Force and Laws of Motion

In physics, force is a push or pull acting on an object that can change its state of motion or shape. A force can cause a stationary object to move, change the speed of a moving object, alter its direction, or even deform it. It is a vector quantity, meaning it has both magnitude (strength) and direction. The SI unit of force is the newton (N), named after Sir Isaac Newton, who formulated the laws of motion. A force can arise from direct physical contact (contact forces, like friction or muscular force) or from a distance (non-contact forces, like gravitational, magnetic, or electrostatic force). Forces play a fundamental role in understanding and predicting how objects interact and move in our everyday lives and in the universe.

Effect of Force

A force can produce one or more of the following effects on an object:

1. Change the State of Rest or Motion
   • A force can make a stationary object start moving.
   • Example: Pushing a parked car to make it roll.
2. Change the Speed of an Object
   • A force can increase or decrease the speed of a moving object (acceleration or deceleration).
   • Example: Pressing the accelerator to speed up a car or applying brakes to slow it down.
3. Change the Direction of Motion
   • A force can alter the path of a moving object without changing its speed.
   • Example: Hitting a moving ball with a bat to change its direction.
4. Change the Shape or Size of an Object
   • A force can deform objects by stretching, compressing, bending, or breaking them.
   • Example: Squeezing a sponge or molding clay.
5. Stop a Moving Object
   • A force opposite to the motion can bring a moving object to rest.
   • Example: Friction between brakes and wheels stopping a bicycle.

Major types of force

1. Contact Forces are the forces that occur only when two objects are physically touching each other. Some of these are :
   • Muscular force (pushing or pulling an object)
   • Frictional force (resistance between two surfaces)
   • Tension in a rope or string
   • Normal reaction force from a surface
2. Non-contact Forces can act without any physical contact between objects.
   • Examples:
     • Gravitational force (Earth attracting objects)
     • Electrostatic force (attraction or repulsion between charged particles)
     • Magnetic force (attraction or repulsion between magnets

Balanced and Unbalanced Forces:

1. Balanced Forces

• Definition: When two or more forces act on an object and their net force is zero.
• Effect on Motion:
• No change in the object’s state of rest or motion.
• The object remains stationary or continues moving at the same speed in the same direction.
• Example:
• A book lying on a table – gravity pulls it down, and the table pushes it up with equal force.
• Two people pushing a box from opposite sides with equal force, so it doesn’t move

2. Unbalanced Forces

• Definition: When two or more forces act on an object and their net force is not zero.
• Effect on Motion:
• Causes the object to accelerate, decelerate, start moving, stop moving, or change direction.
• Example:
• A football being kicked – the force of the kick is greater than the resistance of the ball.
• Tug-of-war where one team pulls harder than the other, causing movement toward that side.

Comparison Table

Feature	Balanced Force	Unbalanced Force
Net Force	Zero	Not zero
Motion	No change in state of motion	Causes change in motion
Effect on Object	Object stays at rest or moves uniformly	Object accelerates, decelerates, or changes direction
Example	Book on a table	Kick to a stationary ball

Newton’s Laws of Motion

Sir Isaac Newton formulated three fundamental laws that explain the relationship between force and motion.

1. Newton’s First Law of Motion – .Law of Inertia

Inertia is the tendency of objects to resist change in their state of motion. Force is needed to change motion (start, stop, or change direction). Newton`s first law states that “An object will remain at rest or continue to move in a straight line at a constant speed unless acted upon by an external unbalanced force”.

Examples:

• A book on a table stays at rest until pushed.
• A moving car stops when brakes are applied due to friction.

2. Newton’s Second Law of Motion – .Law of Acceleration

According to this law, greater force leads to greater acceleration. Heavier objects require more force to accelerate. .The law states that “The rate of change of momentum of an object is directly proportional to the applied force and takes place in the direction of the force”. For example, kicking a football with more force makes it move faster.

Formula for law of acceleration is F=ma ,where: F = force (in Newtons, N), m = mass (in kilograms, kg) and a = acceleration (in meters per second squared, m/s²). Newton's second law of motion, when expressed in terms of momentum, states that the net force acting on an object is equal to the rate of change of its momentum. This can be written as: F = dp/dt, where F is the net force, p is the momentum, and t is time. In simpler terms, a force causes a change in momentum over time

Mass and Inertia

Mass is the measure of the amount of matter in a body. It tells us “how much stuff” is in an object. The SI unit of mass is kilogram (kg). It is a scalar quantity since it has magnitude only buy no direction. Mass is constant because it does not change with location (on Earth, Moon, or in space). The formula of mass is m=F/a. If a bag of rice has a mass of 10 kg . It remains same on whether it is on Earth or the Moon. Inertia is the property of a body by virtue of which it resists any change in its state of rest or of uniform motion in a straight line. It has a relation with mass as the mass is a measure of inertia. The greater the mass, the greater the inertia. Inertia of rest is the resistance to change from rest .Example: A passenger leans backward when a bus starts suddenly. Inertia of motion is the resistance to change in motion. .Example: A passenger leans forward when a bus stops suddenly. Inertia of direction is the resistance to change in direction of motion. .Example: Mud flying off a rotating tire tangentially.

Differences Between Mass and Inertia

Mass	Inertia
Measure of amount of matter in a body.	Property of resisting change in motion.
Scalar quantity with SI unit kg.	No unit — it's a qualitative property.
Constant everywhere.	Depends on the mass of the body.
Can be measured directly (using balance).	Cannot be measured directly — inferred from mass.

Momentum of a body

Momentum is the quantity of motion possessed by a moving body. It is the product of the body’s mass and its velocity. Therefore, the formula for momentum is p=m.v. where p = momentum , m=mass and v= velocity . It is a vector quantity as it has both magnitude and direction (same as velocity). The SI Unit of momentum is kilogram metre per second (kg⋅m/s). If an object has more mass or more velocity , it possess more momentum. So, momentum changes if either mass or velocity changes. If a body is at rest then v=0 and the momentum is also zero. Momentum plays a very important role in daily life. Even a small bullet is able to kill a person when it is fired from a gun because of its momentum due to great velocity. A person get injured severely when hit by a moving vehicle because of momentum of vehicle due to mass and velocity. S.I unit of momentum is kgm/s. It is mathematically expressed as p=kg X m/s because SI unit of mass = kg, and SI unit of velocity = m/s.

Calculation of Momentum

Example 1

What will be the momentum of a stone having mass of 100 grams when it is thrown with a velocity of 20 m/s?

Solution:

· Given: Mass m=100 grams and the velocity v=20 m/s.

&· Now convert  mass into kilograms 100/1000 which gives 0.1kg

&· Apply the momentum formula p=m.v

&· p=0.1x20

&· = 2 meter/second

Example 2:

Calculate the momentum of a stone weighing 100 grams when it is hit on a glass at the velocity of 5 m/s.

Solution:

Given: Mass (m=100 gm) and the velocity( v=5 5 m/s)

Now convert mass of stone from grams to kilograms which gives 100/1000=0.1 kg

Use the momentum formula p=m.v

=0.1x5

=o.5 meter/second is the momentum of stone

3. Newton’s Third Law of Motion – .Action and Reaction

Forces always occur in pairs. The two forces act on different objects, so they do not cancel each other. 3d law states .that “.For every action, there is an equal. and opposite reaction”. Examples: A swimmer pushes water backward, and water pushes the swimmer forward or when you jump off a boat, the boat moves backward

Application of third law of motion in Real Life

1. Walking or Running
   • Action: Your foot pushes the ground backward.
   • Reaction: The ground pushes your foot forward, moving you ahead.
2. Rowing a Boat
   • Action: The oar pushes water backward.
   • Reaction: The water pushes the boat forward.
3. Jumping off a Small Boat
   • Action: You push the boat backward with your legs.
   • Reaction: The boat pushes you forward into the water.
4. Recoil of a Gun
   • Action: The bullet is pushed forward when the gunpowder explodes.
   • Reaction: The gun is pushed backward (recoil).
5. Launching a Rocket
   • Action: The rocket expels hot gases downward.
   • Reaction: The gases push the rocket upward.
6. Bird Flying
   • Action: The bird’s wings push air downward.
   • Reaction: The air pushes the bird upward and forward.
7. Swimming
   • Action: A swimmer pushes water backward with their hands and feet.
   • Reaction: The water pushes them forward

Textbook Solutions-1

Question 1. Which of the following has more inertia

1. A rubber ball and a stone of the same size
2. A bicycle and a train
3. A five rupees coin and a one rupee coin

Answer: A train

Question 2. In the following example, try to identify the number of times the velocity of the ball changes:“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team.” Also, identify the agent supplying the force in each case.

Answer: In the event 1action is  First player kicks the football to a teammate. Here the ball goes from rest (or some speed) to a new speed and direction. In the event 2, second player kicks the ball towards the goal led to the change in the velocity change and direction. In the third event , goalkeeper collects the football which means the ball’s velocity reduces to zero when caught. In the 4^th event also velocity changes as the goalkeeper kicks the ball towards his teammate.

Summary Table

Event	Velocity Change?	Agent Supplying Force
1. First player kicks ball to teammate	Yes	First player’s leg
2. Second player kicks ball to goal	Yes	Second player’s leg
3. Goalkeeper collects ball	Yes	Goalkeeper’s hands/body
4. Goalkeeper kicks ball to teammate	Yes	Goalkeeper’s leg

Question 3. Explain why some of the leaves may get detached from a tree when we vigorously shake its branch.

Answer :Reason for this phenomenon is the inertia of leaves .Leaves are initially at rest with respect to the branch. When you shake the branch suddenly, the branch moves but the leaves tend to stay in their original position because of their inertia of rest. This relative motion between the moving branch and the stationary leaves applies a force at the point of attachment. If this force is strong enough to overcome the attachment strength (like the stalk holding the leaf), the leaves detach and fall.

Question 4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall back when it accelerates from rest?

Answer: In the case 1 - we fall forward when Bus Stops Suddenly. This is because while the bus is moving, your whole body is in motion along with it. When the bus brakes suddenly, the lower part of your body (in contact with the bus floor) stops with the bus. But the upper part of your body continues to move forward due to inertia of motion. This resulted in fall forward.

In the case 2 , when the bus is at rest, your whole body is at rest. When the bus accelerates suddenly, the lower part of your body (in contact with the bus) starts moving forward with the bus. The upper part of your body tends to remain at rest due to inertia of rest. The result is you fall backward.

Question 5. If action is always equal to the reaction, explain how a horse can pull a cart.
Answer:

Newton’s Third Law says, For every action, there is an equal and opposite reaction. So when a horse pulls on the cart, the cart pulls back on the horse with equal force. It might seem like these forces cancel each other — so how does the cart move?. The Key Idea here is that the action and reaction act on different bodies. The pull of the horse on the cart and the pull of the cart on the horse are equal and opposite, but they act on different bodies. Forces on different bodies do not cancel each other. The horse pushes backward on the ground with its hooves which reeult in the ground pushes the horse forward with an equal and opposite force (reaction). This forward force on the horse is what moves both the horse and the cart forward.

Question 6. Explain Why is it difficult for a firemen to hold a hose, which ejects large amounts of water at a high velocity.
Answer: A fire hose ejects a large quantity of water at high speed to put out a fire. Firefighters often struggle to hold it steady because, according to Newton’s third law of motion, action is water pushed forward out of the hose at high velocity and the reaction is the water pushes the hose backward with an equal and opposite force. This reaction force is called recoil force. It is hard to hold the hose because of large mass flow rate and the high velocity of water/ second . The faster the water is ejected, the greater its momentum change per second.

Question 7. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 ms^-1, calculate the initial recoil velocity of the rifle.

Answer:

• Given, mas of the gun M = 4 kg, mass of bullet M = 50 g = 50 × 10^-3 Initial velocity of the bullet is v = 35ms^-1.
• Let recoil velocity of the gun be vms^-1. Before firing the bullet, both gun and bullet were in the state of inertia , therefore, the total momentum of gun and bullet=
• After firing, the momentum of bullet = mv, and the momentum of gun = Mv and the total momentum of bullet and gun after firing = mv + Mv.
• Since, no external force applied to the system, the total momentum after firing equals the total momentum before firing which is expressed as mv + Mv = 0 OR Mv = -mv.
• Now apply the formula f1 v=mv/M
• = 50x10^-3x35/4
• =-0.4375 ms^-1
• The recoil velocity is always expressed in negative sign hence the net velocity is =-0.4375 ms^-1

Question 8. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2ms^-1 and 1ms^-1 respectively. They collide and after the collision, the first object moves at a velocity of 1.67ms^-1. Determine the velocity of the second object.

Given:
m1 = 100 g = 0.10 kg, m2 = 200 g = 0.20 kg, u1 = 2 m/s, u2 = 1 m/s, v1 = 1.67 m/s
Find: v2 (final velocity of second object)

Before collision:
p_initial = m1 × u1 + m2 × u2
p_initial = (0.10 × 2) + (0.20 × 1)
p_initial = 0.20 + 0.20 = 0.40 kg·m/s

After collision:
p_final = m1 × v1 + m2 × v2
p_final = (0.10 × 1.67) + (0.20 × v2)
p_final = 0.167 + 0.20 × v2

Equating p_initial and p_final:
0.167 + 0.20 × v2 = 0.40
0.20 × v2 = 0.40 – 0.167
0.20 × v2 = 0.233
v2 = 0.233 ÷ 0.20
v2 = 1.165 ms-¹

Let velocity of second object after collision is v₂ms^-1 Since there is no external force on the system.

 

Textbook Solutions-2

Question 1. An object experiences a zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no provide a reason.
Answer: Yes, it is possible. According to Newton’s First Law of Motion, if the external unbalanced force on an object is zero, the object will maintain its state of motion — either at rest or moving with constant velocity in a straight line. Conditions for the velocity are the magnitude which must remain constant (no speeding up or slowing down) and the direction should remain unchanged (the object moves in a straight line).The reason for this is that a non-zero, constant velocity means there is no change in momentum, so no net (unbalanced) force is required.

Question 2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer:

When a carpet is beaten with a stick, the carpet moves suddenly due to the force applied. The dust particles embedded in the carpet have the inertia of rest, meaning they tend to remain in their original position (at rest) even when the carpet moves. Because of this, the dust gets separated from the moving carpet and falls out. This is based on the first law of motion -Newton.

Question 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:

It is advised to tie luggage kept on the roof of a bus with a rope to prevent it from falling off when the bus starts suddenly, stops suddenly, or takes a sharp turn. When the bus starts suddenly, the luggage tends to remain at rest due to inertia of rest and may slide backward. When the bus stops suddenly, the luggage tends to keep moving forward due to inertia of motion and may slide forward. During a turn, the luggage tends to continue moving in a straight line due to inertia of direction, which can make it slide sideways.

Question 4. A batsman hits a cricket ball which then rolls on level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because.

Answer: The ball slows to a stop because frictional force between the ball and the ground, along with air resistance, acts in the direction opposite to its motion. These forces are unbalanced external forces that gradually reduce the ball’s velocity to zero. If there were no friction and no air resistance, the ball would keep moving at a constant speed in a straight line, as stated by Newton’s First Law of Motion.

Question 5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tons
(Hint: 1 ton = 1000 kg.)

Given:
Initial velocity (u) = 0 m/s
Distance (s) = 400 m
Time (t) = 20 s
Mass (m) = 7 tons = 7000 kg

Step 1: Find acceleration using equation of motion:
s = u·t + (1/2)·a·t²
400 = 0 × 20 + (1/2)·a·(20²)
400 = (1/2)·a·400
400 = 200·a
a = 400 ÷ 200
a = 2 m/s²

Step 2: Find force using Newton’s Second Law:
F = m·a
F = 7000 × 2
F = 14000 N

Answer: Acceleration = 2 m/s² and Force = 14000 N (Newtons)

Question 6. A stone of 1 kg is thrown with a velocity of 20ms^-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Answer: Let mass of the stone m = 1 kg, initial velocity u = 20 ms^-1 and the ^final velocity v = 0 (Therefore the stone comes to rest distance travelled S = 50 m). From third equation of motion
v² = u² + 2as
(0)² = (20)² + 2a(50)
100 a = -400
∴ a = -4ms^-2

The negative sign indicates there is relation in the motion of stone.
Force of friction between stone and ice = Force required to stop the stone
= ∴ ma = 1 × -4 =

-4N or 4N.

Question 7. An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000N then calculate
a) the net accelerating force. b) The acceleration of the train and
c) the force of wagon 1 on wagon 2

Given

• Engine mass = 8000 kg
• Wagon mass = 2000 kg each, number of wagons = 5 → total wagon mass = 5 × 2000 = 10000 kg
• Total train mass = 8000 + 10000 = 18000 kg
• Pulling force by engine = 40000 N
• Total track/frictional resistance = 5000 N

1. a) Net accelerating force:
   Net force = Pulling force − Friction = 40000 − 5000 = 35000 N
2. b) Acceleration of the train:
   F = m a ⇒ a = F / m = 35000 / 18000 = 35/18 ≈94 m/s²
3. c) Force of wagon 1 on wagon 2:
   Consider the trailing set (wagons 2–5). Its mass = 4 × 2000 = 8000 kg.
   Contact force must (i) accelerate these wagons and (ii) overcome their share of friction.

Assume the given 5000 N friction is distributed proportional to mass.
Trailing mass fraction = 8000 / 18000 = 4/9
Trailing friction = 5000 × (4/9) = 20000/9 ≈ 2222.22 N

Required contact force:
F1→2 = (mass of wagons 2–5) × a + (their friction)
= 8000 × (35/18) + 20000/9
= 140000/9 + 20000/9
= 160000/9 ≈ 17777.78 N

Answers:
a) Net accelerating force = 35000 N
b) Acceleration = 35/18 ≈ 1.94 m/s²
c) Force of wagon 1 on wagon 2 ≈ 1.78 × 104 N (about 17778 N)

Question 8. An automobile vehicle has a mass of 1500 Kg. what must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms^-2
Answer: Given the (Mass) m = 1500 Kg and the acceleration a = -1.7 ms^-2
From Newton’s second law of motion
F = ma
= 1500 × (-1.7)
= -2550 N.

Question 9. What is the momentum of an object of mass m, moving with a velocity V?
(a) (mv)²
(b) mv²
(c) 1/2 mv²
(d) mv.

>Answer:
(d) P = mv.

Question 10. Using a horizontal force of 200N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer:The cabinet will move across the floor with constant velocity if there is no net external force of 200N that should be applied on the cabinet in opposite direction.
Thus, the frictional force = 200N
(Frictional force always acts in the direction opposite to the direction of motion)

Question 11. Two objects, each of mass 1.5 kg are moving in the same straight line but in opposite directions. The velocity of each object is 2.5ms^-1 before the collision during which they stick together is what will be the velocity of combined object after collision?

Given:
Mass of object 1, m1 = 1.5 kg
Mass of object 2, m2 = 1.5 kg
Velocity of object 1 before collision, u1 = +2.5 m/s
Velocity of object 2 before collision, u2 = -2.5 m/s (negative because opposite direction)

Principle: Conservation of linear momentum

Before collision, total momentum:
p_initial = m1 × u1 + m2 × u2
= (1.5 × 2.5) + (1.5 × -2.5)
= 3.75 - 3.75 = 0

Let v be the velocity of the combined mass after collision and the total mass after collision = m1 + m2 = 3.0 kg

After collision, total momentum:
p_final = (m1 + m2) × v = 3.0 × v

By conservation of momentum:
p_initial = p_final
0 = 3.0 × v

Therefore: v = 0 m/s and the combined object will remain at rest after the collision.

Question 12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer: The student’s logic is incorrect because the two forces mentioned in Newton’s third law act on different objects, not on the same object. When you push on the truck, your hands exert a force on the truck, and the truck exerts an equal and opposite force on your hands. These forces are an action–reaction pair but they do not cancel out because they are applied to different bodies. The reason the truck does not move is that the force you apply is not enough to overcome the friction between the truck’s tyres and the ground (and possibly its large inertia due to its massive mass). If your applied force exceeded the frictional force, the truck would move.

Question 13. A hockey ball of mass 200 g travelling at 10ms^-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5ms^-1 Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick. Answer:Given:
Mass of hockey ball = 200 g = 0.2 kg, Initial velocity = 10 m/s and the final velocity = -5 m/s (negative because it moves in the opposite direction after being hit)

Change in momentum = m × (v - u)
= 0.2 × (-5 - 10)
= 0.2 × (-15)
= -3 kg·m/s. therefore, change in momentum is -3 kg·m/s. The negative sign indicates a reversal in the direction of motion.

Question 14. A bullet of mass 10 g travelling horizontally with a velocity of 150ms’1 strikes a stationary wooden block and comes to rest in 0.03s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.

Given:
Mass of bullet, m = 10 g = 0.01 kg
Initial velocity, u = 150 m/s
Final velocity, v = 0
Time taken to stop, t = 0.03 s

Step 1: Find acceleration
v = u + a t
0 = 150 + a(0.03)
a = -150 / 0.03 = -5000 m/s²

Step 2: Find distance of penetration
s = u t + (1/2) a t²
s = 150(0.03) + (1/2)(-5000)(0.03)²
s = 4.5 - 2.25 = 2.25 m

Step 3: Find magnitude of force
F = m a
F = 0.01 × (-5000) = -50 N
Magnitude = 50 N

Answer: Distance of penetration = 2.25 meters and magnitude of force = 50 N

Question 15. An object of mass 1Kg travelling in straight line with a velocity of 10ms’1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact Also calculates the velocity of the combined object.
Answer:
Mass of object m₁ = 1 Kg,
Velocity, u₁ = 10ms^-1
Mass of wooden block m₂ = 5Kg,
velocity u₂ = 0 (since wooden block is rest)
= 1 × 10 + 5 × 0 = 10 Kgms^-1
According to law of conservation of momentum
Momentum after impact = Momentum before impact
(m₁ + m₂)v = 10
Where v = velocity of combined object
(1 + 5) v = 10
v = 10/6 = 1.67ms^-1.

Question 16. An object of mass 100 Kg is accelerated uniformly from a velocity of 5ms^-1 to 8ms^-1 in 6s. Calculate the initial and final momentum of the object. Also find the magnitude of the force exerted on the object.
Given: Mass, m = 100 kg, Initial velocity, u = 5 m/s, Final velocity, v = 8 m/s and Time, t = 6 s

Step 1: Initial momentum
p1 = m × u
p1 = 100 × 5 = 500 kg·m/s

Step 2: Final momentum
p2 = m × v
p2 = 100 × 8 = 800 kg·m/s

Step 3: Force exerted
F = (p2 − p1) / t
F = (800 − 500) / 6
F = 300 / 6 = 50 N

Answer:Initial momentum = 500 kg·m/s, Final momentum = 800 kg·m/s and Force exerted = 50 N

Question 17. Akhtar, Kiran, and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the momentum of the motor car (because the change in the velocity of the insect was much more than that of the motor car). Akthar said that since the motor car was moving with larger velocity, it exerted a larger force on the insect. And as a result, the insect died. Rahul while putting an entirely new explanation said that both the motor car and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Answer: According to Newton’s third law of motion, when two bodies interact, the force exerted by the first body on the second is equal in magnitude and opposite in direction to the force exerted by the second body on the first. Therefore, when the insect hits the moving motor car:

• Rahul’s statement is correct — the insect and the motor car exert equal and opposite forces on each other, and both experience an equal magnitude of change in momentum (impulse), but in opposite directions.
• Kiran’s statement is incorrect — the change in momentum is the same for both, but since the insect’s mass is very small, this change causes a large change in velocity for the insect, while the motor car’s large mass results in a negligible change in velocity.
• Akhtar’s statement is misleading — the car does not exert a “larger” force; the force is the same on both, but the insect suffers greater damage due to its small mass and inability to withstand the large acceleration caused by the force.

Question 18. How much momentum will a dumb-bell of mass 10 Kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10ms^-2.
Given: m = 10 Kg, u = 0, acceleration s = 80 cm = 0.8 m.
Velocity a = 10 ms^-2
v² = u² + 2as
v² = 0 + 2 × 10 × 0.8 =16
∴ v = √16 = 4ms^-1
Momentum of dumb-bell just before it touches the floor
P = mv = 10 × 4
= 40 Kgms^-1.
When the dumbbell touches the floor its velocity becomes zero. Thus the total momentum of the dumbbell is transferred to the floor. Hence the momentum transferred to floor = 40 Kgms^-2.