NCERT Solutions Class 9 Science Chapter 9 Force and Laws of Motion

NCERT Solutions Notes Important Questions MCQ Quiz – 1 MCQ Quiz – 2

In-Text Questions Solutions

In Text-Questions Page 118

  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:- (a) Mass of a stone is more than the mass of a rubber ball of same size. Hence, inertia of a stone is greater than that of a rubber ball of same size.
    (b) A train has much greater mass than that of a bicycle, so the train will have more inertia.
    (c) A five-rupee coin has more mass than a one-rupee coin, so five-rupee coin will have greater 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 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 football changes four times.
    First, when a player kicks the football to another player, second when that player kicks the football to the goalkeeper. Third when the goalkeeper collects the football. Fourth, when the goalkeeper kicks the football.
    Agent supplying the force:
    → First case – First player
    → Second case – Second player
    → Third case – Goalkeeper
    → Fourth case – Goalkeeper
  3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
    Ans:- When a tree is vigorously shaken, the branches of the tree come in motion but the leaves tend to continue in their state of rest due to inertia of rest. As a result of this, leaves get separated from the branches of the tree and hence fall down.
  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:- When the bus is moving, our body is also in motion, but due to sudden brake, the lower part of our body comes to rest as soon as the bus stops. But the upper part of our body continues to be in motion and hence we fall in forward direction due to inertia of motion.

    When the bus is stationary our body is at rest but when the bus accelerates, the lower part of our body being in contact with the floor of the bus comes in motion, but the upper part of our body remains at rest due to inertia of rest. Hence we fall in backward direction.

In Text-Questions Page 126

  1. If action is always equal to the reaction, explain how a horse can pull a cart.
    Ans:-  The third law of motion states that action is always equal to the reaction but they act on two different bodies.
    In this case the horse exerts a force on the ground with its feet while walking, the ground exerts an equal and opposite force on the feet of the horse, which enables the horse to move forward and the cart is pulled by the horse.
  2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
    Ans:- The water that is ejected out from the hose in the forward direction comes out with a large momentum and equal amount of momentum is developed in the hose in the opposite direction and hence the hose is pushed backward. It becomes difficult for a fireman to hold a hose which experiences this large momentum.
  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 rifle, m1 = 4 kg
    Mass of bullet, m2 = 50 g = 0.05 kg
    When the rifle and velocity is at rest,
    velocity of rifle, u1 = 0
    velocity of bullet, u2 = 0
    After firing,
    velocity of bullet, v2 = 35 m/s
    Recoil velocity of rifle, v1 = ?
    According to the law of conservation of momentum,
    ⇒ m1u1 + m2u2 = m1v1 + m2v2
    ⇒ 4 x 0 + 0.05 x 0 = 4 x v1 + 0.05 x 35
    ⇒ 0 = 4 v1 + 1.75
    -1.75/4 = v1
    ⇒ v1 = -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.
m1 = 100 g = 0.1 kg
m2 = 200 g = 0.2 kg
u1 = 2 m/s
u2 = 1 m/s
After collision,
v1 = 1.67 m/s
v2 = ?
From law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
⇒ (0.1 x 2) + (0.2 x 1) = (0.1 x 1.67) +(0.2 x v2)
⇒ 0.2 + 0.2 = 0.167 + 0.2 v2
⇒ 0.4 – 0.167 = 0.2 v2
0.233/0.2 = v2
⇒ v2 = 1.165 m/s
So, the velocity of the object is 1.165 m/s.

Chapter End Exercises

  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.
    When an object experiences a net zero external unbalanced force, in accordance with second law of motion its acceleration is zero. If the object was initially in a state of motion, then in accordance with the first law of motion, the object will continue to move in same direction with same speed. It means that the object may be travelling with a non-zero velocity but the magnitude as well as direction of velocity must remain unchanged or constant throughout.
  2. When a carpet is beaten with a stick, dust comes out of it. Explain.
    Here, both carpet and the dust herein are at rest. When the carpet is beaten with a stick, the carpet is comes in motion. Due to inertia of rest, the dust particles tend to remain at rest. As a result, the dust particles fall down.
  3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
    In moving vehicle like bus, the motion is not uniform, the speed of vehicle varies and it may apply brake suddenly or takes sudden turn. The luggage will resist any change in its state of rest or motion, due to inertia and this luggage has the tendency to fall sideways, forward or backward. To avoid the fall of the luggage, it is tied with the 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.

    (c) The cricket ball comes to rest after covering a short distance, because there is a force on the ball, opposing the motion. This force is due to resistance of air and also due to friction between the ball and the ground.
  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.)
    u = 0 m/s
    s = 400 m
    t = 20 s
    m = 7 metric tonnes = 7000 kg
    a = ?
    F = ?
    s = ut + 1/2 at2
    ⇒ 400 = 0 x 20 + 1/2 a (20)2
    ⇒ 400 = 200a
    ⇒ a = 400/200 = 2 m/s2
    F = ma
    ⇒ F = 7000 x 2 = 14000 N

6. A stone of 1 kg is thrown with a velocity of 20 m s-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?
m = 1 kg
u = 20 m/s
v = 0
s = 20 m
F = ?
2as = v2 – u2
⇒ 2a x 50 = 02 – 202
⇒ 100a = -400
⇒ a = -400/100 = -4 m/s2
Force of friction, F
= ma
= 1 x (-4) = -4 N
Negative sign indicates that force of friction is opposing the motion of the ball.

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

(a) Net accelerating force
= 40000 – 5000 = 35000 N

(b) Total mass of train = 8000 + (5 x 2000) = 18000 kg
F = ma
⇒ 35000 = 18000a
⇒ a = 35000/18000 = 1.94 m/s2
So, acceleration of the train = 1.94 m/s2

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?
Ans:- mass = 1500 kg
a = -1.7 ms-2
F = ?
F = m x a
= 1500 x (-1.7)
= -2550 N
The force between the vehicle and road is -2550 N

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

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?
For a body to move with constant velocity, the forces (horizontal force and frictional force) acting on it should be balanced. So, the friction force should be equal in magnitude to the horizontal force.
Therefore, friction force that will be exerted on the cabinet = 200 N

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 they stick together. What will be the velocity of the combined object after collision?
Ans:- Let the two objects be A and B
Mass of object A, m1 = 1.5 kg
Mass of object B, m2 = 1.5 kg
Velocity of object A before collision,
u1 = 2.5 ms-1
Velocity of object B before collision
u2 = -2.5 ms-1
Let velocity of combined object after collision be ‘v’.
From law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
⇒ 1.5 x 2.5 + 1.5 x (-2.5) = 1.5v + 1.5v
⇒ 0 = 3v
⇒ v = 0
So, velocity of combined object after collision = 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:- Student’s justification is not correct. Two equal and opposite forces cancel each other if they act on the same body. According to the third law of motion, action and reaction forces are equal and opposite but they both act on different bodies. Hence, they cannot cancel each other.
The mass of truck is too large and hence its inertia is too high. The small force exerted on the truck cannot move it and the truck remains at rest. For the truck to attain motion, an external large amount of unbalanced force need to be exerted on it.

13. A hockey ball of mass 200 g travelling at 10 m s-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s-1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
m = 200 g = 0.2 kg
u = 10 m/s
v = -5 m/s
Change in momentum
= mv – mu
= m(v – u)
= 0.2(-5 – 10)
= 0.2 x (-15) = -3 kg m/s
So, change in momentum of hockey ball is 3 kg m/s.

14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s-1 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.
Here, m = 10 g = 0.01 kg
u = 150 m/s
v = 0 m/s
t = 0.03 s
Using, v = u + at
⇒ 0 = 150 + a(0.03)
⇒ a = -150/0.03 = -5000 m/s2
2as = v2 – u2
⇒ 2 x (-5000)s = 02 – 1502
⇒ -10000s = -22500
⇒ s = 22500/10000 = 2.25 m
So, the distance of penetration of the bullet into the block = 2.25 m

Now, F = ma
⇒ F = 0.01 x (-5000) = -50 N
So, magnitude of force = 50 N

15. An object of mass 1 kg travelling in a straight line with a velocity of 10 m s-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, calculate the velocity of the combined object.
m1 = 1 kg and u1 = 10 m/s
m2 = 5 kg and u2 = 0 m/s
Let the final velocity of the combined object be ‘v’.
Total momentum just before the impact
= m1u1 + m2u2
= 1 x 10 + 5 x 0
= 10 + 0 = 10 kg m/s
According to law of conservation of momentum,
Total momentum after the impact = Total momentum before the impact
So, Total momentum after the impact = 10 kg m/s
Or, m1v + m2v = 10
Or, (m1 + m2)v = 10
Or, (1 + 5)v = 10
Or, 6v = 10
Or, v = 10/6 = 1.67 m/s
So, the velocity of the combined object is 1.67 m/s.

16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s-1 to 8 m s-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
m = 100 kg
u = 5 m/s
v = 8 m/s
t = 6 s
Initial momentum of the object = mu = 100 x 5 = 500 kg m/s
Final momentum of the object = mv = 100 x 8 = 800 kg m/s
a = (v-u)/t = (8-5)/6 = 3/6 = 1/2 m/s2
F = ma = 100 x 1/2 = 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:-  Rahul gave the correct reasoning and explanation that both the motorcar and the insect experienced the same force and a change in their momentum. As per the law of conservation of momentum.
When 2 bodies collide:
Initial momentum before collision = Final momentum after collision
i.e., mu1+ mu= mv1+ mv2
The equal force is exerted on both the bodies but, because the mass of insect is very small it will suffer greater change in velocity.

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-1.
m = 10 kg
s = 80 cm = 0.8 m
a = 10 m/s2
u = 0
Using, v2 – u2 = 2as
⇒ v2 – 02 = 2 x 10 x 0.8
⇒ v2 = 16
⇒ v = 4 m/s
Momentum transferred to the floor, p
= mv = 10 x 4 = 40 kg m/s


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