Laws of Motion Question Answers

Laws of Motion Question Answers: NCERT Class 11 Physics

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Chapter
4 — Laws of Motion

NCERT Book
Physics Parts I and II

Language
English

Edition
Academic session 2026-27

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Exercise 1

Q 1:

Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed
(b) a cork of mass 10 g floating on water
(c) a kite skillfully held stationary in the sky
(d) a car moving with a constant velocity of 30 km/h on a rough road
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.

(a) Zero net force

The rain drop is falling with a constant speed. Hence, it acceleration is zero. As per Newton’s second law of motion, the net force acting on the rain drop is zero.

(b) Zero net force

The weight of the cork is acting downward. It is balanced by the buoyant force exerted by the water in the upward direction. Hence, no net force is acting on the floating cork.

(c) Zero net force

The kite is stationary in the sky, i.e., it is not moving at all. Hence, as per Newton’s first law of motion, no net force is acting on the kite.

(d) Zero net force

The car is moving on a rough road with a constant velocity. Hence, its acceleration is zero. As per Newton’s second law of motion, no net force is acting on the car.

(e) Zero net force

The high speed electron is free from the influence of all fields. Hence, no net force is acting on the electron.

Q 12:

A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms^-1. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.

(a) Vertically downward

(b) Parabolic path

(a) At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground.

(b) At the mean position, the velocity of the bob is 1 m/s. The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only. Hence, it will follow a parabolic path.

Q 18:

Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms^-1collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

Mass of each ball = 0.05 kg

Initial velocity of each ball = 6 m/s

Magnitude of the initial momentum of each ball, p_i= 0.3 kg m/s

After collision, the balls change their directions of motion without changing the magnitudes of their velocity.

Final momentum of each ball, p_f= -0.3 kg m/s

Impulse imparted to each ball = Change in the momentum of the system

= pf- pi = -0.3 - 0.3 = -0.6 kg m/s

The negative sign indicates that the impulses imparted to the balls are opposite in direction.

Q 19:

A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms^-1, what is the recoil speed of the gun?

Mass of the gun, M= 100 kg

Mass of the shell, m= 0.020 kg

Muzzle speed of the shell, v = 80 m/s

Recoil speed of the gun = V

Both the gun and the shell are at rest initially.

Initial momentum of the system = 0

Final momentum of the system = mv - MV

Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.

According to the law of conservation of momentum:

Final momentum = Initial momentum

mv - MV= 0

∴ V = mv / M

= 0.020 x 80 / 100 x 1000

= 0.016 m/s

Q 2:

A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,

(a) during its upward motion,

(b) during its downward motion,

Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.

0.5 N, in vertically downward direction, in all cases

Acceleration due to gravity, irrespective of the direction of motion of an object, always acts downward. The gravitational force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton's second law of motion as:

F = m x a

Where,

F = Net force

m= Mass of the pebble = 0.05 kg

a = g = 10 m/s²

∴F = 0.05 x 10 = 0.5 N

The net force on the pebble in all three cases is 0.5 N and this force acts in the downward direction.

If the pebble is thrown at an angle of 45° with the horizontal, it will have both the horizontal and vertical components of velocity. At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble.

Q 22:

If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:

(a) the stone moves radially outwards,

(b) the stone flies off tangentially from the instant the string breaks,

(b)

When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.

Q 29:

Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of

(a) the force on the 7^th coin (counted from the bottom) due to all the coins on its top,

(b) the force on the 7^th coin by the eighth coin,

(a) Force on the seventh coin is exerted by the weight of the three coins on its top. Weight of one coin = mg

Weight of three coins = 3mg

Hence, the force exerted on the 7th coin by the three coins on its top is 3mg. This force acts vertically downward.

(b) Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top.

Weight of the eighth coin = mg

Weight of the ninth coin = mg

Weight of the tenth coin = mg

Total weight of these three coins = 3mg

Hence, the force exerted on the 7^th coin by the eighth coin is 3mg. This force acts vertically downward.

(c) The 6th coin experiences a downward force because of the weight of the four coins (7^th, 8^th, 9^th, and 10^th) on its top.

Therefore, the total downward force experienced by the 6^th coin is 4mg.

As per Newton's third law of motion, the 6th coin will produce an equal reaction force on the 7^thcoin, but in the opposite direction. Hence, the reaction force of the 6^th coin on the 7^th coin is of magnitude 4mg. This force acts in the upward direction.

Q 3:

Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
(c) just after it is dropped from the window of a train accelerating with 1 m s 2,
(d) lying on the floor of a train which is accelerating with 1 m s 2, the stone being at rest relative to the train. Neglect air resistance throughout.

(a)1 N; vertically downward

Mass of the stone, m = 0.1 kg

Acceleration of the stone, a = g = 10 m/s²

As per Newton’s second law of motion, the net force acting on the stone,

F = ma = mg

= 0.1 × 10 = 1 N

Acceleration due to gravity always acts in the downward direction.

(b)1 N; vertically downward

The train is moving with a constant velocity. Hence, its acceleration is zero in the direction of its motion, i.e., in the horizontal direction. Hence, no force is acting on the stone in the horizontal direction.

The net force acting on the stone is because of acceleration due to gravity and it always acts vertically downward. The magnitude of this force is 1 N.

(c)1 N; vertically downward

It is given that the train is accelerating at the rate of 1 m/s².

Therefore, the net force acting on the stone, F' = ma = 0.1 × 1 = 0.1 N

This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force F,' stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations.

Therefore, the net force acting on the stone is given only by acceleration due to gravity.

F = mg = 1 N

This force acts vertically downward.

(d)0.1 N; in the direction of motion of the train

The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train.

Acceleration of the train, a = 0.1 m/s²

The net force acting on the stone will be in the direction of motion of the train. Its magnitude is given by:

F = ma

= 0.1 × 1 = 0.1 N

Q 30:

An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?

Speed of the aircraft, v = 720 km/h = 720 x 5/18 =200 m/s

Acceleration due to gravity, g = 10 m/s²

Angle of banking, θ = 15°

For radius r, of the loop, we have the relation:

tanθ = v² / rg

r = v² / gtanθ

= 200 x 200 / 10 tan15

= 4000 / 0.268

= 14925.37 m

= 14.92 km

Q 31:

A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 10⁶ kg. What provides the centripetal force required for this purpose - The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?

Radius of the circular track, r = 30 m

Speed of the train, v = 54 km/h = 15 m/s

Mass of the train, m= 10⁶kg

The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton's third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and rear of the rail.

The angle of banking θ, is related to the radius (r) and speed (v) by the relation:

tanθ = v² / rg

= 15² / 30 x 10 = 225 / 300

θ = tan-1 (0.75) = 36.87°

Therefore,the angle of banking is about 36.87°.

Q 32:

A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

750 N and 250 N in the respective cases; Method (b)

Mass of the block, m = 25 kg

Mass of the man, M = 50 kg

Acceleration due to gravity, g = 10 m/s²

Force applied on the block, F = 25 x 10 = 250 N

Weight of the man, W = 50 x 10 = 500 N

Case (a): When the man lifts the block directly

In this case, the man applies a force in the upward direction. This increases his apparent weight.

∴Action on the floor by the man = 250 + 500 = 750 N

Case (b): When the man lifts the block using a pulley

In this case, the man applies a force in the downward direction. This decreases his apparent weight.

∴Action on the floor by the man = 500 - 250 = 250 N

If the floor can yield to a normal force of 700 N, then the man should adopt the second method to easily lift the block by applying lesser force.

Q 33:

A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey

(a) climbs up with an acceleration of 6 m s^-2

(b) climbs down with an acceleration of 4 m s^-2

(d) falls down the rope nearly freely under gravity?

(Ignore the mass of the rope).

Case (a)

Mass of the monkey, m= 40 kg

Acceleration due to gravity, g= 10 m/s

Maximum tension that the rope can bear, T_max= 600 N

Acceleration of the monkey, a= 6 m/s²upward

UsingNewton's second law of motion, we can write the equation of motion as:

T- mg= ma

∴T= m(g+ a)

= 40 (10 + 6)

= 640 N

Since T > T_max, the rope will break in this case.

Case (b)

Acceleration of the monkey, a= 4 m/s²downward

UsingNewton's second law of motion, we can write the equation of motion as:

mg - T = ma

∴T= m (g- a)

= 40(10 - 4)

= 240 N

Since T < T_max, the rope will not break in this case.

Case (c)

The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e., a= 0.

UsingNewton's second law of motion, we can write the equation of motion as:

T- mg = ma

T- mg = 0

∴T = mg

= 40 x 10

= 400 N

Since T < T_max, the rope will not break in this case.

Case (d)

When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e., a = g

UsingNewton's second law of motion, we can write the equation of motion as:

mg - T= mg

∴T= m(g- g) = 0

Since T < T_max, the rope will not break in this case.

Q 35:

A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms^-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.

(a) Mass of the block, m = 15 kg

Coefficient of static friction, μ = 0.18

Acceleration of the trolley, a = 0.5 m/s²

As per Newton's second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation:

F = ma = 15 x 0.5 = 7.5 N

This force is acted in the direction of motion of the trolley.

Force of static friction between the block and the trolley:

f = μmg = 0.18 x 15 x 10 = 27 N

The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at rest.

When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.

(b) An observer, moving with the trolley, has some acceleration. This is the case of non-inertial frame of reference. The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.

Q 4:

One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is:

(i) T, (ii) T - mv² / l , (iii) T + mv² / l , (iv) 0

T is the tension in the string. [Choose the correct alternative].

(i)

When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T, i.e.,

F = T = mv² / l

Where F is the net force acting on the particle.

Q 5:

A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^–1. How long does the body take to stop?

Retarding force, F = -50 N

Mass of the body, m= 20 kg

Initial velocity of the body, u= 15 m/s

Final velocity of the body, v= 0

UsingNewton's second law of motion, the acceleration (a) produced in the body can be calculated as:

F= ma

-50 = 20 × a

∴ a = -50 / 20 = -2.5 m/s²

Usingthe first equation of motion, the time (t) taken by the body to come to rest can be calculated as:

v= u + at

∴ t = -u /a = -15 / -2.5 = 6s

Q 6:

A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s^–1 to 3.5 m s^–1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

0.18 N; in the direction of motion of the body

Mass of the body, m= 3 kg

Initial speed of the body, u= 2 m/s

Final speed of the body, v= 3.5 m/s

Time, t = 25 s

Using the first equation of motion, the acceleration (a) produced in the body can be calculated as:

v= u + at

∴ a = v-u / t

= (3.5 - 2) / 25 = 1.5 / 25 = 0.06 m/s²

As per Newton's second law of motion, force is given as:

F= ma

= 3 × 0.06 = 0.18 N

Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.

Q 8:

The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.

Initial speed of the three-wheeler, u = 36 km/h = 10 m/s

Final speed of the three-wheeler, v = 0 m/s

Time, t = 4 s Mass of the three-wheeler, m = 400 kg

Mass of the driver, m' = 65 kg

Total mass of the system, M = 400 + 65 = 465 kg

Using the first law of motion, the acceleration (a) of the three-wheeler can be calculated as:

v = u + at

∴ a = v - u / t = 0-10/4 = -2.5 m/s²

The negative sign indicates that the velocity of the three-wheeler is decreasing with time.

Using Newton’s second law of motion, the net force acting on the three-wheeler can be calculated as:

F = Ma = 465 × (–2.5)

= –1162.5 N

The negative sign indicates that the force is acting against the direction of motion of the three-wheeler.

Q 9:

A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s^–2. Calculate the initial thrust (force) of the blast.

Mass of the rocket, m = 20,000 kg

Initial acceleration, a = 5 m/s²

Acceleration due to gravity, g = 10 m/s²

Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation:

F – mg = ma

F = m (g + a)

= 20000 × (10 + 5)

= 20000 × 15 = 3 × 10⁵ N

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