# CBSE Class 9 Science Physics Force And Law Of Motion Notes Set C

Download CBSE Class 9 Science Physics Force And Law Of Motion Notes Set C in PDF format. All Revision notes for Class 9 Physics have been designed as per the latest syllabus and updated chapters given in your textbook for Physics in Standard 9. Our teachers have designed these concept notes for the benefit of Grade 9 students. You should use these chapter wise notes for revision on daily basis. These study notes can also be used for learning each chapter and its important and difficult topics or revision just before your exams to help you get better scores in upcoming examinations, You can also use Printable notes for Class 9 Physics for faster revision of difficult topics and get higher rank. After reading these notes also refer to MCQ questions for Class 9 Physics given our website

## Force And Law Of Motion Class 9 Physics Revision Notes

Class 9 Physics students should refer to the following concepts and notes for Force And Law Of Motion in standard 9. These exam notes for Grade 9 Physics will be very useful for upcoming class tests and examinations and help you to score good marks

### Force And Law Of Motion Notes Class 9 Physics

FORCE AND LAWS OF MOTION

FORCE

A force is anything that can cause a change to objects. Forces can:

• change the shape of an object

• move or stop an object

• change the direction of a moving object.

A force can be classified as either a contact force or a non-contact force.

A contact force must touch or be in contact with an object to cause a change. Examples of contact forces are:

• the force that is used to push or pull things, like on a door to open or close it

• the force that a sculptor uses to turn clay into a pot

• the force of the wind to turn a windmill

A non-contact force does not have to touch an object to cause a change. Examples of noncontact forces are:

• the force due to gravity, like the Earth pulling the Moon towards itself

• the force due to electricity, like a proton and an electron attracting each other

• the force due to magnetism, like a magnet pulling a paper clip towards itself

The unit of force is the newton (symbol N). This unit is named after Sir Isaac Newton who first defined force. Force is a vector quantity and has a magnitude and a direction.

EFFECT OF FORCE:

1. Force can make a stationary body in motion. For example a football can be set to move by kicking it, i.e. by applying a force.

2. Force can stop a moving body – For example by applying brakes, a running cycle or a running vehicle can be stopped.

3. Force can change the direction of a moving object. For example; By applying force, i.e. by moving handle the direction of a running bicycle can be changed. Similarly by moving steering the direction of a running vehicle is changed.

4. Force can change the speed of a moving body – By accelerating, the speed of a running vehicle can be increased or by applying brakes the speed of a running vehicle can be decreased.

5. Force can change the shape and size of an object. For example -– By hammering, a block of metal can be turned into a thin sheet. By hammering a stone can be broken into pieces.

Forces are can also divided into two types:

1. Balanced Forces

2. Unbalanced Forces

BALANCED FORCES

If the resultant of applied forces is equal to zero, it is called balanced forces. Example : - In the tug of war if both the teams apply similar magnitude of forces in opposite directions, rope does not move in either side. This happens because of balanced forces in which resultant of applied forces become zero. Balanced forces do not cause any change of state of an object. Balanced forces are equal in magnitude and opposite in direction. Balanced forces can change the shape and size of an object. For example - When forces are applied from both sides over a balloon, the size and shape of balloon is changed.

UNBALANCED FORCES

If the resultant of applied forces are greater than zero the forces are called unbalanced forces. An object in rest can be moved because of applying balanced forces. Unbalanced forces can do the following:

• Move a stationary object.

• Increase the speed of a moving object.

• Decrease the speed of a moving object.

• Stop a moving object.

• Change the shape and size of an object.

LAWS OF MOTION:

NEWTON’S LAWS OF MOTION:

• Newton's First Law of Motion - Any object remains in the state of rest or in uniform motion along a straight line, until it is compelled to change the state by applying external force.

• Newton's Second Law of Motion - The rate of change of momentum is directly proportional to the force applied in the direction of force.

• Newton's Third Law of Motion - There is an equal and opposite reaction for evrey action

NEWTON’S FIRST LAW OF MOTION:

Any object remains in the state of rest or in uniform motion along a straight line, until it is compelled to change the state by applying external force. Explanation: If any object is in the state of rest, then it will remain in rest untill a exernal force is applied to change its state. Similarly an object will remain in motion untill any external force is applied over it to change its state. This means all objects resist to in changing their state. The state of any object can be changed by applying external forces only.

NEWTON’S FIRST LAW OF MOTION IN EVERYDAY LIFE:

a. A person standing in a bus falls backward when bus is start moving suddenly. This happens because the person and bus both are in rest while bus is not moving, but as the bus starts moving the legs of the person start moving along with bus but rest portion of his body has tendency to remain in rest. Because of this person falls backward; if he is not alert.

b. A person standing in a moving bus falls forward if driver applies brakes suddenly. This happens because when bus is moving, the person standing in it is also in motion along with But when driver applies brakes the speed of bus decreases suddenly or bus comes in the state of rest suddenly, in this condition the legs of the person which are in the contact with bus come in rest while the rest parts of his body have tendency to remain in motion.Because of this person falls forward if he is not alert.

MASS AND INERTIA:

The property of an object because of which it resists to get disturbed its state is called Inertia. Inertia of an object is measured by its mass. Inertia is directly proportional to the mass. This means inertia increases with increase in mass and decreases with decrease in mass. A heavy object will have more inertia than lighter one.

In other words, the natural tendency of an object that resists the change in state of motion or rest of the object is called inertia.

Since a heavy object has more inertia, thus it is difficult to push or pul a heavy box over the ground than lighter one.

1. Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five-rupees coin and a one-rupee coin?

Ans. Inertia is the measure of the mass of the body. The greater is the mass of the body; the greater is its inertia and vice-versa.

(a) Mass of a stone is more than the mass of a rubber ball for the same size. Hence, inertia of the stone is greater than that of a rubber ball.

(b) Mass of a train is more than the mass of a bicycle. Hence, inertia of the train is greater than that of the bicycle.

(c) Mass of a five rupee coin is more than that of a one-rupee coin. Hence, inertia of the five rupee coin is greater than that of the one-rupee coin.

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. Ans. The velocity of the ball changes four times. As a football player kicks the football, its speed changes from zero to a certain value. As a result, the velocity of the ball gets changed. In this case, the player applied a force to change the velocity of the ball. Another player kicks the ball towards the goal post. As a result, the direction of the ball gets changed. Therefore, its velocity also changes. In this case, the player applied a force to change the velocity of the ball. The goalkeeper collects the ball. In other words, the ball comes to rest. Thus, its speed reduces to zero from a certain value. The velocity of the ball has changed. In this case, the goalkeeper applied an opposite force to stop/change the velocity of the ball. The goalkeeper kicks the ball towards his team players. Hence, the speed of the ball increases from zero to a certain value. Hence, its velocity changes once again. In this case, the goalkeeper applied a force to change the velocity of the ball.

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

Ans. Some leaves of a tree get detached when we shake its branches vigorously. This is because when the branches of a tree are shaken, it moves to and fro, but its leaves tend to remain at rest. This is because the inertia of the leaves tend to resist the to and fro motion. Due to this reason, the leaves fall down from the tree when shaken vigorously.

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

Ans. Due to the inertia of the passenger. Every body tries to maintain its state of motion or state of rest. If a body is at rest, then it tries to remain at rest. If a body is moving, then it tries to remain in motion. In a moving bus, a passenger moves with the bus. As the driver applies brakes, the bus comes to rest. But, the passenger tries to maintain his state of motion. As a result, a forward force is exerted on him. Similarly, the passenger tends to fall backwards when the bus accelerates from rest. This is because when the bus accelerates, the inertia of the passenger tends to oppose the forward motion of the bus. Hence, the passenger tends to fall backwards when the bus accelerates forward.

MOMENTUM

Momentum is the power of motion of an object.

The product of velocity and mass is called the momentum. Momentum is denoted by ‘p’.

Therefore, momentum of the object = Mass x Velocity. Or, p = m x v

Where, p = momentum, m = mass of the object and v = velocity of the object.

NEWTON'S SECOND LAW OF MOTION

Newton's second Law of Motion states that The rate of change of momentum is directly proportional to the force applied in the direction of force.

For example; when acceleration is applied on a moving vehicle, the momentum of the vehicle increases and the increase is in the direction of motion because the force is being applied in the direction of motion. On the other hand, when brake is applied on the moving vehicle, the momentum of the vehicle decreases and the decrease is in the opposite direction of motion because the force is being applied in the opposite direction of motion.

Mathematical formulation of Newton’s Second Law of Motion:

Let mass of an moving object = m.

Let the velocity of the object changes from ‘u’ to ‘v’ in the interval of time ‘t’. This means,

Initial velocity of the object = u. Final velocity of the object = v.

We know that momentum (p) = Mass x velocity

Therefore,

Momentum (p) of the object at its initial velocity u = m x u = mu

Momentum (p) of the object at its final velocity v = m x v = mv

The change in momentum = mv – mu

Rate of change of momentum =( mv mu) / ………………… (i)

According to the Newton’s Second Law of motion force is directly proportional to the rate of change of momentum.

This means, Force ∝ Rate of change of moentum

After substituting the value of rate of change of momentum from equation (i) we get.

Force = mv mu / t

⇒ F =m(u) / t

⇒ F = ma ( u / t )

⇒ k.ma ……………. (ii)

where, k is proportionality constant.

Since, 1 unit force is defined as the mass of 1kg object produces the acceleration of 1m/s2

Therefore, 1 unit of Force = k x 1 kg x 1m/s2

Thus k = 1.

By substituting the value of ‘k = 1’ in equation (ii) we get

F = m.a ……………… (iii)

⇒ Force = mass × acceleration

Thus Newton’s Second Law of Motion gives the relation between force, mass and acceleration of an object.

According to the relation obtained above, Newton’s Second Law can be modified as follows:

The product of mass and acceleration is the force acting on the object.

The SI unit of Force: Newton (N)

Since Force = Mass x Acceleration

The unit of mass = kg and The unit of acceleration = m/s2

If force, mass and acceleration is taken as 1 unit.

Therefore,

1 Newton (N) = 1kg x 1m/s2

Thus, Newton (N) = kg m/s2

Equation (iii) can be also written as

F/………………… (iv)

This equation is the form of Newton’s Second Law of Motion.

According to this equation, Newton’s Second Law of Motion can also be stated as follow:

The acceleration produced by a moving body is directly proportional to the force applied over it and inversely proportional to the mass of the object.

From the above relation it is clear that

Acceleration increases with increase in force and vice versa.

Acceleration decreases with increase in mass and vice versa.

That’s why a small vehicle requires less force to attain more acceleration while a heavy vehicle requires more force to get the same acceleration.

NEWTON’S SECOND LAW OF MOTION IN EVERYDAY LIFE:

a. A fielder pulls his hand backward; while catching a cricket ball coming with a great speed, to reduce the momentum of the ball with a little delay. According to Newton’s Second Law of Motion; rate of change of momentum is directly proportional to the force applied in the direction.

bWhile catching a cricket ball the momentum of ball is reduced to zero when it is stopped after coming in the hands of fielder. If the ball is stopped suddenly, its momentum will be reduced to zero instantly. The rate of change in momentum is very quick and as a result, the player’s hand may get injured. Therefore, by pulling the hand backward a fielder gives more time to the change of momentum to become zero. This prevents the hands of fielder from getting hurt.

NUMERICAL

1. What is acceleration produced by a force of 12 Newton exerted on an object of mass 3 kg?

2. What force would be needed to produce an acceleration of 4m/s2 on a ball of mass 6 kg?

3. A force of 5 N gives a mass m1, an acceleration of 8 m/s2, and a mass m2, an acceleration of 24 m/s2. What acceleration would give if both the masses are tied together?

4. Calculate the force required to impart a car a velocity of 30m/s in 10 seconds. The mass of the car is 1500 kg.

5. A motorcycle is moving with a velocity of 90km/h and it takes 5 seconds to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcycle if its mass along with the rider is 200kg.

6. What is the momentum of a man of mass 75kg when he walks with a velocity of 2m/s?

7. What would be the force required to produce an acceleration of 2m/s2 in a body of mass 12 kg? What would be the acceleration it the force were doubled?

8. A man pushes a box of mass 50 kg with a force of 80N.What will be the acceleration of the box? What would be the acceleration if the mass were doubled?

9. A certain force exerted for 1.2 second raises the speed of an object from 1.8m/s to 4.2 m/s. Later, the same force is applied for 2 second. How much does the speed change in 2 second?

10. A constant force acts on an object of mass 5 kg for duration of 2 second. It increases the object’s velocity from 3cm/s to 7m/s. Find the magnitude of the applied force. Now if the force were applied for a duration of 5 seconds, what would be the final velocity of the object?

11. A motorcar is moving with a velocity of 108km/h and it takes 4 seconds to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar it its mass along with the passengers is 1000 kg.

12. A force of 5 N gives a mass m1, an acceleration of 10 m/s2, and a mass m2, an acceleration of 20 m/s2. What acceleration would it give if both the masses were tied together?

13. For how long should a force of 100 N act on a body of mass 20 kg so that it acquires a velocity of 100 m/s?

14. A 150 g ball traveling at 30m/s strikes the palm of a players hand and is stopped in 0.06 sec. Calculate the force exerted by the ball on the hand.

15. A body of mass 1 kg is kept at rest. A constant force of 6.0 N starts acting on it . Find the time taken by the body to move through a distance of 12m.

16. A force of 4 N acts on a body of mass 2 kg for 4 s. Assuming that the body to be initially at rest, find (i) its velocity when the force stops acting (ii) the distance covered in 10 s after the force starts acting.

17. A feather of mass 10 g is dropped from a height. It is found to fall down with a constant velocity. What is the net force acting on it

18. A hockey ball of mass 200g traveling from west to east at 10m/s is struck by a hockey stick. As a result, then ball gets turned back and now has a speed of 5m/s. If the ball and hockey stick were in contact for 0.2 s, calculate (i) initial and final momentum of the ball (ii) rate of change of momentum of the ball (iii) the force exerted by hockey stick on the ball.

19. A stone of mass 500 g is thrown with a velocity of 20m/s across the frozen surface of a lake. It comes to rest after traveling a distance of 0.1 km. Calculate force of friction between the stone and frozen surface of lake.

20. A body starts from rest and rolls down a hill with a constant acceleration. If its travels 400 m in 20 seconds, calculate the force acting on the body if its mass is 10kg.

21. The velocity time graph of a ball of mass 20g moving along a straight line on a long table is given in below figure. How much force does the table exert on the ball to bring it to rest? 22. The speed time graph of a ball of mass 30g moving along a straight line is shown in below figure. Calculate the opposing force that brings the ball to rest. What will be the percentage change in momentum of a body when both its mass and velocity are doubled?

23. A force of 2 N gives a mass m1 an acceleration of 5m/s2 and a mass m2, an accelerated of 7m/s2. What acceleration would be produced if both the masses are tied together?

24. A body of mass 2 kg moving with a velocity of 10m/s is brought to rest in 5 sec. Calculate the stopping force applied.

NEWTON'S THIRD LAW OF MOTION

Newton’s Third Law of Motion states that there is always reaction for every action in opposite direction and of equal magnitude.

Explanation: Whenever a force is applied over a body, that body also applies same force of equal magnitude and in opposite direction.

Example –

(a) Walking of a person - A person is able to walk because of the Newton’s Third Law of Motion. During walking, a person pushes the ground in backward direction and in the reaction the ground also pushes the person with equal magnitude of force but in opposite direction. This enables him to move in forward direction against the push.

(b) Recoil of gun - When bullet is fired from a gun, the bullet also pushes the gun in opposite direction, with equal magnitude of force. This results in gunman feeling a backward push from the butt of gun.

(c) Propulsion of a boat in forward direction – Sailor pushes water with oar in backward direction; resulting water pushing the oar in forward direction. Consequently, the boat is pushed in forward direction. Force applied by oar and water are of equal magnitude but in opposite directions.

CONSERVATION OF MOMENTUM –

Law of Conservation of Momentum – The sum of momenta of two objects remains same even after collision.

In other words, the sum of momenta of two objects before collision and sum of momenta of two objects after collision are equal.

Mathematical Formulation of Conservation of Momentum:

Suppose that, two objects A and B are moving along a straight line in same direction and the velocity of A is greater than the velocity of B. Let the initial velocity of A=uA Let the initial velocity of B= uB Let the mass of A= mA

Let the mass of B=mB

Let both the objects collide after some time and collision lasts for ' t' second.

Let the velocity of A after collision= vA

Let the velocity of B after collision= vB

We know that, Momentum = Mass x Velocity

Therefore,

Momentum of A (FA) before collision = mA x uA

Momentum of B (FB) before collision = mB x uB

Momentum of A after collision = mA x vA

Momentum of B after collision = mB x vB

Now, we know that Rate of change of momentum = mass x rate of change in velocity

⇒ Rate of change of momentum =mass * Change in velocity / time

Therefore, rate of change of momentum of A during collision, FABma (vA uA / t)

Similarly, the rate of change of momentum of B during collision, FABmB ( vB uB / t )

Since, according to the Newton’s Third Law of Motion, action of the object A (force exerted by A) will be equal to reaction of the object B (force exerted by B). But the force exerted in the course of action and reaction is in opposite direction.

Therefore,

FAB = -FBA

ormA( vA uA / t ) = -mBvB uB / t )

⇒ mA(vA uA  ) = -mBvB uB

⇒ mAvA  mAuA  = -mBvB  mBuB

⇒ mAvA  mBvB  mAuA  mBuB ……………. (i)

Above equation says that total momentum of object A and B before collision is equal to the total momentum of object A and B after collision. We observe that the total momentum of the two balls remains unchanged or conserved provided no other external force acts. As a result of this ideal collision experiment, we say that the sum of momenta of the two objects before collision is equal to the sum of momenta after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision.

CONSERVATION OF MOMENTUM – PRACTICAL APPLICATION

• Bullet and Gun – When bullet is fired from a gun, gun recoils in the opposite direction of bullet. The momentum of bullet is equal to momentum of gun. Since, the bullet is has very small mass compared to the gun, hence velocity of bullet is very high compared to the recoil of gun. In the case of firing of bullet, law of conservation of momentum is applied as usual.

• When a cricket ball is hit by bat, the Law of Conservation of Momentum is applied.

NUMERICAL

1. The velocity of a body of mass 10kg increases from 4m/s to 8m/s when a force acts on it for 2s. (a) What is the momentum before and after the force acts? (b) What is the gain in momentum per second? (c) What is the value of the force?

2. A boy pushes a wall with a force of 20N. What is the magnitude and direction of the force experienced by the body?

3. A 20 g bullet is shot from a 5 kg gun with a velocity of 500m/s. What is the speed of the recoil of the gun?

4. A 10 g bullet is shot from a 5 kg gun with a velocity of 400m/s. What is the speed of the recoil of the gun?

5. When two bodies A and B interact with each other, A exerts a force of 10N on B, towards east. What is the force exerted by B on A?

6. A man weighting 60kg runs along the rails with a velocity of 18km/h and jumps into a car of mass 1 quintal standing on the rails. Calculate the velocity with which car will start traveling along the rails.

7. The car A of mass 1500kg, traveling at 25m/s collides with another car B of amss 1000kg traveling at 15m/s in the same direction. After collision, the velocity of car. A becomes 20m/s. Calculate the velocity of car B after collision.

8. A bullet of mass 10g is fired from a gun of mass 6 kg with a velocity of 300m.s. Calculate the recoil velocity of the gun.

9. A bullet of mass 50g is fired from a gun of mass 6 kg with a velocity of 400m.s. Calculate the recoil velocity of the gun.

10. A bullet of mass 10g is moving with a velocity of 400m/s gets embedded in a freely suspended wooden block of mass 900g. What is the velocity acquired by the block?

11. A gun of mass 3 kg fires a bullet of mass 30g. The bullet takes 0.003s to move through the barrel of the gun and acquires a velocity of 100m/s. Calculate (i) the velocity with which the gun recoils (ii) the force exerted on gunman due to recoil of the gun.

12. A heavy car of mass 200kg traveling at 10m/s has a head on collision with a sports car B of mass 500kg. If both cars stop dead on colliding, what was the velocity of car B?

13. A machine gun fires 25h bullet at the rate of 600 bullets per minute with a speed of 200m/s. Calculate the force required to keep the gun in position.

14. A bullet of mass 20g is moving with a velocity of 300m/s gets embedded in a freely suspended wooden block of mass 880g. What is the velocity acquired by the block?

15. A girl of mass 50kg jumps out of a rowing boat of mass 300kg on to the bank with a horizontally velocity of 3m/s. With what velocity does the boat begin to move backwards?

16. A truck of mass 2500kg moving at 15m/s collides with a car of mass 1000kg moving with at 5m/s in the opposite direction. What is their common velocity?

17. A bullet of mass 20 g is fired horizontally with a velocity of 150m/s from a pistol of mass 2kg. What is the recoil velocity of the pistol?

18. A body of mass 60kg running at 3m/s jumps on to a trolley of mass 140kg moving with a velocity of 1.5m/s in the same direction. What is their common velocity?

19. A girl of mass 40kg jumps with a horizontal velocity of 5m/s onto a stationary cart with frictionless wheels. The mass of the cart is 3kg. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction?

20. Two hockey players of opposite teams, while trying to hit a hockey ball on the ground collide and immediately become entangled. One has a mass of 60kg, and was moving with a velocity 5m/s, while the other has a mass 55kg and was moving faster with a velocity of 6m/s towards the first player. In which direction and with what velocity will they move after they become entangled? Assume that the frictional force acting between the feet of the two players and ground is negligible.

INTEXT QUESTIONS PAGE NO. 126

1. If action is always equal to the reaction, explain how a horse can pull a cart.

Ans. A horse pushes the ground in the backward direction. According to Newton’s third law of motion, a reaction force is exerted by the Earth on the horse in the forward direction. As a result, the cart moves forward.

2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Ans. Due to the backward reaction of the water being ejected. When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then a reaction force is exerted on him by the ejecting water in the backward direction. This is because of Newton’s third law of motion. As a result of the backward force, the stability of the fireman decreases. Hence, it is difficult for him to remain stable while holding the hose.

3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s−1. Calculate the initial recoil velocity of the rifle.

Ans. Mass of the rifle, m1 = 4kg

Mass of the bullet, m2 = 50g = 0.05 kg

Recoil velocity of the rifle = v1

Bullet is fired with an initial velocity, v2 = 35m/s

Initially, the rifle is at rest.

Thus, its initial velocity, = 0

Total initial momentum of the rifle and bullet system =(m1 + m2)v = 0

Total momentum of the rifle and bullet system after firing

= m1v1 + m2v2 = 4(v1) + 0.05 x 35 = 4v1 + 1.75

According to the law of conservation of momentum:

Total momentum after the firing = Total momentum before the firing

4v1 + 1.75 = 0 ⇒ v1 = - 1.75/4 = -0.4375ms

The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.

4. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s−1 and 1 m s−1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m s−1. Determine the velocity of the second object.

Ans.

Mass of one of the objects, m1 = 100 g = 0.1 kg

Mass of the other object, m2 = 200 g = 0.2 kg

Velocity of m1 before collision, v1 = 2 m/s

Velocity of m2 before collision, v2 = 1 m/s

Velocity of m1 after collision, v3 = 1.67 m/s

Velocity of m2 after collision = v4

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

∴ m1v1 + m2v2 = m3v3 + m4v4

⇒ (0.1) x 2 + (0.2) x 1 = (0.1) x 1.67 + (0.2) x v4

⇒ 0.4 = 0.167 + 0.2v4

∴ v4 = 1.165 m/s

Hence, the velocity of the second object becomes 1.165 m/s after the collision.

EXERCISE QUESTIONS PAGE NO. 128

1. An object experiences a net 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.

Ans. Yes. Even when an object experiences a net zero external unbalanced force, it is possible that the object is travelling with a non-zero velocity. This is possible only when the object has been moving with a constant velocity in a particular direction. Then, there is no net unbalanced force applied on the body. The object will keep moving with a non-zero velocity. To change the state of motion, a net non-zero external unbalanced force must be applied on the object.

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

Ans. Inertia of an object tends to resist any change in its state of rest or state of motion. When a carpet is beaten with a stick, then the carpet comes to motion. But, the dust particles try to resist their state of rest. According to Newton’s first law of motion, the dust particles stay in a state of rest, while the carpet moves. Hence, the dust particles come out of the carpet.

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

Ans. When the bus accelerates and moves forward, it acquires a state of motion. However, the luggage kept on the roof, owing to its inertia, tends to remain in its state of rest. Hence, with the forward movement of the bus, the luggage tends to remain at its original position and ultimately falls from the roof of the bus. To avoid this, it is advised to tie any luggage kept on the roof of a bus with a rope.

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

(a) the batsman did not hit the ball hard enough.

(b) velocity is proportional to the force exerted on the ball.

(c) there is a force on the ball opposing the motion.

(d) there is no unbalanced force on the ball, so the ball would want to come to rest.

Ans. (c) A batsman hits a cricket ball, which then rolls on a level ground. After covering a short distance, the ball comes to rest because there is frictional force on the ball opposing its motion. Frictional force always acts in the direction opposite to the direction of motion. Hence, this force is responsible for stopping the cricket ball.

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

Ans. Initial velocity, = 0 (since the truck is initially at rest)

Distance travelled, = 400 m

Time taken, = 20 s

According to the second equation of motion: ut + 1/2 at2

Where, Acceleration = a

400 = 0 + 1/2 a(20)2   ⇒ 400 = 1/2 a(400)

⇒ a = 2m/s2

1 metric tonne = 1000 kg (Given)

∴ 7 metric tonnes = 7000 kg

Mass of truck, = 7000 kg

From Newton’s second law of motion: Force, = Mass × Acceleration

ma = 7000 × 2 = 14000 N

Hence, the acceleration of the truck is 2 m/s2 and the force acting on the truck is 14000 N.

6. A stone of 1 kg is thrown with a velocity of 20 m/s 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?

Ans. Initial velocity of the stone, = 20 m/s

Final velocity of the stone, = 0 (finally the stone comes to rest)

Distance covered by the stone, = 50 m

According to the third equation of motion:

v2 = u2 + 2as

Where, Acceleration, a

(0)2 = (20)2 + 2 × × 50

= −4 m/s2

The negative sign indicates that acceleration is acting against the motion of the stone.

Mass of the stone, = 1 kg

From Newton’s second law of motion: Force, = Mass × Acceleration

ma

= 1 × (− 4) = −4 N

Hence, the force of friction between the stone and the ice is −4 N.

7. A 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 5000 N, then calculate:

(a) the net accelerating force;

(b) the acceleration of the train; and

(c) the force of wagon 1 on wagon 2.

Ans.

(a)Force exerted by the engine, = 40000 N

Frictional force offered by the track, F= 5000 N

Net accelerating force, F− F= 40000 − 5000 = 35000 N

Hence, the net accelerating force is 35000 N.

(b)Acceleration of the train = a

The engine exerts a force of 40000 N on all the five wagons. Net accelerating force on the wagons, F= 35000 N

Mass of the wagons, = Mass of a wagon × Number of wagons

Mass of a wagon = 2000 kg

Number of wagons = 5

∴ = 2000 × 5 = 10000 kg

Mass of the engine, m′ = 8000 kg Total mass, M = m′ = 18000 kg From Newton’s second law of motion: FMa

⇒ F= 35000/18000 = 1.944 m/s2

Hence, the acceleration of the wagons and the train is 1.944 m/s2. (c)Mass of all the wagons except wagon 1 is 4 × 2000 = 8000 kg

Acceleration of the wagons = 3.5 m/s2

Thus, force exerted on all the wagons except wagon 1

= 8000 × 3.5 = 28000 N

Therefore, the force exerted by wagon 1 on the remaining four wagons is 28000 N.

Hence, the force exerted by wagon 1 on wagon 2 is 28000 N.

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 m/s2?

Ans. Mass of the automobile vehicle, = 1500 kg Final velocity, = 0 (finally the automobile stops) Acceleration of the automobile, = −1.7 ms−2

From Newton’s second law of motion:

Force = Mass × Acceleration = 1500 × (−1.7) = −2550 N

Hence, the force between the automobile and the road is −2550 N, in the direction opposite to the motion of the automobile.

9. What is the momentum of an object of mass m, moving with a velocity v? (a) (mv)2 (b) mv2 (c) ½ mv2 (d) mv

Ans. (d) mv

Mass of the object = m

Velocity = v

Momentum = Mass × Velocity

Momentum = mv

10. Using a horizontal force of 200 N, 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?

Ans. A force of 200 N is applied in the forward direction. Thus, from Newton’s third law of motion, an equal amount of force will act in the opposite direction. This opposite force is the fictional force exerted on the cabinet. Hence, a frictional force of 200 N is exerted on the cabinet.

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 2.5 m s-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?

Ans. Mass of one of the objects, m1 = 1.5 kg

Mass of the other object, m2 = 1.5 kg

Velocity of m1 before collision, v1 = 2.5 m/s

Velocity of m2, moving in opposite direction before collision, v2 = −2.5 m/s

(Negative sign arises because mass m2 is moving in an opposite direction)

After collision, the two objects stick together.

Total mass of the combined object = m1 + m2

Velocity of the combined object = v

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

m1v1 + mv1 = (m1 + m2) v

1.5(2.5) + 1.5 (−2.5) = (1.5 + 1.5) v

3.75 − 3.75 = 3 v

= 0

Hence, the velocity of the combined object after collision is 0 m/s.

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.

Ans. The truck has a large mass. Therefore, the static friction between the truck and the road is also very high. To move the car, one has to apply a force more than the static friction. Therefore, when someone pushes the truck and the truck does not move, then it can be said that the applied force in one direction is cancelled out by the frictioal force of equal amount acting in the opposite direction. Therefore, the student is right in justifying that the two opposite and equal cancel each other.

13. A hockey ball of mass 200 g travelling at 10 m/s is struck by a hockey stick so as to return it along its original path with a velocity at 5 m/s. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Ans. Mass of the hockey ball, = 200 g = 0.2 kg

Hockey ball travels with velocity, v1 = 10 m/s

Initial momentum = mv1

Hockey ball travels in the opposite direction with velocity, v2 = −5 m/s

Final momentum = mv2

Change in momentum = mv1 − mv2 = 0.2 [10 − (−5)] = 0.2 (15) = 3 kg m s−1

Hence, the change in momentum of the hockey ball is 3 kg m s−1.

14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 s. 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.

Ans. Now, it is given that the bullet is travelling with a velocity of 150 m/s.

Thus, when the bullet enters the block, its velocity = Initial velocity, = 150 m/s

Final velocity, = 0 (since the bullet finally comes to rest) Time taken to come to rest, = 0.03 s

According to the first equation of motion, at

Acceleration of the bullet, a

0 = 150 + (×0.03 s)

= -150/0.03 = -5000s2

According to the third equation of motion:

v2 = u2 + 2as

0 = (150)2 + 2 (−5000) s

=-(150)2/-2(5000) =2250 / 10000= 2.25m

Hence, the distance of penetration of the bullet into the block is 2.25 m.

From Newton’s second law of motion:

Force, = Mass × Acceleration

Mass of the bullet, = 10 g = 0.01 kg

Acceleration of the bullet, = 5000 m/s2

ma = 0.01 × 5000 = 50 N

Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.

15. An object of mass 1 kg travelling in a straight line with a velocity of 10 m/s 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, calculate the velocity of the combined object.

Ans.

Mass of the object, m1 = 1 kg

Velocity of the object before collision, v1 = 10 m/s

Mass of the stationary wooden block, m2 = 5 kg

Velocity of the wooden block before collision, v2 = 0 m/s

∴ Total momentum before collision = mv1 + mv2

= 1 (10) + 5 (0) = 10 kg m s−1

It is given that after collision, the object and the wooden block stick together. Total mass of the combined system = m1 + m2

Velocity of the combined object = v

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

mv1 + mv2 = (m1 + m2) v

1 (10) + 5 (0) = (1 + 5) v

= 10/6 = 5/3 ms

The total momentum after collision is also 10 kg m/s.

Total momentum just before the impact = 10 kg m s−1

Total momentum just after the impact = (m1 + m2) = 6 ´ = 10kgm s

Hence, velocity of the combined object after collision = 5/3 .

16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 s.

Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Ans.

Initial velocity of the object, = 5 m/s

Final velocity of the object, = 8 m/s

Mass of the object, = 100 kg

Time take by the object to accelerate, = 6 s

Initial momentum = mu = 100 × 5 = 500 kg m s−1

Final momentum = mv = 100 × 8 = 800 kg m s−1

Force exerted on the object, mv mu t

⇒ m(u) / t = 800 - 500 / 6 = 300/6 = 50N

Initial momentum of the object is 500 kg m s−1.

Final momentum of the object is 800 kg m s−1.

Force exerted on the object is 50 N.

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 change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a 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 motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.

Ans.

According to the law of conservation of momentum:

Momentum of the car and insect system before collision = Momentum of the car and insect system after collision

Hence, the change in momentum of the car and insect system is zero.

The insect gets stuck on the windscreen. This means that the direction of the insect is reversed. As a result, the velocity of the insect changes to a great amount. On the other hand, the car continues moving with a constant velocity. Hence, Kiran’s suggestion that the insect suffers a greater change in momentum as compared to the car is correct. The momentum of the insect after collision becomes very high because the car is moving at a high speed. Therefore, the momentum gained by the insect is equal to the momentum lost by the car.

Akhtar made a correct conclusion because the mass of the car is very large as compared to the mass of the insect.

Rahul gave a correct explanation as both the car and the insect experienced equal forces caused by the Newton’s action-reaction law. But, he made an incorrect statement as the system suffers a change in momentum because the momentum before the collision is equal to the momentum after the collision.

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 10 m/s2.

Ans. Mass of the dumbbell, = 10 kg

Distance covered by the dumbbell, = 80 cm = 0.8 m

Acceleration in the downward direction, = 10 m/s2

Initial velocity of the dumbbell, = 0

Final velocity of the dumbbell (when it was about to hit the floor) = v

According to the third equation of motion:

v2 = u2 + 2as

v2 = 0 + 2 (10) 0.8

= 4 m/s

Hence, the momentum with which the dumbbell hits the floor is = mv = 10 × 4 = 40 kg m s−1

ADDITIONAL EXERCISE QUESTIONS PAGE NO. 128

1. The following is the distance-time table of an object in motion:

 Time in seconds Distance in metres 1 1 2 8 3 27 4 64 5 125 6 216 7 343

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?

(b)What do you infer about the forces acting on the object?

Ans.

(a) There is an unequal change of distance in an equal interval of time. Thus, the given object is having a non − uniform motion. Since the velocity of the object increases with time, the acceleration is increasing.

(b)According to Newton’s second law of motion, the force acting on an object is directly proportional to the acceleration produced in the object. In the given case, the increasing acceleration of the given object indicates that the force acting on the object is also increasing.

2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s-2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)

Ans.

Mass of the motor car = 1200 kg

Only two persons manage to push the car. Hence, the acceleration acquired by the car is given by the third person alone.

Acceleration produced by the car, when it is pushed by the third person,

= 0.2 m/s2

Let the force applied by the third person be F. From Newton’s second law of motion:

Force = Mass × Acceleration

= 1200 × 0.2 = 240 N

Thus, the third person applies a force of magnitude 240 N.

Hence, each person applies a force of 240 N to push the motor car.

3. A hammer of mass 500 g, moving at 50 m s-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

Ans.

Mass of the hammer, = 500 g = 0.5 kg

Initial velocity of the hammer, = 50 m/s

Time taken by the nail to the stop the hammer, = 0.01 s

Velocity of the hammer, = 0 (since the hammer finally comes to rest)

From Newton’s second law of motion:

Forcem(u) / t = 0.5(0 - 50) / 0.01 = -2500N

The hammer strikes the nail with a force of −2500 N. Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e., +2500 N.

4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

Ans.

Mass of the motor car, = 1200 kg

Initial velocity of the motor car, u = 90 km/h = 25 m/s

Final velocity of the motor car, = 18 km/h = 5 m/s

Time taken, = 4 s

According to the first equation of motion:

at

5 = 25 + a (4)

= − 5 m/s2

Negative sign indicates that its a retarding motion i.e. velocity is decreasing. Change in momentum = mv − mu (vu)

= 1200 (5 − 25) = − 24000 kg m s−1

Force = Mass × Acceleration

= 1200 × − 5 = − 6000 N

Acceleration of the motor car = − 5 m/s2

Change in momentum of the motor car = − 24000 kg m s−1

Hence, the force required to decrease the velocity is 6000 N. (Negative sign indicates retardation, decrease in momentum and retarding force)

5. A large truck and a car, both moving with a velocity of magnitude v, have a head-on collision and both of them come to a halt after that. If the collision lasts for 1 s:

(a) Which vehicle experiences the greater force of impact?

(b) Which vehicle experiences the greater change in momentum? (c) Which vehicle experiences the greater acceleration?

(d) Why is the car likely to suffer more damage than the truck?

Ans.

Let the mass of the truck be and that of the car be m. Thus, m

Initial velocity of both vehicles, v

Final velocity of both vehicles, v’ = 0 (since the vehicles come to rest after collision)

Time of impact, = 1 s

(a) From Newton’s second law of motion, the net force experienced by each vehicle is given by the relation:

Fcarm(v '- v) / t = -mv

Ftruck (v'- v) / t = -Mv

Since the mass of the truck is greater than that of the car, it will experience a greater force of impact.

(b) Initial momentum of the car = mv

Final momentum of the car = 0

Change in momentum = mv

Initial momentum of the truck = Mv

Final momentum of the truck = 0

Change in momentum = Mv

Since the mass of the truck is greater than that of the car, it will experience a greater change in momentum.

(c) From the first equation of motion, acceleration produced in a system is independent of the mass of the system. The initial velocity, the final velocity, and the time of impact remain the same in both cases. Hence, both the car and the truck experience the same amount of acceleration.

(d)According to Newton’s third law of motion, for every action there is an equal and opposite reaction that acts on different bodies. Since the truck experiences a greater force of impact (action), this larger impact force is also experienced by the car (reaction). Thus, the car is likely to suffer more damage than the truck.

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