 24 \text{ hours/day} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} \)
\( T = 27.3 \times 24 \times 60 \times 60 \text{ s} \)
\( T = 2.35872 \times 10^6 \text{ s} \)

Now, calculate the angular velocity \( \omega \):
\( \omega = \frac{2\pi}{T} \)
\( \omega = \frac{2\pi}{2.35872 \times 10^6 \text{ s}} \)
\( \omega \approx 2.66 \times 10^{-6} \text{ rad/s} \)

The tension (centripetal force) \( F \) is given by:
\( F = m r \omega^2 \)
\( F = (7.34 \times 10^{22} \text{ kg}) \times (3.84 \times 10^8 \text{ m}) \times (2.66 \times 10^{-6} \text{ rad/s})^2 \)
\( F = (7.34 \times 10^{22}) \times (3.84 \times 10^8) \times (7.0756 \times 10^{-12}) \)
\( F \approx 2.0 \times 10^{20} \text{ N} \)
The tension in this invisible string, representing the gravitational force, is approximately \( 2.0 \times 10^{20} \text{ N} \). This enormous force keeps the Moon in orbit around the Earth.
In simple words: We calculate how strong the pull must be to keep the Moon moving in its circle around the Earth. This pull is the same as the centripetal force needed, found by using the Moon's mass, its distance from Earth, and how fast it goes around.

🎯 Exam Tip: When dealing with orbital motion, remember that gravitational force often provides the necessary centripetal force. Ensure units are consistent (SI units are best) for all calculations.

 

Question 12. Two bodies of masses 15kg and 10kg are connected with light string kept on a smooth surface. A horizontal force F = 500N is applied to a 15kg as shown in the figure. Calculate the tension acting in the string.
Answer:
Given:
Mass 1 \( m_1 = 15 \text{ kg} \)
Mass 2 \( m_2 = 10 \text{ kg} \)
Applied force \( F = 500 \text{ N} \)

First, find the acceleration of the system. Since the surface is smooth, there is no friction.
Total mass of the system \( M_{total} = m_1 + m_2 = 15 \text{ kg} + 10 \text{ kg} = 25 \text{ kg} \)
According to Newton's second law, \( F = M_{total} \times a \)
\( a = \frac{F}{M_{total}} = \frac{500 \text{ N}}{25 \text{ kg}} = 20 \text{ ms}^{-2} \)

Now, consider the free body diagram for mass \( m_2 \) (10 kg block). The only horizontal force acting on \( m_2 \) is the tension \( T \) in the string, pulling it forward.
Using Newton's second law for \( m_2 \):
\( T = m_2 \times a \)
\( T = 10 \text{ kg} \times 20 \text{ ms}^{-2} \)
\( T = 200 \text{ N} \)
The tension acting in the string is \( 200 \text{ N} \). This tension is responsible for pulling the second block.
In simple words: First, we find how fast both blocks move together using the total force and their combined mass. Then, we look at only the second block and see that the string's pull (tension) is the only thing making it move, so we can find the tension.

🎯 Exam Tip: For connected bodies, calculate the system's overall acceleration first. Then, use that acceleration with the individual mass to find internal forces like tension.

 

Question 13. People often say "for every action there is an equivalent opposite reaction". Here they meant 'action of a human'. Is it correct to apply Newton's third law to human actions? What is mean by 'action' in Newton's third law? Give your arguments based on Newton's laws.
Answer:
Newton's third law states that for every action, there is an equal and opposite reaction. This law applies to *physical forces* between objects, not to human thoughts or psychological actions. It is incorrect to apply Newton's third law to human actions in a metaphorical or psychological sense.

In Newton's third law, 'action' refers to a force exerted by one object on another. For example, when you push a wall (action), the wall pushes back on you with an equal and opposite force (reaction). This principle is fundamental to understanding how objects interact mechanically. The law requires two distinct objects interacting through force, and the forces must be of the same type (e.g., both gravitational, both contact forces).
In simple words: Newton's third law is about real pushes and pulls between things, like when your hand pushes a door. It's not about what people think or feel.

🎯 Exam Tip: Clearly distinguish between physical forces and metaphorical actions when discussing Newton's third law. Emphasize that the action-reaction pair acts on different bodies.

 

Question 14. A car takes a turn with velocity \( 50ms^{-1} \) on the circular road of the radius of curvature 10m. Calculate the centrifugal force experienced by a person of mass 60kg inside the car?
Answer:
Given:
Velocity of car \( V = 50 \text{ ms}^{-1} \)
Radius of curvature \( r = 10 \text{ m} \)
Mass of person \( m = 60 \text{ kg} \)

Centrifugal force is a pseudo force that appears to act outwards on an object in a rotating frame of reference. Its magnitude is the same as the centripetal force.
\( F_{centrifugal} = \frac{mV^2}{r} \)
\( F_{centrifugal} = \frac{(60 \text{ kg}) \times (50 \text{ ms}^{-1})^2}{10 \text{ m}} \)
\( F_{centrifugal} = \frac{60 \times 2500}{10} \)
\( F_{centrifugal} = 6 \times 2500 \)
\( F_{centrifugal} = 15000 \text{ N} \)
The centrifugal force experienced by the person is \( 15000 \text{ N} \). This large outward force is what makes passengers feel pushed outwards during a sharp turn.
In simple words: When a car turns fast, a person inside feels a push outwards. We use the person's mass, the car's speed, and the curve's tightness to find how strong this outward push (centrifugal force) is.

🎯 Exam Tip: Remember that centrifugal force is a pseudo force, experienced in a rotating frame of reference. Its calculation is identical to centripetal force, \( \frac{mv^2}{r} \).

 

Question 15. A long stick rests on the surface. A person standing 10m away from the stick with what minimum speed an object of mass 0.5 kg should he threw so that it hits the stick. (Coefficient of kinetic friction is 0.7)
Answer:
Given:
Distance to stick \( s = 10 \text{ m} \)
Mass of object \( m = 0.5 \text{ kg} \)
Coefficient of kinetic friction \( \mu_k = 0.7 \)
Acceleration due to gravity \( g = 9.8 \text{ ms}^{-2} \)

First, calculate the frictional force acting on the object. This force will cause the object to slow down.
Normal force \( N = mg \)
Frictional force \( F_f = \mu_k N = \mu_k mg \)
\( F_f = 0.7 \times 0.5 \text{ kg} \times 9.8 \text{ ms}^{-2} \)
\( F_f = 3.43 \text{ N} \)

This frictional force causes a deceleration (negative acceleration).
\( F_f = ma \)
\( a = \frac{F_f}{m} = \frac{3.43 \text{ N}}{0.5 \text{ kg}} \)
\( a = 6.86 \text{ ms}^{-2} \)

We need to find the initial speed \( u \) required for the object to travel \( 10 \text{ m} \) and come to rest (\( v = 0 \)).
Using the kinematic equation \( v^2 - u^2 = 2as \):
\( 0^2 - u^2 = 2 \times (-6.86 \text{ ms}^{-2}) \times 10 \text{ m} \)
(The acceleration is negative because it opposes motion)
\( -u^2 = -137.2 \)
\( u^2 = 137.2 \)
\( u = \sqrt{137.2} \)
\( u \approx 11.71 \text{ ms}^{-1} \)
The object should be thrown with a minimum speed of approximately \( 11.71 \text{ ms}^{-1} \). This speed ensures the object overcomes friction to reach the stick.
In simple words: We find how much friction will slow the object down. Then, we calculate the starting speed needed for the object to travel 10 meters and stop right at the stick, despite the slowing effect of friction.

🎯 Exam Tip: Remember that kinetic friction acts opposite to the direction of motion and causes deceleration. Always include the negative sign for deceleration in kinematic equations.

11th Physics Guide Laws Of Motion Additional Important Questions And Answers

I. Multiple Choice Questions:

 

Question 1. The concept "force causes motion" was given by _
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer: (b) Aristotle
In simple words: The idea that a continuous force is always needed to keep something moving was first put forward by Aristotle.

🎯 Exam Tip: Understand the historical progression of physics concepts. Aristotle's ideas on motion were later challenged and replaced by Galileo and Newton.

 

Question 2. A body of 5 kg is moving with a velocity of 20m/s. If a force of 100 N is applied on it for 10s in the same direction as its velocity, what will now be the velocity of the body _
(a) 2000 m/s
(b) 220 m/s
(c) 240 m/s
(d) 260 m/s
Answer: (b) 220 m/s
In simple words: We find the initial momentum of the body and add the extra momentum gained from the applied force over time. Then, we use the total momentum to find the new, higher speed.

🎯 Exam Tip: This problem uses the impulse-momentum theorem. Remember that impulse (Force x time) equals the change in momentum (mass x change in velocity).

 

Question 3. The inability of objects to move on their own or change their state of motion is called as –
(a) force
(b) momentum
(c) inertia
(d) impulse
Answer: (c) inertia
In simple words: Inertia is the natural tendency of an object to resist changes in its state of motion, meaning it wants to stay still if it's still, or keep moving at the same speed and direction if it's already moving.

🎯 Exam Tip: Inertia is directly related to an object's mass. The more massive an object, the greater its inertia.

 

Question 4. A force vector applied on a mass is represented as \( \vec {f} = 6\vec { i } – 8\vec { j } + 10\vec { k } \) and accelerates with \( 1ms^{-2} \), what will be the mass of the body in kg \( \_ \)
(a) \( 10\sqrt{2} \)
(b) 20
(c) \( 2\sqrt{10} \)
(d) 10
Answer: (a) \( 10\sqrt{2} \)
In simple words: We first find the total strength of the force vector. Then, knowing how much it makes the object speed up, we can figure out the object's mass using the simple formula Force = Mass x Acceleration.

🎯 Exam Tip: To find the magnitude of a force vector \( \vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k} \), use the formula \( |\vec{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2} \). Then apply Newton's second law \( F=ma \).

 

Question 5. A particle of mass m is at rest at the origin at time t = 0. It is subjected to force \( F(t)= F_0e^{-bt} \) in the x direction. Its speed v(t) is depicted by which of the following
(a)
(b)
(c)
(d)
Answer: (b)
In simple words: The force starts strong and then quickly becomes very weak. This means the object will speed up rapidly at first and then its speed will increase more and more slowly, eventually reaching a maximum constant speed as the force nearly disappears.

🎯 Exam Tip: When force decreases exponentially, acceleration also decreases exponentially. Speed increases rapidly at first and then levels off, approaching a constant value.

 

Question 6. If the brake is applied in the moving bus suddenly, passengers move forward is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer: (a) Inertia of motion
In simple words: When the bus suddenly stops, the passengers keep moving forward because their bodies want to continue in the state of motion they were already in, due to inertia.

🎯 Exam Tip: Inertia of motion is the tendency of an object to resist changes in its state of motion, causing it to continue moving when a force tries to stop it.

 

Question 7. Two masses \( m_1 \) and \( m_2 \) are connected by a light string passing over a smooth pulley. When set free \( m_1 \) moves down by 2m in 2s the ratio of \( m_1/m_2 \) is _
(a) 9/7
(b) 11/9
(c) 13/11
(d) 15/13
Answer: (b) 11/9
In simple words: We first find how fast the system is speeding up. Then, using that speed-up value in the formula for masses connected over a pulley, we can work out the ratio of their masses.

🎯 Exam Tip: For Atwood machine problems, first calculate the acceleration using kinematics, then apply Newton's second law for the system to find the relationship between masses.

 

Question 8. Two blocks are connected by a string as shown in the figure. The upper block is hung by another string. A force 'F' applied on the upper string produces an acceleration of \( 2ms^{-2} \). in the upward direction in both the blocks. If T and T¹ in both parts of the string then _
(a) T = 84N; T¹ = 50N
(b) T = 70N; T¹ = 70N
(c) T = 84N; T¹ = 60N
(d) T = 70N; T¹ = 60N
Answer: (c) T = 84N; T¹ = 60N
In simple words: We consider the forces acting on each block separately. Starting from the bottom block, we find the tension in the lower string that pulls it up. Then, we use this tension, along with the weight of the upper block, to find the tension in the top string that pulls both up.

🎯 Exam Tip: When dealing with multiple connected blocks, draw free-body diagrams for each block and apply Newton's second law. Work from the outermost object inwards or vice-versa, depending on which forces are known.

 

Question 9. Four identical blocks each of mass m linked by threads as shown. IF the system moves with constant acceleration under the influence of force F, the Tension T2 _
(a) F/2
(b) F/2
(c) 2F
(d) F/4
Answer: (b) F/2
In simple words: We find the acceleration of the entire system using the total force and total mass. Then, we look at how many blocks are being pulled by Tension T2 and use that to calculate the force T2.

🎯 Exam Tip: In multi-block systems with a common acceleration, the tension in an intermediate string is equal to the mass it pulls multiplied by the system's acceleration.

 

Question 10. Rate of change of momentum of an object is equal to –
(a) acceleration
(b) work done
(c) force
(d) impulse
Answer: (c) force
In simple words: This is Newton's second law, which says that how quickly an object's motion changes is directly related to the force applied to it.

🎯 Exam Tip: Newton's second law is defined as \( F = \frac{dp}{dt} \) (rate of change of momentum), which simplifies to \( F=ma \) for constant mass.

 

Question 11. A block of mass 10 kg is pushed on a smooth inclined plane of inclination 30°, so that it has an acceleration \( 2ms^{-2} \). The applied force is \( \_ \)
(a) 50 N
(b) 60 N
(c) 70 N
(d) 80 N
Answer: (c) 70 N
In simple words: We calculate the part of gravity that pulls the block down the slope. Since the block is also speeding up, we add the force needed for that acceleration to the force of gravity to find the total push required.

🎯 Exam Tip: When analyzing motion on an inclined plane, always resolve forces parallel and perpendicular to the incline. For smooth surfaces, only the gravitational component along the incline and the applied force contribute to acceleration.

 

Question 12. Two blocks of masses 6 kg and 3 kg are connected by the string as shown over a _ _ _ . The acceleration of the system is _
(a) \( 4 ms^{-2} \)
(b) \( 2 ms^{-2} \)
(c) Zero
(d) \( 6 ms^{-2} \)
Answer: (c) Zero
In simple words: We check the forces pulling each side of the pulley. If the pull from the block on the slope is the same as the pull from the hanging block, then the system stays balanced and does not speed up.

🎯 Exam Tip: In systems involving inclined planes and hanging masses, compare the component of gravity along the incline with the gravitational force on the hanging mass to determine the net force and acceleration. Assuming \( g=10 \text{ m/s}^2 \) simplifies calculations if not specified.

 

Question 13. The acceleration of the systems shown in the figure is _
(a) \( 1 ms^{-2} \)
(b) \( 2 ms^{-2} \)
(c) \( 3 ms^{-2} \)
(d) \( 4 ms^{-2} \)
Answer: (a) \( 1 ms^{-2} \)
In simple words: We figure out how much gravity pulls each block down its own slope. Then, we find the difference between these pulls, which is the net force. We divide this net force by the total mass of both blocks to get the system's speed-up rate.

🎯 Exam Tip: For systems on double inclined planes, calculate the gravitational component along each incline for both masses. The net force is the difference between these components, and the acceleration is then found by dividing by the total mass.

 

Question 14. A uniform rope of length L is pulled by constant force P as shown, The Tension in the rope at a distance x from the end where the force is applied _
(a) P
(b) \( P(a - \frac{x}{L}) \)
(c) \( Px/L \)
(d) \( P(1 + \frac{x}{L}) \)
Answer: (b) \( P(a - \frac{x}{L}) \)
In simple words: When a rope is pulled, the tension at any point depends on the mass of the rope *beyond* that point. So, the tension is the force needed to pull the remaining part of the rope, which is less than the total force P. The term 'a' in the given option would typically be '1' for a standard rope pulled from one end.

🎯 Exam Tip: For a rope pulled by a force, the tension is not uniform. The tension at a point \( x \) from the pulling end is \( T(x) = P(1 - \frac{x}{L}) \), as it only needs to accelerate the portion of the rope of length \( (L-x) \).

 

Question 15. The law which is valid in both inertial and non-inertial frame is –
(a) Newton's first law
(b) Newton's second law
(c) Newton's third law
(d) none
Answer: (c) Newton's third law
In simple words: Newton's third law, which talks about action-reaction pairs, holds true no matter if you are in a steady (inertial) or speeding up/slowing down (non-inertial) frame of reference.

🎯 Exam Tip: While Newton's first and second laws need modification (introduction of pseudo forces) in non-inertial frames, the third law about mutual interaction forces remains universally valid.

 

Question 16. One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards centre) is _
(a) T
(b) \( T - \frac{m v^{2}}{r} \)
(c) \( T + \frac{m v^{2}}{r} \)
(d) 0
Answer: (a) T
In simple words: When an object moves in a circle, the force pulling it towards the center (centripetal force) is the net force. In this case, the string's tension is the only force doing that job.

🎯 Exam Tip: For uniform circular motion, the net force on the object is the centripetal force, which must be provided by some physical force like tension, gravity, or friction.

 

Question 17. A block is kept on a frictionless inclined surface with the angle of inclination \( \alpha \). The incline is given an acceleration 'a' to keep the block stationary. Then a is equal to _
(a) g
(b) g tan \( \alpha \)
(c) g / tan \( \alpha \)
(d) g cosec \( \alpha \)
Answer: (b) g tan \( \alpha \)
In simple words: For the block to stay still on the moving slope, the part of gravity pulling it down the slope must be exactly cancelled by the "fake" force (pseudo force) acting on it because the slope is speeding up. This balance helps us find the needed acceleration 'a'.

🎯 Exam Tip: To solve problems in accelerating frames, introduce a pseudo force equal to \( -ma \) acting opposite to the acceleration of the frame. Then, apply Newton's laws as if the frame were inertial.

 

Question 18. The action and reaction forces acting on –
(a) same body
(b) different bodies
(c) either same or different bodies
(d) none of the above
Answer: (b) different bodies
In simple words: For every push or pull (action) by one object, there is an equal and opposite push or pull (reaction) on a *different* object. They never act on the same object.

🎯 Exam Tip: A common misconception is that action and reaction forces cancel each other out. This is incorrect because they always act on different objects.

 

Question 19. If an elevator moving vertically up with an acceleration g, the force entered on the floor by a passenger of mass M is _
(a) mg
(b) 1/2 mg
(c) 2 mg
(d) 2 mg
Answer: (d) 2 mg
In simple words: When an elevator moves upwards and speeds up, the floor has to push the passenger up not only against gravity but also to make them accelerate. So, the passenger feels heavier, and the force on the floor is double their normal weight.

🎯 Exam Tip: For an accelerating elevator, the apparent weight \( R \) is given by \( R = m(g \pm a) \). Use \( +a \) for upward acceleration and \( -a \) for downward acceleration. If \( a=g \) and upwards, \( R = m(g+g) = 2mg \).

 

Question 20. A man of weight 80 kg, stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of \( 5 ms^{-2} \). What would be the reading on the scale? [g = 10 ms¯²].
(a) 400 N
(b) 1200 N
(c) 800 N
(d) zero
Answer: (b) 1200 N
In simple words: When the lift is speeding up upwards, the weighing scale not only supports the man's normal weight but also provides an extra push to make him accelerate. This extra push makes him feel heavier, so the scale shows a bigger number.

🎯 Exam Tip: Remember to add the acceleration of the lift to gravity when the lift accelerates upwards, and subtract it when it accelerates downwards, to find the apparent weight.

 

Question 21. A person is standing in an elevator in which situation he find his weight more than the actual weight _
(a) the elevator moves upwards with constant acceleration
(b) the elevator moves downwards with constant acceleration
(c) the elevator moves upwards with uniform velocity
(d) the elevator moves downwards with uniform velocity
Answer: (a) the elevator moves upwards with constant acceleration
In simple words: A person feels heavier than usual when the elevator speeds up as it moves upwards, because the floor has to push harder to lift them and also make them go faster.

🎯 Exam Tip: Apparent weight increases when an elevator accelerates upwards or decelerates downwards. It decreases when an elevator accelerates downwards or decelerates upwards.

 

Question 22. Newton's second law gives –
(b) \( \overrightarrow{\mathrm{F}}=\frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}} \)
(c) \( \overrightarrow{\mathrm{F}}=m \vec{a} \)
(d) all the above
Answer: (d) all the above
In simple words: Newton's second law can be shown in different ways, either by talking about how momentum changes over time or by showing that force equals mass times acceleration. Both expressions mean the same thing for most situations.

🎯 Exam Tip: The most general form of Newton's second law is \( \vec{F} = \frac{d\vec{P}}{dt} \). When mass is constant, this simplifies to \( \vec{F} = m\vec{a} \).

 

Question 23. A mass of 1 kg is suspended by a string. Another string C is connected to its lower end (as in the figure). If a sudden jerk is given to c, then _
(a) The portion AB of the string break
(b) The portion BC of the string will break
(c) The mass will be rotating
(d) none of the above
Answer: (b) The portion BC of the string will break
In simple words: When a quick, sharp tug is given to the bottom string, the heavy mass at the top cannot respond fast enough due to its inertia. So, the sudden force breaks the weak string at the bottom instead of moving the whole mass.

🎯 Exam Tip: This is an example of inertia. A sudden force acts over a very short time, and the mass cannot overcome its inertia quickly, so the bottom string breaks.

 

Question 24. In the above questions, if the string c is stretched slowly then _
(a) The portion AB of the string break
(b) The portion BC of the string will break
(c) The mass will be rotating
(d) none of the above
Answer: (a) The portion AB of the string break
In simple words: If you pull the bottom string slowly, the weight of the mass has enough time to add tension to the top string. Eventually, the top string breaks because it has to hold both the mass's weight and the slow pull from below.

🎯 Exam Tip: When force is applied slowly, the entire system has time to react. The tension in the top string (AB) will be the sum of the weight of the mass plus the applied force, making it the most likely to break.

 

Question 25. As shown in the figure, the tension in the horizontal card is 30N. The weight w and the tension in the string OA in Newton are
(a) 30, \( \sqrt{3} \), 30
(b) 30\( \sqrt{3} \), 60
(c) 60\( \sqrt{3} \), 30
(d) none of the options
Answer: (b) 30\( \sqrt{3} \), 60
In simple words: The horizontal string has a tension of 30N. Since the string OA is at 30 degrees to the vertical, its horizontal component balances this 30N, making the tension in OA 60N. The vertical component of OA balances the weight 'w', which comes out to be 30\( \sqrt{3} \)N.

🎯 Exam Tip: Always resolve forces into their horizontal and vertical components when dealing with equilibrium problems on inclined or angled strings. Ensure the angle used (with horizontal or vertical) is consistent.

 

Question 26. A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms-1. How long does the body take to stop?
(a) 0.75 s
(b) 1.33 s
(c) 6 s
(d) 35 s
Answer: (c) 6 s
In simple words: A force of 50 N slows down a 20 kg object. First, find how fast it is slowing down (acceleration). Then, use that to calculate how long it takes for the object to completely stop from its initial speed of 15 m/s.

🎯 Exam Tip: Remember that a retarding force means negative acceleration, and 'to stop' implies the final velocity is zero. Use the kinematic equation \( v = u + at \) to find the time.

 

Question 27. A block of mass 10 kg is suspended through two light spring balances as in figure.
(a) both scales will read 10 kg
(b) both scales will read 5 kg
(c) the upper scale will read 10 kg and the lower zero
(d) the reading may be anything but their sum will be 10 kg
Answer: (a) both scales will read 10 kg
In simple words: When spring balances are connected in series, the same tension runs through both. So, each balance will show the total weight of the mass hanging below them, which is 10 kg in this case.

🎯 Exam Tip: Understand that in a series connection, the force is transmitted equally through each component. This is different from a parallel connection where the force might be distributed.

 

Question 28. Two blocks A and B of masses 2m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as in figure. The magnitude of the acceleration of A and B after the string is cut, are respectively.
(a) g/2, g/2
(b) g/2, g
(c) g, g
(d) g/2, g/2
Answer: (b) g/2, g
In simple words: When the string holding block B is cut, block B instantly falls with gravity (acceleration 'g'). Block A is still connected to the spring, which stretches to support the total weight (3m) before cutting. After cutting, the spring only supports block A (2m) and will cause it to accelerate upwards, with acceleration \( g/2 \) (after some oscillations).

🎯 Exam Tip: Immediately after a string is cut, tension becomes zero. However, a spring's force does not change instantaneously. Calculate the spring force before cutting and then apply Newton's second law right after the cut.

 

Question 29. Choose the correct statement
(a) The frictional forces are dependent on the roughness of the surface.
(b) The kinetic friction is proportional to normal reaction
(c) The friction is independent of area of contact
(d) All statements are correct
Answer: (d) All statements are correct
In simple words: All the given statements are true about friction. Friction depends on how rough a surface is, kinetic friction changes with the normal force pushing surfaces together, and it does not depend on the size of the contact area.

🎯 Exam Tip: Remember the fundamental laws of friction: it opposes relative motion, depends on surface properties and normal force, and is largely independent of contact area (for rigid bodies).

 

Question 30. When an object of mass m slides on a frictionless surface inclined at an angle \( \theta \), then the normal force exerted by the surface is-
(a) g cos \( \theta \)
(b) mg cos \( \theta \)
(c) g sin \( \theta \)
(d) mg tan \( \theta \)
Answer: (b) mg cos \( \theta \)
In simple words: On a slope, the normal force (the push from the surface) is not equal to the object's full weight. Instead, it balances only the part of the weight that pushes directly into the slope, which is \( mg \cos \theta \).

🎯 Exam Tip: Always resolve the gravitational force into components parallel and perpendicular to the inclined plane. The normal force will always balance the perpendicular component of gravity.

 

Question 31. Two cars of unequal masses are similar types. If they are moving at the same initial speed, the minimum stopping distance
(a) is smaller for the heavier car
(b) is smaller for the lighter car
(c) is same for both car
(d) depends on volume of the car
Answer: (c) is same for both car
In simple words: If two similar cars brake, their stopping distance only depends on their initial speed and the friction from the road, not their mass. This is because both the braking force (friction) and the car's inertia (resistance to change in motion) are proportional to mass.

🎯 Exam Tip: For similar cars, the coefficient of friction is typically the same. Since friction force (\( \mu N \)) and inertia (mass \( m \)) both scale with mass, the mass cancels out in the stopping distance calculation (\( F = ma \), so \( \mu mg = ma \implies a = \mu g \)).

 

Question 32. An ice block is kept on an inclined plane of angle of 30°. The coefficient of kinetic friction between the block and the inclined plane is \( \frac{1}{\sqrt{3}} \). The acceleration of the block is
(a) zero
(b) 2 ms-2
(c) 1.5 ms-2
(d) 5 ms-2
Answer: (a) zero
In simple words: The ice block is on a slope. The force pulling it down the slope is \( mg \sin \theta \), and the friction opposing it is \( \mu_k mg \cos \theta \). Since \( \theta = 30^\circ \) and \( \mu_k = \frac{1}{\sqrt{3}} \), the friction force is equal to the component of gravity pulling it down the slope. This means the net force is zero, so there is no acceleration.

🎯 Exam Tip: Compare the component of gravity down the incline (\( mg \sin \theta \)) with the maximum static friction (\( \mu_s mg \cos \theta \)) or kinetic friction (\( \mu_k mg \cos \theta \)). If they are equal, the net force is zero and acceleration is zero (or constant velocity if already moving).

 

Question 33. Starting from rest a body slides down at 45° inclined plane in twice the time, it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is
(a) 0.33
(b) 0.25
(c) 0.75
(d) 0.80
Answer: (c) 0.75
In simple words: When there is no friction, an object slides down a slope faster. With friction, it slides slower, taking twice as long to cover the same distance. By comparing these times, we can calculate how much friction is present, giving us the friction coefficient.

🎯 Exam Tip: Use the equation of motion \( s = ut + \frac{1}{2}at^2 \). For motion from rest (\( u=0 \)), \( s = \frac{1}{2}at^2 \). The acceleration without friction is \( g \sin \theta \), and with friction it's \( g(\sin \theta - \mu \cos \theta) \).

 

Question 34. A uniform metal chain if placed on a rough table such that the one end of the chain hangs down over the edge of the table. When one-third of its length hangs over the edge, the chain starts sliding. Then the coefficient of static friction is
(a) 3/4
(b) 1/4
(c) 2/3
(d) 1/2
Answer: (d) 1/2
In simple words: When one-third of the chain hangs off the table, its weight creates a pulling force. This force becomes just enough to overcome the friction from the two-thirds of the chain remaining on the table. By balancing these forces, we can find the friction coefficient.

🎯 Exam Tip: For a uniform chain, the mass is proportional to its length. The weight of the hanging part provides the pulling force, and the weight of the part on the table determines the normal force for friction.

 

Question 35. If two masses m\( _1 \) and m\( _2 \) (m\( _1 \) > m\( _2 \)) tied to string moving over a frictionless pulley, then the acceleration of masses -
(a) \( \frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}} g \)
(b) \( \frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)} g \)
(c) \( \frac{2 m_{1} m_{2}}{m_{1}+m_{2}} g \)
(d) \( \frac{m_{1} m_{2}}{2 m_{1} m_{2}} g \)
Answer: (a) \( \frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}} g \)
In simple words: When two unequal masses are connected by a string over a pulley, the heavier mass pulls the system, causing acceleration. The formula shows that this acceleration depends on the difference in masses and the total mass, scaled by gravity.

🎯 Exam Tip: This setup is an Atwood machine. Remember to apply Newton's second law (\( F = ma \)) to each mass separately, considering the tension in the string and gravitational forces.

 

Question 36. While walking on ice one should take small steps to avoid slipping. This is because smaller steps ensure
(a) large friction
(b) smaller friction
(c) larger normal force
(d) smaller normal force
Answer: (c) larger normal force
In simple words: Taking smaller steps on ice means you put more of your weight straight down onto a smaller area with each step. This increases the normal force, which helps increase the friction available to prevent slipping.

🎯 Exam Tip: On slippery surfaces, the frictional force (which prevents slipping) is directly proportional to the normal force. Increasing the normal force, by stepping more vertically, maximizes the available friction.

 

Question 37. A box is placed on inclined plane and has to be pushed down. The angle of inclination is
(a) equal to angle of friction
(b) more than angle of friction
(c) equal to angle repose
(d) less than angle of repose
Answer: (d) less than angle of repose
In simple words: If you have to push a box down a slope, it means the slope is not steep enough for the box to slide on its own. This happens when the slope's angle is less than the angle of repose, which is the angle where it would just start to slide.

🎯 Exam Tip: The angle of repose is the maximum angle a slope can have before an object on it starts to slide spontaneously due to gravity overcoming static friction. If the angle is less than the angle of repose, an external push is needed to initiate motion.

 

Question 38. Three masses in contact is as shown above. If force F is applied to mass m\( _1 \) then the contact force acting on mass m\( _2 \) is –
(a) \( \frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}} \)
(b) \( \frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)} \)
(c) \( \frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)} \)
(d) \( \frac{m_{3} F}{\left(m_{1}+m_{2}+m_{3}\right)} \)
Answer: (c) \( \frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)} \)
In simple words: When a force is applied to one mass in a connected system, that force makes all the masses accelerate together. The contact force between mass m\( _1 \) and m\( _2 \) is essentially the force needed to accelerate m\( _2 \) and m\( _3 \).

🎯 Exam Tip: First, find the acceleration of the entire system using \( F_{net} = (m_1+m_2+m_3)a \). Then, consider the forces acting on the part of the system that mass \( m_2 \) is pushing (i.e., \( m_2 \) and \( m_3 \)) to find the contact force.

 

Question 39. A block B is placed on block A. The mass block B is less than the mass of block friction that exists between the blocks whereas the ground on which block A is placed is taken to be smooth. A horizontal force 'F' increasing linearly with time beings to act on B. The acceleration aA and aB of blocks A and B respectively are plotted against Y. The correctly plotted graph is
(a)
(b)
(c)
(d)
Answer: (c)
In simple words: Initially, both blocks move together because of friction, so their accelerations are the same. As the applied force increases, it eventually overcomes the maximum static friction between the blocks. At this point, block B (where the force is applied) will start to accelerate more rapidly than block A, which is still being dragged by the kinetic friction.

🎯 Exam Tip: For problems involving blocks stacked on top of each other, remember that they move together until the applied force exceeds the maximum static friction between them. Once this limit is reached, they start sliding relative to each other.

 

Question 40. Two blocks of masses 6 kg and 3 kg are connected as in figure coefficient of friction between m\( _1 \) and m\( _2 \) and between m\( _1 \) and surface is 0.5 and 0.4 respectively. The maximum horizontal force to can be applied to the mass m\( _1 \) so that they move without separation is
(a) 41 N
(b) 61 N
(c) 81 N
Answer: (c) 81 N
In simple words: We want to find the largest force that can push the bottom block without the top block sliding off. This means we need to consider the friction between the two blocks and the friction between the bottom block and the ground, ensuring both blocks move together.

🎯 Exam Tip: To solve this, first find the maximum acceleration for which the top block does not slide relative to the bottom block (determined by friction between m1 and m2). Then, use this acceleration for the entire system (m1+m2) and calculate the applied force needed, considering the friction between the bottom block and the ground.

 

II. Additional Questions:

 

Question 1. Identify the internal and external forces acting on the following systems.
(a) Earth alone as a system
(b) Earth and sun as a system
(c) Our body as a system while walking
(d) our body and earth as a system
Answer:
(a) Earth alone as a system:

The Earth orbits the Sun due to the Sun's gravitational attraction. If we consider only the Earth as our system, then the gravitational force from the Sun acting on Earth is an external force. Other celestial bodies like the Moon also exert external forces on the Earth.
(b) Earth and Sun as a system:
In this combined system, the gravitational force between the Sun and Earth is an internal force. It forms an action-reaction pair within the system.
(c) Our body as a system while walking:
When we walk, we push the Earth backward with our feet. According to Newton's third law, the Earth pushes our body forward. If our body is considered the system, the force exerted by the Earth on our body is an external force that propels us forward.
(d) Our body and Earth as a system:
In this larger system, the forces our body exerts on Earth and Earth exerts on our body are both internal forces. These forces form action-reaction pairs within the combined system. The system as a whole would only be accelerated by forces from outside it (e.g., if an external rocket pushed the entire Earth-person system).
In simple words: An external force comes from outside the chosen system, while an internal force acts between parts inside the system. When you walk, the ground pushing you forward is external if you only consider your body. But if you consider your body and the Earth together, your push on the ground and the ground's push on you are internal.

🎯 Exam Tip: Clearly define the boundaries of your "system" before identifying forces. Forces between objects *within* the system are internal, and forces from objects *outside* the system are external. Only external forces can change the total momentum of a system.

 

Question 2. When a cricket player catches the ball he pulls his hands gradually is the directions of the ball's motion. Why?
Answer: When a cricket player catches a ball, he pulls his hands back slowly in the direction the ball is moving. This action increases the time during which the ball's momentum changes to zero. By extending the time of impact, the average force exerted on his hands is reduced, preventing injury. This is a practical application of the impulse-momentum theorem, where a longer impact time results in a smaller force for the same change in momentum.
In simple words: A cricket player pulls his hands back when catching a ball to increase the time it takes for the ball to stop. A longer stopping time means less force is felt on the hands, which helps to avoid getting hurt.

🎯 Exam Tip: Remember the impulse-momentum theorem: \( \text{Impulse} = F_{avg} \times \Delta t = \Delta p \). To keep the change in momentum \( \Delta p \) constant, if you increase \( \Delta t \), the average force \( F_{avg} \) must decrease.

 

Question 3. An impulse is applied to a moving object with a force at an angle of 20° w.r.t. velocity vector, what is the angle between the impulse vector and change in momentum vector?
Answer: The impulse vector and the change in momentum vector are always in the same direction. This is a fundamental principle derived from Newton's second law, which states that impulse is equal to the change in momentum. Therefore, the angle between them is always 0 degrees, regardless of the angle between the force and velocity vectors. The impulse acts in the direction of the net force, which causes the change in momentum.
In simple words: The push (impulse) on an object always happens in the same direction as the way its movement changes (change in momentum). So, the angle between them is always zero.

🎯 Exam Tip: Always remember that impulse is defined as the change in momentum, both of which are vector quantities. Therefore, they must have the same direction.

 

Question 4. Why a high jumper after crossing the bar made to fall on a spongy floor instead of the cemented floor.
Answer: A high jumper falls onto a spongy mat rather than a hard cemented floor to reduce the impact force upon landing. The spongy mat increases the time it takes for the jumper's body to come to a stop. By increasing the collision time, the average force experienced by the jumper's body is significantly reduced, which helps prevent serious injuries. This is a direct application of the impulse-momentum principle. The mats are designed to deform and absorb energy effectively.
In simple words: High jumpers land on soft mats because it makes the stopping time longer. A longer stop time means less force is pushed back onto their body, which keeps them safe from getting hurt.

🎯 Exam Tip: Relate this directly to the concept of impulse. A large change in momentum over a short time results in a large force, while the same change over a longer time results in a smaller, safer force.

 

Question 5. Obtain an expression for centrifugal force acting on a man on the surface of the earth.
Answer: The Earth rotates about its axis with an angular velocity \( \omega \). A man standing on the Earth's surface in a rotating frame of reference experiences a centrifugal force. This force acts outwards, away from the axis of rotation. The centrifugal force is given by \( F_c = mr\omega^2 \), where \( m \) is the mass of the man, \( r \) is the perpendicular distance of the man from the Earth's axis of rotation, and \( \omega \) is the Earth's angular velocity. If \( \lambda \) is the latitude of the man on Earth, then \( r = R \cos \lambda \), where \( R \) is the radius of the Earth. So, the centrifugal force can be expressed as \( F_c = m(R \cos \lambda)\omega^2 \). This force slightly reduces the man's effective weight, especially at the equator where \( \lambda = 0 \).
In simple words: As the Earth spins, a person on its surface feels a force pushing them outwards, called the centrifugal force. This force depends on their mass, how fast the Earth spins, and how far they are from the Earth's spinning axis. It's strongest at the equator and zero at the poles.

🎯 Exam Tip: Remember that centrifugal force is a pseudo force, which means it appears in a rotating frame of reference. It is always directed radially outwards from the axis of rotation.

 

Question 6. A stone when thrown on a glass window smashes the window pane to pieces, but a bullet from the gun passes through, by making a clean hole. Why?
Answer: When a stone is thrown at a glass window, it hits with a relatively low speed. The impact force is distributed over a larger area and for a longer duration, allowing the glass to deform and fracture, causing it to smash into pieces. In contrast, a bullet travels at a very high speed. When it strikes the glass, the impact force is concentrated over a very small area for an extremely short duration. The glass in the immediate vicinity of the bullet's path accelerates so rapidly that it breaks away, creating a clean hole, while the surrounding glass remains largely undisturbed. This difference in behavior is due to the varying speeds and contact times, which affect how stress waves propagate through the material. This concept relates to inertia and how quickly forces can propagate through a material.
In simple words: A slow stone spreads its force out and breaks the whole window. A fast bullet pushes a tiny spot so quickly that only that small spot breaks, leaving a clean hole, because the rest of the glass doesn't have time to react.

🎯 Exam Tip: Focus on the concepts of impulse, force distribution, and time of contact. A larger time of contact or distributed force leads to greater overall damage, while a concentrated force over a very short time leads to localized damage.

 

Question 7. China wares are wrapped in a straw paper before packing why?
Answer: China wares are wrapped in straw paper before packing to protect them from breaking during transportation. The straw paper acts as a cushioning material, which increases the impact time if the package is dropped or experiences a sudden jerk. By increasing the time over which the collision occurs, the force experienced by the delicate china wares is significantly reduced, preventing them from shattering. This principle is based on the impulse-momentum theorem, where a longer duration of impact results in a smaller average force for the same change in momentum. The soft packaging helps to absorb the kinetic energy slowly.
In simple words: China dishes are wrapped in straw paper to protect them. The soft paper makes any bumps last longer, which reduces the force on the dishes and stops them from breaking.

🎯 Exam Tip: The key idea here is to minimize the impact force by maximizing the time of impact. Cushioning materials serve this purpose by deforming and extending the interaction time.

 

Question 8. Why it is necessary to bend knees while jumping from greater height?
Answer: When jumping from a greater height, it is necessary to bend the knees upon landing to increase the time taken to bring the body to rest. Bending the knees allows the body to absorb the impact over a longer duration, which significantly reduces the average force exerted on the legs and joints. If the knees are kept stiff, the body would stop very quickly, resulting in a large impact force that could cause severe injury. This action is a protective mechanism based on the impulse-momentum principle. The body's kinetic energy is gradually converted into potential energy stored in the muscles and absorbed as heat.
In simple words: Bending your knees when you land from a jump makes the stopping time longer. A longer stop means less force on your legs, which prevents you from getting hurt.

🎯 Exam Tip: This is another application of the impulse-momentum theorem. A controlled deceleration over a longer period results in a smaller force, reducing the risk of injury.

 

Question 9. Why is it difficult to drive a nail into a wooden block without supporting it?
Answer: It is difficult to drive a nail into a wooden block without supporting it because of Newton's third law of motion. When you hit the nail with a hammer, the nail exerts a force on the wooden block. If the block is unsupported, it will accelerate away from the hammer with the nail, instead of the nail penetrating the wood. There is no external reaction force to hold the block in place. When the block is supported, the support provides an equal and opposite reaction force, allowing the nail to be driven into the block efficiently. The support essentially grounds the block, giving the nail a firm target.
In simple words: It's hard to hammer a nail into a block if the block isn't held still. This is because the block just moves away with the nail instead of letting the nail go in. You need to hold it steady so the nail can push into the wood.

🎯 Exam Tip: Think about action-reaction pairs. For the nail to go *into* the wood, the wood needs to offer a substantial reaction force. Without support, the entire block moves, reducing the effective penetration force.

 

Question 10. Why is static friction called a self-adjusting force?
Answer: Static friction is called a self-adjusting force because its magnitude varies to match the applied force, up to a certain maximum limit. When a small external force is applied to an object at rest, the static friction force will increase to exactly oppose that force, keeping the object stationary. If the applied force increases, the static friction also increases proportionally until it reaches its maximum possible value, known as limiting friction. Beyond this limit, the object starts to move. This adaptive nature is why it's described as "self-adjusting." The intermolecular bonds between the surfaces adjust to resist the applied stress.
In simple words: Static friction changes its strength to exactly match any push you give an object, as long as the object doesn't start moving. It "adjusts" itself to keep things still until your push becomes too strong.

🎯 Exam Tip: Remember that static friction prevents motion, and its value is not constant but can be anywhere from zero up to a maximum value, \( f_{s,max} = \mu_s N \).

 

Question 11. Carts with rubber tyres are easier to fly than those with iron wheels. Why?
Answer: Carts with rubber tires are easier to move than those with iron wheels because rubber tires provide a much smaller coefficient of rolling friction compared to iron wheels. Rolling friction is generally much less than sliding friction. Rubber tires can also deform slightly, increasing the contact area and providing better grip without increasing the overall resistance significantly on many surfaces. Iron wheels, especially on rough surfaces, experience greater resistance due to their rigid nature and higher friction coefficient with the ground. This helps in efficient movement and reduced effort.
In simple words: Carts with rubber tires are easier to pull because rubber has less friction when rolling than iron does. Rubber also grips the ground better without making it harder to move.

🎯 Exam Tip: Focus on the difference between rolling friction (which is generally much smaller) and sliding friction, and how the material properties of the wheels (e.g., rubber's deformability) affect friction with the ground.

 

Question 12. Why ball bearings are used in machinery?
Answer: Ball bearings are used in machinery to reduce friction between moving parts. They achieve this by converting sliding friction into rolling friction. When ball bearings are placed between two surfaces that slide past each other, they roll instead of slide. Since rolling friction is significantly smaller than sliding friction, the energy loss due to friction is greatly reduced, making the machinery more efficient and durable. This also reduces wear and tear on the moving parts, extending their lifespan. The balls spread the load evenly, further improving performance.
In simple words: Ball bearings help machines work smoothly by changing the rubbing friction into rolling friction. Rolling friction is much weaker, so less energy is wasted, and the machine parts last longer.

🎯 Exam Tip: The primary advantage of ball bearings is the conversion of a high-friction sliding motion into a low-friction rolling motion, leading to improved efficiency and reduced heat generation in machines.

 

Question 14. A long rope is hanging, passing over a pulley. Two monkeys of equal weight climb up from the opposite ends of the rope. One of them climbs up more rapidly relative to rope. Which monkey will reach the top first? Pulley is frictionless and the rope is mass less and inextensible.
Answer: In this scenario, since the rope is massless and inextensible, and the pulley is frictionless, the tension throughout the rope is uniform. If both monkeys have equal weight, their individual actions to climb the rope will exert forces that are internal to the monkey-rope-monkey system. Even if one monkey climbs faster relative to the rope, the net external force on the *system* of two monkeys plus the rope remains zero. Therefore, their acceleration relative to the ground will be the same, and if they start at the same height, they will reach the top simultaneously. The faster monkey will simply cover more rope length relative to the slower monkey, but this does not change their absolute motion if the pulley system remains balanced. This is a classic example illustrating relative motion and internal forces in a system.
In simple words: If two monkeys of the same weight climb a rope over a pulley, even if one climbs faster *on the rope itself*, they will both reach the top at the same time. This is because they are part of a balanced system, and their individual efforts don't change how the whole system moves relative to the ground.

🎯 Exam Tip: This problem highlights the concept of center of mass. For a system with no external forces, the center of mass remains constant. If the monkeys have equal mass, their combined center of mass remains fixed if they climb at the same rate relative to the ground.

 

Question 15. A light string passing over a smooth pulley connects two blocks of masses m\( _1 \) and m\( _2 \). If the acceleration of the system is g/8. Find the ratio of the two masses.
Answer: Let the acceleration of the system be \( a = g/8 \). For two masses connected by a string over a frictionless pulley (Atwood machine), the acceleration is given by the formula: \( a = \frac{m_1 - m_2}{m_1 + m_2} g \). Substituting the given acceleration, we have \( \frac{g}{8} = \frac{m_1 - m_2}{m_1 + m_2} g \). Dividing both sides by \( g \), we get \( \frac{1}{8} = \frac{m_1 - m_2}{m_1 + m_2} \). Cross-multiplying gives \( m_1 + m_2 = 8(m_1 - m_2) \). Expanding this, \( m_1 + m_2 = 8m_1 - 8m_2 \). Rearranging the terms to find the ratio: \( m_2 + 8m_2 = 8m_1 - m_1 \implies 9m_2 = 7m_1 \). Thus, the ratio \( \frac{m_1}{m_2} = \frac{9}{7} \). The ratio of the two masses is 9:7.
In simple words: We are given the acceleration of two masses connected over a pulley. Using the special formula for this setup, we can work backward to find the ratio of the masses that would cause that specific acceleration.

🎯 Exam Tip: For Atwood machine problems, remember the acceleration formula \( a = \frac{|m_1 - m_2|}{m_1 + m_2} g \). Be careful with algebra when rearranging to find the mass ratio.

 

Question 16. Briefly explains how a horse is able to pull a cart?
Answer: A horse pulls a cart by pushing against the ground with its hooves. According to Newton's third law, the ground then exerts an equal and opposite reaction force on the horse, pushing the horse forward. This forward force from the ground on the horse is what enables the horse to move. The horse is connected to the cart by a rope or harness, which transmits the forward force to the cart as tension. The horse needs to exert a force on the ground that is greater than the combined resistive forces (like friction) acting on both the horse and the cart. The horse's legs act like levers, efficiently transferring power to the ground. The cart then moves because of the tension in the harness.
In simple words: A horse pulls a cart by pushing the ground backward. The ground then pushes the horse forward, and this forward push is transferred through a rope to the cart, making it move.

🎯 Exam Tip: This is a classic example of Newton's Third Law in action. Remember that the action-reaction pair for propulsion is always between the moving object and the surface it's pushing against (e.g., foot-ground, tire-road, propeller-water).

 

Question 17. A man of mass m is standing on the floor of a lift. Find his apparent weight when the lift is (i) moving upwards with uniform acceleration 'a' (ii) moving downwards with uniform acceleration a (iii) at rest or moving with uniform velocity (v) (iv) falling freely. Take acceleration due to gravity as g.
Answer:
(i) When the lift moves upwards with an acceleration 'a':
The net upward force on the man is \( R - mg \), where R is the apparent weight and \( mg \) is his actual weight. This net force causes the acceleration. So, \( R - mg = ma \).
Solving for R, his apparent weight is \( R = m(g+a) \). This means the man feels heavier than his actual weight. This sensation of increased weight is similar to what you feel at the start of a roller coaster ride.

(ii) When the lift moves downwards with an acceleration 'a':
The net downward force on the man is \( mg - R \). This net force causes the downward acceleration. So, \( mg - R = ma \).
Solving for R, his apparent weight is \( R = m(g-a) \). This means the man feels lighter than his actual weight. This is why you might feel a slight lift from your seat when an elevator starts going down quickly.

(iii) When the lift is at rest or moving with uniform velocity (constant speed):
In this case, the acceleration 'a' is zero. The net force on the man is also zero. So, \( R - mg = m \times 0 \), which simplifies to \( R - mg = 0 \).
Therefore, his apparent weight is \( R = mg \), which is equal to his actual weight. He feels normal. This is the most common experience we have in an elevator, feeling our true weight.

(iv) When the lift is falling freely:
When the lift falls freely, its acceleration 'a' is equal to the acceleration due to gravity 'g'. The net downward force on the man is \( mg - R \). So, \( mg - R = mg \).
This means the apparent weight \( R \) is \( 0 \). The man experiences complete weightlessness. Astronauts in space feel weightless because they are constantly in free fall around the Earth, similar to this scenario.
In simple words: A person's weight in a lift changes based on how the lift moves. If it speeds up going up, they feel heavier; if it speeds up going down, they feel lighter. If it moves at a steady speed or stays still, they feel their normal weight. If it falls freely, they feel weightless.

🎯 Exam Tip: Clearly define the direction of acceleration and the forces (actual weight and normal reaction) for each case to correctly apply Newton's second law.

 

Question 18. Derive the law of conservation of linear momentum from Newton's third law of motion.
Answer:
Let's consider two objects, A and B, with masses \( m_1 \) and \( m_2 \). They are moving in the same direction along a straight line with initial velocities \( u_1 \) and \( u_2 \) respectively, where \( u_1 > u_2 \).
After a short time interval \( \Delta t \), the two objects collide. After the collision, their final velocities are \( v_1 \) and \( v_2 \).

During the collision, object A exerts a force \( \vec{F}_{AB} \) on object B, and object B exerts a force \( \vec{F}_{BA} \) on object A.
According to Newton's third law of motion, these forces are equal in magnitude and opposite in direction:
\[ \vec{F}_{BA} = -\vec{F}_{AB} \quad (1) \]
From Newton's second law of motion, the force acting on an object is equal to the rate of change of its linear momentum:
\[ \vec{F}_{AB} = \frac{d\vec{P}_B}{dt} \quad (2) \]
\[ \vec{F}_{BA} = \frac{d\vec{P}_A}{dt} \quad (3) \]
Here, \( \vec{P}_A \) is the linear momentum of object A and \( \vec{P}_B \) is the linear momentum of object B.
Substitute equations (2) and (3) into equation (1):
\[ \frac{d\vec{P}_A}{dt} = -\frac{d\vec{P}_B}{dt} \]
Rearrange the equation:
\[ \frac{d\vec{P}_A}{dt} + \frac{d\vec{P}_B}{dt} = 0 \]
This means the total rate of change of momentum for the system is zero:
\[ \frac{d}{dt}(\vec{P}_A + \vec{P}_B) = 0 \]
If the rate of change of the total momentum of the system is zero, it implies that the total linear momentum of the system remains constant.
So, the total linear momentum \( \vec{P}_{total} = \vec{P}_A + \vec{P}_B \) is a constant vector.

Alternatively, using the concept of impulse (which is the change in momentum):
The impulse exerted on object A by B is \( \vec{F}_{BA} \Delta t \), which equals the change in momentum of A:
\[ \vec{F}_{BA} \Delta t = m_1\vec{v}_1 - m_1\vec{u}_1 \]
The impulse exerted on object B by A is \( \vec{F}_{AB} \Delta t \), which equals the change in momentum of B:
\[ \vec{F}_{AB} \Delta t = m_2\vec{v}_2 - m_2\vec{u}_2 \]
Since \( \vec{F}_{BA} = -\vec{F}_{AB} \), we can write:
\( m_1\vec{v}_1 - m_1\vec{u}_1 = -(m_2\vec{v}_2 - m_2\vec{u}_2) \)
Now, expand the equation:
\( m_1\vec{v}_1 - m_1\vec{u}_1 = -m_2\vec{v}_2 + m_2\vec{u}_2 \)
Rearrange the terms to group initial and final momenta:
\( m_1\vec{v}_1 + m_2\vec{v}_2 = m_1\vec{u}_1 + m_2\vec{u}_2 \)
This equation shows that the total linear momentum of the system after the collision (final momentum) is equal to the total linear momentum of the system before the collision (initial momentum). This derivation confirms the law of conservation of linear momentum when no external forces act on the system. When objects interact, their individual momenta might change, but the total momentum of the system stays the same. The universe always maintains a balance of momentum.
In simple words: The law of conservation of momentum comes from Newton's third law. It means that in a system where no outside forces act, the total pushing power (momentum) of all objects combined stays the same before and after they interact, like in a collision. One object might slow down while another speeds up, but their combined momentum remains constant.

🎯 Exam Tip: Remember to clearly state Newton's third law, then apply Newton's second law to relate force and momentum change for both interacting bodies, leading to the conclusion that total momentum is conserved.