# NCERT Class 9 Science Notes & Solutions – Chapter 9 – Forces and Laws of Motion

## Exercises Questions & Answers - NCERT - Forces and Laws of Motion

#### In-Text Questions & Answers | NCERT Class 9 Science Chapter 9 | Forces and Laws of Motion

##### In-Text Questions & Answers | NCERT Class 9 Science Chapter 9 | Forces and Laws of Motion | Page Nos. 118

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

Read Q4 in detail for mass-inertia relation.
a) A stone is more heavier than a rubber ball of same size. So, inertia of the stone is more than the rubber ball.
b) Mass of a train is very much more than a bicycle. So, inertia of a train is more than a bicycle.
c) A five-rupee coin is heavier than a one-rupee coin. So, it has more inertia.

###### 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 collect the football & kicks it towards a player of his own team”.Also identify the agent supplying the force in each case.

The velocity changes four times. Details are as follows –
Case – I: when the first player kicks the football to the second player. In this case, the agent supplying the force is the first player.
Case – II: when the second player kicks the football to the goalkeeper. In this case, the agent supplying the force is the second player.
Case – III: when the goalkeeper stops the football. In this case, the agent supplying the force is the goalkeeper.
Case – IV: when the goalkeeper kicks the football towards a player of his own team. In this case, the agent supplying the force is the goalkeeper.

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

At first, both the tree & leaves are in rest. When the branches of a tree are shaken, it moves to & fro. But it leaves tend to remain at rest due to inertia. This inertia of rest of leaves tends to opposes the ‘to & fro’ motion. For 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 & fall backwards when it accelerates from rest?

Firstly, when the bus suddenly applies brake, passengers tend to fall forwards. It is due to inertia of motion. During braking, bus starts slowing down, also lower body parts of passenger response simultaneously as it is attached to the floor of the bus, but the upper body part remains in state of motion due to inertia. So passenger tends to fall forwards. Due to this safety belt is given in private car to tie our body with seat to avoid such case.

Secondly, when the bus starts moving, passengers tend to fall backwards. It is due to inertia of rest. When the bus starts, it’s state changes from rest to moving with lower body part of passenger, but upper body part remains in rest and acts against the motion of the bus due to inertia. So the passengers tend to fall backwards.

###### In-Text Questions & Answers | NCERT Class 9 Science Chapter 9 | Forces and Laws of Motion | Math of Second Law of Motion

1. Velocity of an object of mass 10 Kg changes from 20 m/s to 40 m/s after applying a force over a time period 10 second. What is rate of change in momentum?

m = 10 Kg, u = 20 m/s, v = 40 m/s, t = 10 s, initial momentum = mu = 10×20 Kgm/s = 200Kgm/s, final momentum = mv = 10X40 Kgm/s = 400 Kgm/s, So rate of change in momentum = (mv – mu)/t  = (400 – 200)/ 10 = 200/10 = 20 kg m/s2.

2. What will be acceleration if 150 Newton force is applied on an object of 10Kg mass?

F = ma , here F = 150 N , m = 10Kg , a = ?
So, 150 = 10xa ,
⇒ a = 150/10 = 15 m/s2.

###### 3. A car of mass 500 Kg attains speed of  30 m/s after 10 seconds of starting of engine. What is the force applied by engine?

We know F= ma,
⇒ F = m [ (v-u)/t ],
⇒ F = ( mv – mu )/t. Here m = 500 Kg,  v = 30 m/s, u = 0, t = 10 s.
So  F = ( 500×30 – 500×0) / 10  = (15000-0)/10 = 15000/10 = 1500 N.

###### 4. A force acting on an object of mass 20 Kg makes changing in velocity from 15 m/s to 25 m/s over 5 second. What is the value of force? Find final velocity if same force is applied for total 10 second?

F = ( mv – mu )/t, ⇒ F = ( 20.25 – 20.15 )/5 = 200/5 = 40 N.
We know  v = u + at, here  a = ( 25 – 15 )/5 = 2 m/s2 .
So final velocity after 10 second = v = u + at = 15 + 2.10 = 15 + 20 = 35 m/s.

We may also find final velocity by using  equation F = (mv – mu)/t,
⇒ Ft = mv – mu,
⇒ mv = mu + Ft,
So, v = (mu + Ft )/m = u + (Ft/m).
[ If  ‘v’ is in Km/hr, then v = (35 x 3600) m/hr = 126,000 m/hr = 126 Km/hr,  1 hr = 60×60 = 3600 s].

###### 5. Which force is more – i ) Force accelerating a mass of 1 Ton at 10 m/s2 or ii) Force accelerating a mass of 500 Kg at 30 m/s2.

Case 1: F = ma = 1000 Kg x 10 m/s2= 10000 N  [ 1 Ton = 1000Kg].
Case 2: F = ma = 500 Kg x 30 m/s2 = 15000 N. So force in case 2 is more.

6. A bus with speed 72 Km/hr suddenly tries to stop by applying brake and it finally stops after 10 seconds. Find magnitude of force exerted by brakes. Weight of bus is 2 Ton.

2 Ton = 2 x 1000 = 2000 Kg ;  u = 72 Km/hr = 72000/3600 m/s = 20 m/s ; v = 0 ; t = 10 s,
Now F = (mv – mu)/t = (2000×0  – 2000×20)/10 =   4000 N . () indicates that brake force acts against velocity of mass. So magnitude of force is 4000 N.

###### In-Text Questions & Answers | NCERT Class 9 Science Chapter 9 | Forces and Laws of Motion | Page Nos. 126-127

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

In this case, horse pushes the ground in backward direction. Against this push, as per Newton’s third law of motion, a reaction force is exerted by the ground on the horse in the forward direction. So the horse along with cart moves in the forward direction.

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

When large amount of water ejects from a hose at a high velocity, it gets associated with a great force & pushes the hose in backward direction with equal force. Because as per Newton’s third law of motion, every action has  has an equal and opposite reaction. So it is difficult for the fireman to hold the hose.

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

We know  m1u1  + m2u2  =  m1v1 + m2v2. In this case:- m1 = mass of rifle = 4 kg , u1 = velocity of rifle before firing = 0, v1 = velocity of rifle after firing = ?
m2 = mass of bullet =50 g = 0.05 kg , u2 = velocity of bullet before firing = 0, v2=velocity of bullet after firing.
m1u1 + m2u2  =  m1v1 + m2v2 ,
⇒ 4 x 0  +  0.05 x 0   =  4 x v1  +   0.05 x 35,
⇒ 4 x v1 =  ( – )0.05 x 35,
therefore,  v1 =  ( – ) 0.4375 m/s. [ – indicates movement of rifle is opposite of bullet movement].

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

Now  m1 u1  + m2u2   =  m1v1 + m2v2 . As mass of both object is in gram, we do not need it to convert in kg. Now here –
m1 = 100 g , u1 = 2 m/s,  v1 = 1.67 m/s, m2 = 200 g, u2 = 1 m/s, v2 = ?
m1 u1  + m2u2   =  m1v1 + m2v2
⇒ 100 x 2 +   200 x 1   =  100 x 1.67   +   200 x v2
⇒ 200 x v2 = 400 – 167

So, v2 =  1.165  m/s.

##### Exercises Questions & Answers | NCERT Class 9 Science Chapter 9 | Forces and Laws of Motion | Page Nos. 128 - 129

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 & direction of the velocity. If no, provide reason.

Off course, it is possible. When the object moves with a constant velocity in a particular direction, there is no net unbalanced force acting on the object. The object moves with a non-zero velocity. If we want to change the state of motion of the object, we need to apply a net non-zero external unbalanced force.

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

When we beat a carpet with a stick, it comes to motion. But the dust particles try to resist its state of rest as per the law of inertia. Therefore, the dust particles comes out of the carpet.

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

When the bus starts, it changes state from rest to moving. But the luggage kept on the roof remains in rest and acts against the motion of the bus due to inertia. So it tends to fall from the roof of the bus. Therefore, the luggage kept on the roof is tied 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.

The answer is c. There is a frictional force between the ball & the ground. This force opposes the movement of the ball. As a result the ball stops after covering some distance.

5. A truck starts from rest and rolls down a hill with constant acceleration. It travels a distance of 400 m in 20 s. Find the acceleration. Find the force acting on it if truck weighs 7 tons.

At first apply S = ut  + ½ at2. Here u=0, t=20 s, a =?
So, 400 = 0.20  + ½xax202
⇒ 400 = 200a
a = 2 ms-2.
We know 1 Ton = 1000 Kg , so 7 Tons = 7000 Kg. Also Force F = ma.
Here m = 7000 Kg,  a = 2 ms-2, F = 7000 x 2 N = 14,000 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 of 50 m. What is the force of friction between the stone and the ice?

Friction force acts opposite of stone movement and it is equal to the force applied to stone during the time of throwing. So Force F = ma. Here m = 1 Kg , F =? & a =?
At first find acceleration a. Also we know 2as = v2 – u2.
In case of negative acceleration it is  -2as = v2 – u2
So, -2 x a x 50 = 02 – 202 [ Here S=Distance= 50 m , Initial velocity = u = 20 ms-1]
100a = -400
a = -4 ms-2
Therefore, Force F = ma = 1 x -4 N = -4 N.
(-) indicates that friction force acts opposite of stone movement.

###### 7. A 8,000Kg engine pulls a train of 5 wagons, each of 2,000 Kg, along a horizontal track. If the engine exerts a force of 40,000 N and the track offers friction force of 5,000 N, then calculate – a) the net accelerating force and b) the acceleration of the train.

Total weight of train = engine + wagons = 8,000  + 5×2,000 Kg = 18,000 Kg.
a) Friction force opposes against movement of train , so net accelerating force = 40,000 – 5,000 N = 35,000 N.
b) We get, net accelerating force = 35,000 N and total weight of train = 18,000 Kg.
we know, F = ma
a = F/m = 35,000/18,000 = 1.944 ms-2.

###### 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 ?

Friction force between vehicle and road opposes movement of vehicle. And this force should be equal to the force that is exerted by vehicle engine. Force exerted by vehicle engine is obtained by applying  F = ma equation .
So, force between the vehicle and road = 1500 x 1.7 N =2,550 N. And it is acting against movement of vehicle.

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

Momentum = mass x velocity = mxv = mv, so answer is d.

10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at constant velocity. What is the friction force that will be exerted on the cabinet?

There is a static friction between the cabinet & the floor. When we apply a force of 200 N to move a wooden cabinet in the forward direction, an equal force is exerted in the opposite direction. This opposite force is the static frictional force. This incident satisfies the Newton’s third law of motion.

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 before the collision during which the stick together, what will be velocity of the combined object after collision?

m1 u1  + m2u2  = m1v1 + m2v2
In this case –   m1v1 + m2v2 = m1 u1 + m2u2
⇒ (m1 + m2)v = m1 u1 + m2u2
⇒ (1.5 + 1.5) v = (1.5 x 2.5) + {1.5 x (–) 2.5}   [∵ (–) velocity in opposite directions]
So, ( 1.5 + 1.5 ) v = 0
Therefore, v = 0 ms-1.

###### 12. According to the third law of motion when we push on an object, the object pushes back on us with an equal & 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 & equal forces cancel each other. Comment on this logic & explain why the truck does not move.

The truck is heavy & has large mass. The static friction between the truck & the road is very high. When we push the truck, we need to apply a force more than the static friction to move the truck. But the force applied by us in the direction of the truck is cancelled out by the static frictional force of equal amount acting in the opposite direction. So, the student is right in justifying that two opposite & equal force 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 magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Momentum = mass x velocity. Before striking of the ball –
i) momentum = 0.2 x 10 = 2 kgms-1 [200 g = 0.2 kg].

After striking the ball moves at 5 m/s  in opposite direction. So momentum after striking of the ball –
ii) momentum = 0.2 x 5 = 1 kgms-1.

As a result, we realize that force applied by the hockey stick is first nullified momentum of 2 kgms-1  in forward direction and induced reverse momentum of 1 kgms-1. So magnitude of change of momentum is sum of  i  and  ii  i.e. (2 kgms-1 +1 kgms-1) = 3 kgms-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 block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Velocity reaches from 150 m/s to zero in 0.03 s. We know,  v = u – at  in case of negative acceleration. Here v = 0, u = 150 m/s , t = 0.03 s, a = ?
So, v = u – at, ⇒  0 = 150 – a x 0.03, ⇒ a = 150/0.03 = 5,000 ms-2.
i)  2as = v2 – u2. In case of negative acceleration we may write this as:  -2as = v2 – u2 . So distance covered in 0.03 s by the bullet i.e. penetration of the bullet:
s = (v2 – u2) / 2a   = [ ( 02 –  1502 ) / 2 x 5000  = 2.25 m.
ii) Force exerted on bullet by wooden block = F = ma = 0.01 x 5000 N = 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. Also calculate the velocity of the combined object.

m1 u1  + m2u2   =  m1v1 + m2v2
⇒  m1 u1  + m2u2  =  (m1 + m2)v   ( ∵ both objects stick together).
Here , m1 = 1 kg,  u1  = 10 m/s, m2 = 5 kg,   u2 = 0, (m1 + m2) = 1+5 kg = 6 kg, v = ?
i) total momentum just before the impact = m1 u1 + m2u2  = 1×10   +   5×0  kgms-1 = 10 kgms-1.
ii) velocity of the combined object:  (m1 + m2)v  =  m1 u1  + m2u2
⇒ v  = (m1 u1  + m2u2  )/ (m1 + m2)
So, v =  10/6 = 5/3   ms-1.

###### 16. An object of mass 100 kg is accelerated uniformly from velocity of 5 m/s to 8 m/s in 6 s. Calculate the initial and final momentum momentum of the object. Also find the magnitude of force exerted on the object.

Initial momentum = 100 x 5 kgms-1 = 500 kgms-1.
Final momentum = 100 x 8 kgms-1 = 800 kgms-1.
v = u + at , ⇒ a = (v-u)/t ,  ⇒ a = (8-5)/6  ms-2 = 0.5 ms-2.
Therefore, force exerted on the object = F = ma = 100 x  0.5 N = 50 N.

17. Akhtar, Kiran & Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield & got stuck on the windscreen. Akhtar & Kiran started pondering over the situation. Kiran suggested the insect suffered a greater change in momentum as compared to the change in the momentum of the motorcar ( because the change in the velocity of the insect was much more than that of the motorcar. Akhtar said the since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. As a result insect died. Rahul while putting an entirely new explanation said the both the motorcar & the insect experienced the same force & a change in their momentum. Comment on these suggestion.

As per the law of conservation of momentum ⇒

###### Momentum of (car + insect) before collision = Momentum of (car + insect) after collision

The car continues moving with a constant velocity both prior & after the collision. As the insect got stuck on the car, its direction is reversed. So its velocity changes to a great amount.

Kiran’s comment ⇒ It is correct. As the car is moving at a high speed, so the momentum of the insect becomes very high. Therefore, the change in momentum of the insect compared to the car is high.

Akthar’s comment ⇒ It is also correct. As the mass of the car is very very high than the insect, so it exerted a larger force on the insect. Also the momentum gained by the insect is equal to the momentum lost by the car.

Rahul’s comment ⇒ It is partially correct. As per Newton’s third law of motion, every action has  has an equal and opposite reaction. So both the car & the insect experienced equal forces. The incorrect part is that the system suffers a change in momentum because the momentum before & after the collision is equal.

###### 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 ms-2.

80 cm = 0.8 m.  We know , 2as = v2 – u2. Here , S = 0.8 m, u=0, a = 10 ms-2, v =?
So – 2 x 10 x 0.8 = v2 – 02, ⇒ v2 = 16 , ⇒ v = 4 ms-1.
So momentum induced by dumb-bell of mass 10 kg on floor = m x v = 10 x 4 kgms-1 = 40 kgms-1.

###### Additional Exercises Questions & Answers | NCERT Class 9 Science Chapter 9 | Forces and Laws of Motion | Page Nos. 130
A1. Distance-Time of an Object

Time in seconds

Distance in Meters

0
0
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?

It is a non-uniform motion. Because there is unequal change of distance in an equal interval of time. So the velocity as well as acceleration is increasing with time.

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

We know, F = ma where F = force, m = mass, a = acceleration. In this case, acceleration is increasing with time. So the force acting on object is also increasing.

###### A2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motor car can be pushed by three person to produce an acceleration of 0.2 ms-2. What force does each person push the motor car (assume that all persons push the motorcar with same muscular effort).

F = ma. So F = 1200 x 0.2 N = 240 N. This force is additionally applied by the third person to create acceleration. As all persons push the motorcar with same muscular effort, so each person push the motor car with 240 N force.

A3. A hammer of mass 500 g, moving at 50 m/s, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is force of the nail on the hammer?

As nail stops the hammer, force by the nail on the hammer is equal to the force by the the hammer on the nail.  Here m = 500 g = 0.5 kg, u = 50 m/s, t = 0.01 s, v = 0, a = ? Now –
In case of negative acceleration,  v = u – at, ⇒ 0 = 50 – a x 0.01,  ⇒ a = 5000 ms-2.
So force by the nail on the hammer = F = ma = 0.5 x 5000  N = 2500 N.

###### A4. A motorcar of mass 1200 kg is moving along a straight line with 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 magnitude of the force applied.

Initial velocity = u = 90 km/h  = 90000/3600 m/s = 25 m/s.
Final velocity = v = 18 km/h = 18000/3600 m/s = 5 m/s. Time = t = 4 s.
Firstly, acceleration = a = (25-5)/4 ms-2 = 5 ms-2.
Secondly, change in momentum = initial momentum – final momentum = [ (1200 x 25) – (1200 x 5) ] kgms-1 = 24000 kgms-1.
Thirdly, Force = ma = 1200 x 5 N = 6000 N.

Forces and Laws of Motion – To Be Continued – Forces and Laws of Motion