9th standard Chapter 7 Motion NCERT Solution
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 object.
Answer:
Yes, it is possible. If an object is moving with a constant velocity in a straight line, the net external unbalanced force on it is zero. For this to happen, no external unbalanced force should act on the object.
2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer:
When the carpet is beaten, it moves suddenly but the dust particles remain at rest due to inertia, so they separate from the carpet and fall down.
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:
Luggage on the roof of a bus tends to keep moving when the bus stops suddenly due to inertia of motion. To prevent it from falling off and causing injury, it 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.
Answer:
(c) there is a force on the ball opposing the motion.
5. A truck starts from rest and rolls down a hill with a uniform 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 tonnes (Hint: 1 tonne = 1000 kg.)
Answer:
Given: s = 400 m, t = 20 s, u = 0, m = 7000 kgUsing s = ut + ½at²,
400 = 0 + ½a(20)²
400 = 200a
a = 2 m/s²
Force, F = ma = 7000 × 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?
Answer:
Given: m = 1 kg, u = 20 m/s, v = 0, s = 50 mUsing v² = u² + 2as:
0 = 400 + 2a(50)
0 = 400 + 100a
a = -4 m/s²
Force of friction = F = ma = 1 × (-4) = -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, and
(b) the acceleration of the train.
Answer:
Total mass = 8000 + 5 × 2000 = 18000 kgNet force = 40000 N - 5000 N = 35000 N
(a) Net accelerating force: 35000 N
(b) Acceleration: a = 35000/18000 = 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 m s-2?
Answer:
Force, F = ma = 1500 × (-1.7) = -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
Answer:
(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?
Answer:
If the cabinet moves at constant velocity, the friction force will be equal and opposite to the applied force, so friction force = 200 N.
11. 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 equal and opposite forces act on different bodies. The truck does not move because the force applied by us is not sufficient to overcome the friction between the truck and the ground.
12. 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 of 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.
Answer:
m = 0.2 kg, u = 10 m/s, v = -5 m/s (opposite direction)Change in momentum = m(v - u) = 0.2 × (-5 - 10) = 0.2 × (-15) = -3 kg m/s
13. 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.
Answer:
m = 0.01 kg, u = 150 m/s, v = 0, t = 0.03 sUsing a = (v-u)/t = (0 - 150)/0.03 = -5000 m/s²
Using s = ut + ½at² = 150 × 0.03 + ½(-5000)(0.03)²
s = 4.5 - 2.25 = 2.25 m
14. 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.
Answer:
Initial momentum = 1 × 10 + 5 × 0 = 10 kg m/sLet v = final velocity after collision,
Total mass = 6 kg
Using conservation of momentum:
10 = 6v ⇒ v = 10/6 = 1.67 m/s
Total momentum after = 6 × 1.67 = 10 kg m/s
15. 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.
Answer:
Initial momentum = 100 × 5 = 500 kg m/sFinal momentum = 100 × 8 = 800 kg m/s
Change in momentum = 800 - 500 = 300 kg m/s
Force = Change in momentum / time = 300 / 6 = 50 N
16. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windscreen 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.
Answer:
✅ Rahul's explanation is correct according to Newton's Third Law: both the insect and motorcar experience equal and opposite forces for the same time, so the change in momentum is equal in magnitude but opposite in direction.✅ Kiran is correct that the insect's velocity change is much greater than that of the car due to its smaller mass.
✅ Akhtar is incorrect as the force exerted on both is equal; the insect dies because, despite equal force, its acceleration and velocity change are much higher due to its low mass.
Thus, the insect suffers more damage, but forces and momentum changes are equal in magnitude, confirming Newton's Third Law.
17. 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 s-2.
Answer:
h = 0.8 m, g = 10 m/s², m = 10 kgUsing v² = u² + 2gh = 0 + 2 × 10 × 0.8 = 16 ⇒ v = 4 m/s
Momentum transferred = m × v = 10 × 4 = 40 kg m/s
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