Chapter 8: Force and Laws of Motion Exercise & In-Text Questions & Answers
1Which 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?
Answer & Reference:
(a) The stone has more inertia than the rubber ball of the same size because it has more mass.
(b) The train has more inertia than the bicycle because it has more mass.
(c) The five-rupees coin has more inertia than the one-rupee coin because it has more mass.
(Page 91, Section 8.3)
2In the following example, try to identify the number of times the velocity of the ball changes: "A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team". Also identify the agent supplying the force in each case.
Answer & Reference:
The velocity of the ball changes four times:
1. When the first player kicks the ball (agent: first player)
2. When the second player kicks the ball (agent: second player)
3. When the goalkeeper stops the ball (agent: goalkeeper)
4. When the goalkeeper kicks the ball (agent: goalkeeper)
(Page 91, Section 8.3)
3Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer & Reference:
When we shake the branch vigorously, the branch comes into motion but the leaves tend to remain at rest due to inertia. This causes the leaves to detach from the branch.
(Page 91, Section 8.3)
4Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Answer & Reference:
When a moving bus brakes to a stop, our body tends to continue in the same state of motion due to inertia, causing us to fall forward. When the bus accelerates from rest, our feet move with the bus but the upper part of our body tends to remain at rest due to inertia, causing us to fall backward.
(Page 90, Section 8.2)
5An 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 & Reference:
Yes, it is possible. According to Newton's first law of motion, an object will continue to move with a constant velocity (which could be non-zero) if no net external unbalanced force acts on it. The condition is that the object must already be moving with a constant velocity in a straight line.
(Page 89-90, Section 8.2)
6When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer & Reference:
When the carpet is beaten, it comes into motion. However, the dust particles tend to remain at rest due to inertia. This causes the dust particles to separate from the carpet and come out.
(Page 91, Section 8.3)
7Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer & Reference:
When a bus starts or stops suddenly, the luggage on the roof tends to continue in its state of rest or motion due to inertia. Tying it with a rope provides the necessary force to prevent it from falling off.
(Page 90, Section 8.2)
8A 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 & Reference:
(c) there is a force on the ball opposing the motion.
The ball slows down and stops due to the frictional force acting opposite to its direction of motion.
(Page 88-89, Section 8.1)
9A 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 tonnes (Hint: 1 tonne = 1000 kg).
Answer & Reference:
Given: u = 0, s = 400 m, t = 20 s, m = 7000 kg
Using s = ut + ½at²
400 = 0 + ½ × a × (20)²
400 = 200a
a = 2 m/s²
Force F = ma = 7000 × 2 = 14000 N
(Page 92-93, Section 8.4.1)
10A 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 & Reference:
Given: m = 1 kg, u = 20 m/s, v = 0, s = 50 m
Using v² = u² + 2as
0 = (20)² + 2a(50)
0 = 400 + 100a
a = -4 m/s²
Force F = ma = 1 × (-4) = -4 N
The negative sign indicates the force opposes motion.
(Page 92-93, Section 8.4.1)
11An 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 & Reference:
Total mass = 8000 + (5 × 2000) = 18000 kg
(a) Net accelerating force = Applied force - Friction = 40000 - 5000 = 35000 N
(b) Acceleration a = F/m = 35000/18000 = 1.94 m/s²
(Page 92-93, Section 8.4.1)
12An 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 & Reference:
Given: m = 1500 kg, a = -1.7 m/s²
Force F = ma = 1500 × (-1.7) = -2550 N
The negative sign indicates the force opposes motion.
(Page 92-93, Section 8.4.1)
13What 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 & Reference:
(d) mv
Momentum is defined as the product of mass and velocity.
(Page 92, Section 8.4)
14Using 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 & Reference:
Since the cabinet moves with constant velocity, the net force acting on it is zero. Therefore, the friction force must be equal and opposite to the applied force, i.e., 200 N.
(Page 88-89, Section 8.1)
15According 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 & Reference:
The student's logic is incorrect. The action and reaction forces act on different bodies (we push the truck, the truck pushes us), so they don't cancel each other. The truck doesn't move because the force we apply is not enough to overcome the static friction between the truck and the road.
(Page 95, Section 8.5)
16A 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 magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer & Reference:
Given: m = 0.2 kg, u = 10 m/s, v = -5 m/s (opposite direction)
Initial momentum = mu = 0.2 × 10 = 2 kg m/s
Final momentum = mv = 0.2 × (-5) = -1 kg m/s
Change in momentum = Final - Initial = -1 - 2 = -3 kg m/s
Magnitude = 3 kg m/s
(Page 92, Section 8.4)
17A 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.
Answer & Reference:
Given: m = 0.01 kg, u = 150 m/s, v = 0, t = 0.03 s
Acceleration a = (v - u)/t = (0 - 150)/0.03 = -5000 m/s²
Distance s = ut + ½at² = 150×0.03 + ½×(-5000)×(0.03)² = 4.5 - 2.25 = 2.25 m
Force F = ma = 0.01 × (-5000) = -50 N
Magnitude of force = 50 N
(Page 92-93, Section 8.4.1)
18An 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 & Reference:
Given: m₁ = 1 kg, u₁ = 10 m/s, m₂ = 5 kg, u₂ = 0
Total momentum before impact = m₁u₁ + m₂u₂ = 1×10 + 5×0 = 10 kg m/s
Total momentum after impact = (m₁ + m₂)v = 6v
By conservation of momentum: 6v = 10 → v = 10/6 = 1.67 m/s
Total momentum after impact = 10 kg m/s
(Page 92, Section 8.4)
19An 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 & Reference:
Given: m = 100 kg, u = 5 m/s, v = 8 m/s, t = 6 s
Initial momentum = mu = 100 × 5 = 500 kg m/s
Final momentum = mv = 100 × 8 = 800 kg m/s
Acceleration a = (v - u)/t = (8 - 5)/6 = 0.5 m/s²
Force F = ma = 100 × 0.5 = 50 N
(Page 92-93, Section 8.4.1)
20Akhtar, 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. 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 & Reference:
Rahul's explanation is correct. According to Newton's third law, both the motorcar and the insect experience equal and opposite forces. The change in momentum is also equal in magnitude but opposite in direction. However, because the insect has a much smaller mass, it experiences a much larger acceleration (or deceleration), which causes its death.
(Page 95, Section 8.5)
21How 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 & Reference:
Given: m = 10 kg, h = 0.8 m, a = 10 m/s², u = 0
Using v² = u² + 2gh → v² = 0 + 2×10×0.8 = 16 → v = 4 m/s
Momentum transferred = mv = 10 × 4 = 40 kg m/s
(Page 92, Section 8.4)
22The following is the distance-time table of an object in motion:
Time in seconds
Distance in metres
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?
(b) What do you infer about the forces acting on the object?
Answer & Reference:
(a) The distance is proportional to the cube of time (s ∝ t³). This indicates that acceleration is not constant but increasing with time.
(b) Since acceleration is increasing, the net force acting on the object must also be increasing with time (F = ma).
(Page 92-93, Section 8.4.1)
23Two 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.)
Answer & Reference:
When two persons push at uniform velocity, their force balances friction. When three persons push, the net force = 3F - 2F = F (since two persons' force equals friction).
F = ma = 1200 × 0.2 = 240 N
Therefore, each person pushes with a force of 240 N.
(Page 92-93, Section 8.4.1)
24A 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?
Answer & Reference:
Given: m = 0.5 kg, u = 50 m/s, v = 0, t = 0.01 s
Acceleration a = (v - u)/t = (0 - 50)/0.01 = -5000 m/s²
Force F = ma = 0.5 × (-5000) = -2500 N
The negative sign indicates the force opposes motion.
(Page 92-93, Section 8.4.1)
25A 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.
Answer & Reference:
Given: m = 1200 kg, u = 90 km/h = 25 m/s, v = 18 km/h = 5 m/s, t = 4 s
Acceleration a = (v - u)/t = (5 - 25)/4 = -5 m/s²
Change in momentum = m(v - u) = 1200(5 - 25) = -24000 kg m/s
Force F = ma = 1200 × (-5) = -6000 N
Magnitude of force = 6000 N
(Page 92-93, Section 8.4.1)