Force and Laws of Motion - Questions and Answers

Text Questions

Exercise Questions

Additional Exercises

Text Questions

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?

(a) Stone has more inertia than a rubber ball of the same size because stone has greater mass.

(b) Train has more inertia than a bicycle because train has greater mass.

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.

The velocity of the football changes four times:

First player kicks the football (force applied by first player)
Second player kicks the football towards goal (force applied by second player)
Goalkeeper collects/stops the football (force applied by goalkeeper)
Goalkeeper kicks the football towards his own team player (force applied by goalkeeper)

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

When we vigorously shake a branch of a tree, the branch comes into motion but the leaves tend to remain at rest due to their inertia. This causes the leaves to get detached from the branch and 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?

When a moving bus brakes to a stop, our body tends to continue in its state of motion due to inertia, so we fall forward. When the bus accelerates from rest, our body tends to remain at rest due to inertia, so we fall backwards.

Exercise Questions

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.

Yes, it is possible for an object to be travelling with a non-zero velocity when it experiences a net zero external unbalanced force. According to Newton's first law of motion, an object continues to be in a state of rest or of uniform motion in a straight line unless acted upon by an unbalanced force. So if an object is already moving with constant velocity and no unbalanced force acts on it, it will continue to move with the same velocity.

The condition is that the object must be moving with a constant velocity (constant speed in a straight line).

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

When a carpet is beaten with a stick, the carpet is set into motion but the dust particles tend to remain at rest due to their inertia. When the carpet moves, the dust particles get detached and come out.

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

When the bus starts moving, stops suddenly, or takes a turn, the luggage tends to remain in its state of rest or motion due to inertia. If not tied properly, the luggage may fall off. Tying the luggage with a rope provides the necessary force to overcome this inertia and keep the luggage in place.

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 ball comes to rest due to the frictional force acting between the ball and the ground, which opposes the motion of the 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 tonnes (Hint: 1 tonne = 1000 kg).

Given: Initial velocity u = 0 m/s, Distance s = 400 m, Time t = 20 s

Using the equation of motion: s = ut + ½ at²

400 = 0 × 20 + ½ × a × (20)²

400 = ½ × a × 400

400 = 200a

a = 400/200 = 2 m/s²

Mass of truck = 7 tonnes = 7 × 1000 = 7000 kg

Force F = ma = 7000 × 2 = 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?

Given: Mass m = 1 kg, Initial velocity u = 20 m/s, Final velocity v = 0 m/s, Distance s = 50 m

Using v² = u² + 2as

0² = 20² + 2 × a × 50

0 = 400 + 100a

100a = -400

a = -4 m/s²

Force of friction F = ma = 1 × (-4) = -4 N

The negative sign indicates that the force opposes the motion.

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.

Total mass of train = Mass of engine + Mass of 5 wagons = 8000 + (5 × 2000) = 8000 + 10000 = 18000 kg

Force exerted by engine = 40000 N

Friction force = 5000 N

(a) Net accelerating force = Force exerted by engine - Friction force = 40000 - 5000 = 35000 N

(b) Acceleration a = F/m = 35000/18000 = 1.944 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/s²?

Given: Mass m = 1500 kg, Acceleration a = -1.7 m/s²

Force F = ma = 1500 × (-1.7) = -2550 N

The negative sign indicates that the force is acting opposite to the direction of motion.

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

(d) mv

Momentum p = mass × velocity = m × v

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?

When an object moves with constant velocity, the net force acting on it is zero (according to Newton's first law of motion). Therefore, the friction force must be equal and opposite to the applied force.

Friction force = 200 N (opposite to the direction of motion)

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.

The student's logic is incorrect because action and reaction forces act on different objects, so they don't cancel each other. When we push the truck, we apply a force on the truck (action), and the truck applies an equal and opposite force on us (reaction). These forces act on different bodies and cannot cancel each other.

The truck doesn't move because the force we apply is not sufficient to overcome the static friction between the truck's tires and the road. Once the applied force exceeds the maximum static friction, the truck will start moving.

12. 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.

Mass of ball m = 200 g = 0.2 kg

Initial velocity u = 10 m/s

Final velocity v = -5 m/s (negative sign indicates opposite direction)

Initial momentum p₁ = mu = 0.2 × 10 = 2 kg m/s

Final momentum p₂ = mv = 0.2 × (-5) = -1 kg m/s

Change in momentum = p₂ - p₁ = -1 - 2 = -3 kg m/s

Magnitude of change in momentum = 3 kg m/s

13. 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.

Mass of bullet m = 10 g = 0.01 kg

Initial velocity u = 150 m/s

Final velocity v = 0 m/s

Time t = 0.03 s

Acceleration a = (v - u)/t = (0 - 150)/0.03 = -5000 m/s²

Distance of penetration s = ut + ½ at² = 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

14. 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.

Mass of object m₁ = 1 kg, Initial velocity u₁ = 10 m/s

Mass of block m₂ = 5 kg, Initial velocity u₂ = 0 m/s

Total momentum before impact = m₁u₁ + m₂u₂ = 1 × 10 + 5 × 0 = 10 kg m/s

According to law of conservation of momentum, total momentum after impact = total momentum before impact = 10 kg m/s

Total mass after sticking together M = m₁ + m₂ = 1 + 5 = 6 kg

Let V be the velocity of combined object

Momentum after impact = MV = 6V

6V = 10

V = 10/6 = 1.67 m/s

15. 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.

Mass m = 100 kg

Initial velocity u = 5 m/s

Final velocity v = 8 m/s

Time t = 6 s

Initial momentum p₁ = mu = 100 × 5 = 500 kg m/s

Final momentum p₂ = mv = 100 × 8 = 800 kg m/s

Acceleration a = (v - u)/t = (8 - 5)/6 = 3/6 = 0.5 m/s²

Force F = ma = 100 × 0.5 = 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 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.

Rahul's explanation is correct according to Newton's third law of motion. When the insect hits the windshield, the insect exerts a force on the car and the car exerts an equal and opposite force on the insect. Both experience the same magnitude of force.

However, due to the large difference in masses, the effect of this force is different. The car, having much larger mass, experiences negligible acceleration, while the insect, having very small mass, experiences a very large acceleration which causes it to die.

The change in momentum is the same for both (in magnitude) because force is rate of change of momentum, and the force and time of impact are the same for both.

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/s².

Mass m = 10 kg

Height h = 80 cm = 0.8 m

Acceleration a = 10 m/s²

Initial velocity u = 0 m/s

Using v² = u² + 2as

v² = 0 + 2 × 10 × 0.8 = 16

v = 4 m/s

Momentum transferred to floor = mv = 10 × 4 = 40 kg m/s

Additional Exercises

A1. The following is the distance-time table of an object in motion:
Time in seconds: 0, 1, 2, 3, 4, 5, 6, 7
Distance in metres: 0, 1, 8, 27, 64, 125, 216, 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?

(a) The distance is proportional to the cube of time (s ∝ t³), which means the acceleration is increasing with time.

(b) According to Newton's second law (F = ma), if acceleration is increasing, the net force acting on the object must also be increasing.

A2. 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/s². With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)

When two persons push the car at uniform velocity, net force is zero, so the force applied by two persons equals the frictional force.

Let force applied by each person = F

Then, frictional force = 2F

When three persons push, net force = 3F - 2F = F

This net force produces acceleration: F = ma = 1200 × 0.2 = 240 N

So each person pushes with a force of 240 N.

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 the force of the nail on the hammer?

Mass of hammer m = 500 g = 0.5 kg

Initial velocity u = 50 m/s

Final velocity v = 0 m/s

Time t = 0.01 s

Acceleration a = (v - u)/t = (0 - 50)/0.01 = -5000 m/s²

Force F = ma = 0.5 × (-5000) = -2500 N

Magnitude of force = 2500 N

A4. 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.

Mass m = 1200 kg

Initial velocity u = 90 km/h = 90 × 1000/3600 = 25 m/s

Final velocity v = 18 km/h = 18 × 1000/3600 = 5 m/s

Time t = 4 s

Acceleration a = (v - u)/t = (5 - 25)/4 = -20/4 = -5 m/s²

Initial momentum p₁ = mu = 1200 × 25 = 30000 kg m/s

Final momentum p₂ = mv = 1200 × 5 = 6000 kg m/s

Change in momentum = p₂ - p₁ = 6000 - 30000 = -24000 kg m/s

Force F = ma = 1200 × (-5) = -6000 N

Magnitude of force = 6000 N