Chapter 4Laws of Motion

NCERT Solutions Class 11 Physics Chapter 4 – Laws of Motion

NCERT Solutions for Class 11 Physics Chapter 4 (Laws of Motion) mark the beginning of Dynamics, the study of why objects move. While previous chapters focused on describing motion, this chapter introduces the concept of Force as the cause of change in motion. It builds on the revolutionary principles laid down by Isaac Newton that govern everything from a falling apple to the orbit of satellites.

NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion by ALLEN made by expert faculty makes Newton’s Laws easy with clear Free Body Diagrams and structured problem solving. Solutions focus on the conceptual understanding and numerical accuracy required for JEE and NEET.

This chapter is very basic for all engineering and medical entrance exams. Concepts such as Free Body Diagram (FBD), Conservation of Momentum and Friction are the backbone of traditional mechanics. These solutions provide a rigorous way to solve multi-body problems and understand the invisible forces at work in our everyday life.

1.0Class 11 Physics Chapter 4 : Key Concepts

This chapter focuses on the relationship between the forces acting on a body and its motion. Key lessons include:

Aristotle’s Fallacy vs. Galilean Insight: Understanding that an external force is required to change motion, not to maintain it.
Newton’s First Law (Law of Inertia): Defining inertia as the inherent property of a body to resist change in its state of rest or uniform motion.
Newton’s Second Law: Defining Force as the rate of change of momentum: $F = \frac{d p}{d t} = m a$
Momentum and Impulse: Understanding the impact of a large force acting for a short time $(I = F \cdot Δ t = Δ p)$ .
Newton’s Third Law: Every action has an equal and opposite reaction. Understanding that action and reaction act on different bodies.
Conservation of Momentum: In the absence of an external force, the total momentum of a system remains constant. (Crucial for explosion and recoil problems).
Equilibrium of a Particle: Solving problems where the vector sum of all forces is zero $(\sum F = 0) .$
Friction:

Static Friction $(f_{s})$ : A self-adjusting force.
Kinetic Friction $(f_{k})$ : The force opposing motion during sliding.
Laws of Limiting Friction: $f_{s} \leq μ_{s} N .$

Circular Motion (Dynamics):

Centripetal Force: The net force required for circular motion $(F_{c} = \frac{m v ^{2}}{R}) .$
Banking of Roads: Calculating the optimum speed to avoid skidding on turns.

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 4

SOLVED EXAMPLES

An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a constant rate of $100 m s^{- 2}$ . What is the acceleration of the astronaut the instant after he is outside the spaceship? (Assume that there are no nearby stars to exert gravitational force on him.) Sol. Since there are no nearby stars to exert gravitational force on him and the small spaceship exerts negligible gravitational attraction on him, the net force acting on the astronaut, once he is out of the spaceship, is zero. By the first law of motion the acceleration of the astronaut is zero.
A bullet of mass 0.04 kg moving with a speed of $90 m s^{- 1}$ enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet? Sol. The retardation 'a' of the bullet (assumed constant) is given by

$a = \frac{- u ^{2}}{2 s} = \frac{- 90 \times 90}{2 \times 0.6} ms^{- 2} = - 6750 ms^{- 2}$

The retarding force, by the second law of motion, is F = ma

$= 0.04 kg \times 6750 m s^{- 2} = 270 N$

The actual resistive force, and therefore, retardation of the bullet may not be uniform. The answer therefore, only indicates the average resistive force.

3. The motion of a particle of mass m is described by $y = ut + \frac{1}{2} gt^{2}$ . Find the force acting on the particle. Sol. We know

$y = ut + \frac{1}{2} gt^{2}$

Now,

$v = \frac{dy}{dt} = u + gt$

acceleration, $a = \frac{dv}{dt} = g$ Then the force is given by

$F = ma = mg$

Thus the given equation describes the motion of a particle under acceleration due to gravity and y is the position coordinate in the direction of g.

A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of $12 m s^{- 1}$ . If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball) Sol. Change in momentum

$= 0.15 \times 12 - (- 0.15 \times 12) = 3.6 N s,$

Impulse $= 3.6 N s$ , in the direction from the batsman to the bowler. This is an example where the force on the ball by the batsman and the time of contact of the ball and the bat are difficult to know, but the impulse is readily calculated.

5. Two identical billiard balls strike a rigid wall with the same speed but at different angles, and get reflected without any change in speed, as shown in Figure. What is (i) the direction of the force on the wall due to each ball? (ii) the ratio of the magnitudes of impulses imparted to the balls by the wall ?

Sol. An instinctive answer to (i) might be that the force on the wall in case (a) is normal to the wall, while that in case (b) is inclined at 30° to the normal. This answer is wrong. The force on the wall is normal to the wall in both cases. How to find the force on the wall? The trick is to consider the force (or impulse) on the ball due to the wall using the second law, and then use the third law to answer (i). Let u be the speed of each ball before and after collision with the wall, and m the mass of each ball. Choose the x and y axes as shown in the figure, and consider the change in momentum of the ball in each case : Case (a)

$(p_{x})_{initial} = m u, (p_{x})_{final} = - m u, (p_{y})_{initial} = 0 (p_{u})_{final} = 0$

Impulse is the change in momentum vector. Therefore,

$x-component of impulse = - 2 m u y-component of impulse = 0$

Impulse and force are in the same direction. Clearly, from above, the force on the ball due to the wall is normal to the wall, along the negative x-direction. Using Newton's third law of motion, the force on the wall due to the ball is normal to the wall along the positive x-direction. The magnitude of force cannot be ascertained since the small time taken for the collision has not been specified in the problem. Case (b)

$(p_{x})_{initial} = m u cos 3 0^{\circ}, (p_{y})_{initial} = - m u sin 3 0^{\circ} (p_{x})_{final} = - m u cos 3 0^{\circ}, (p_{y})_{final} = - m u sin 3 0^{\circ}$

Note, while $p_{x}$ changes sign after collision, $p_{y}$ does not. Therefore,

$x-component of impulse = - 2 m u cos 3 0^{\circ} y-component of impulse = 0$

The direction of impulse (and force) is the same as in (a) and is normal to the wall along the negative x direction. As before, using Newton's third law, the force on the wall due to the ball is normal to the wall along the positive x direction. The ratio of the magnitudes of the impulses imparted to the balls in (a) and (b) is

$\frac{2 mu}{( 2 mu c o s 3 0 ^{\circ} )} = \frac{2}{3} \approx 1.2$

See Figure. A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the midpoint P of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium ? (Take $g = 10 m s^{- 2}$ ). Neglect the mass of the rope.

Sol. Figures (b) and (c) are known as free-body diagrams. Figure (b) is the free-body diagram of W and Fig. (c) is the free-body diagram of point P.

Consider the equilibrium of the weight W . Clearly, $T_{2} = 6 \times 10 = 60 N$ .

Consider the equilibrium of the point P under the action of three forces - the tensions $T_{1}$ and $T_{2}$ , and the horizontal force 50 N. The horizontal and vertical components of the resultant force must vanish separately :

$T_{1} cos θ = T_{2} = 60 N T_{1} sin θ = 50 N$

which gives that

$tan θ = \frac{5}{6} or θ = tan^{- 1} (\frac{5}{6}) = 4 0^{\circ}$

Note the answer does not depend on the length of the rope (assumed massless) nor on the point at which the horizontal force is applied.

Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary, given that the co-efficient of static friction between the box and the train's floor is 0.15.

Sol. Since the acceleration of the box is due to the static friction,

$ma = f_{s} \leq μ_{s} N = μ_{s} mg$

i.e. $a \leq μ_{s} g$

$∴ a_{m a x} = μ_{s} g = 0.15 \times 10 m s^{- 2} = 1.5 m s^{- 2}$

See Figure. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle $θ = 1 5^{\circ}$ with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface?

Sol. The forces acting on a block of mass m at rest on an inclined plane are (i) the weight mg acting vertically downwards (ii) the normal force N of the plane on the block, and (iii) the static frictional force $f_{s}$ opposing the impending motion.

In equilibrium, the resultant of these forces must be zero. Resolving the weight mg along the two directions shown, we have

$mg sin θ = f_{s}, mg cos θ = N$

As $θ$ increases, the self-adjusting frictional force $f_{s}$ increases until at $θ = θ_{max}, f_{s}$ achieves its maximum value, $(f_{s})_{max} = μ_{s} N$ . Therefore,

$tan θ_{m a x} = μ_{s} or θ_{m a x} = tan^{- 1} (μ_{s})$

When $θ$ becomes just a little more than $θ_{max}$ , there is a small net force on the block and it begins to slide. Note that $θ_{max}$ depends only on $μ_{s}$ and is independent of the mass of the block. For $θ_{max} = 1 5^{\circ}$ ,

$μ_{s} = tan 1 5^{\circ} = 0.27$

What is the acceleration of the block and trolley system shown in a Figure, if the coefficient of kinetic friction between the trolley and the surface is 0.04 ? What is the tension in the string? (Take $g = 10 m s^{- 2}$ ). Neglect the mass of the string.

Sol. As the string is inextensible, and the pully is smooth, the 3 kg block and the 20 kg trolley both have same magnitude of acceleration. Applying second law to motion of the block Figure,

$30 - T = 3 a$

Apply the second law to motion of the trolley Figure,

$T - f_{k} = 20 a .$

Now

$f_{k} = μ_{k} N,$

Here

$μ_{k} = 0.04,$

$N = 20 \times 10$

$N = 200 N .$

Thus the equation for the motion of the trolley is

$T - 0.04 \times 200 = 20 a Or T - 8 = 20 a .$

These equations give $a = \frac{22}{23} ms^{- 2} = 0.96 ms^{- 2}$ and $T = 27.1 N$ .

A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The co-efficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn? Sol. On an unbanked road, frictional force alone can provide the centripetal force needed to keep the cyclist moving on a circular turn without slipping. If the speed is too large, or if the turn is too sharp (i.e. of too small a radius) or both, the frictional force is not sufficient to provide the necessary centripetal force, and the cyclist slips. The condition for the cyclist not to slip is given by

$v^{2} \leq μ_{s} R g$

Now, $R = 3 m, g = 9.8 m s^{- 2}, μ_{s} = 0.1$ . That is, $μ_{s} Rg = 2.94 m^{2} s^{- 2} . v = 18 km / h$ $= 5 ms^{- 1}$ ; i.e., $v^{2} = 25 m^{2} s^{- 2}$ . The condition is not obeyed. The cyclist will slip while taking the circular turn.

A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the (a) optimum speed of the race car to avoid wear and tear on its tyres, and (b) maximum permissible speed to avoid slipping?

Sol. On a banked road, the horizontal component of the normal force and the frictional force contribute to provide centripetal force to keep the car moving on a circular turn without slipping. At the optimum speed, the normal reaction's component is enough to provide the needed centripetal force, and the frictional force is not needed. The optimum speed $v_{o}$ is given by

$v_{o} = (Rg tan θ)^{1/2}$

Here $R = 300 m, θ = 1 5^{\circ}, g = 9.8 m s^{- 2}$ ; we have

$v_{o} = 28.1 m s^{- 1} .$

The maximum permissible speed $v_{max}$ is given by

$v_{m a x} = (Rg \frac{μ _{s} + t a n θ}{1 - μ _{s} t a n θ})^{1/2} = 38.1 ms^{- 1}$

See Figure. A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of $0.1 m s^{- 2}$ . What is the action of the block on the floor (a) before and (b) after the floor yields ? Take $g = 10 m s^{- 2}$ . Identify the action-reaction pairs in the problem. Sol. (a) The block is at rest on the floor. Its freebody diagram shows two forces on the block, the force of gravitational attraction by the earth equal to $2 \times 10 = 20 N$ ; and the normal force R of the floor on the block. By the First Law, the net force on the block must be zero i.e., $R = 20 N$ . Using third law the action of the block (i.e. the force exerted on the floor by the block) is equal to 20 N and directed vertically downwards. (b) The system (block + cylinder) accelerates downwards with $0.1 ms^{- 2}$ . The free-body diagram of the system shows two forces on the system: the force of gravity due to the earth (270 N); and the normal force $R^{'}$ by the floor. Note, the free-body diagram of the system does not show the internal forces between the block and the cylinder. Applying the second law to the system, $270 - R^{'} = 27 \times 0.1 N$ i.e. $R^{'} = 267.3 N$

By the third law, the action of the system on the floor is equal to 267.3 N vertically downward.

Action-reaction pairs For (a) :

(i) The force of gravity (20 N) on the block by the earth (say, action); the force of gravity on the earth by the block (reaction) equal to 20 N directed upwards (not shown in the figure). (ii) The force on the floor by the block (action); the force on the block by the floor (reaction).

For (b):

(i) The force of gravity (270 N) on the system by the earth (say, action); the force of gravity on the earth by the system (reaction), equal to 270 N, directed upwards (not shown in the figure). (ii) The force on the floor by the system (action); the force on the system by the floor (reaction). In addition, for (b), the force on the block by the cylinder and the force on the cylinder by the block also constitute an action-reaction pair. The important thing to remember is that an action-reaction pair consists of mutual forces which are always equal and opposite between two bodies. Two forces on the same body which happen to be equal and opposite can never constitute an action-reaction pair. The force of gravity on the mass in (a) or (b) and the normal force on the mass by the floor are not action reaction pairs. These forces happen to be equal and opposite for (a) since the mass is at rest. They are not so for case (b), as seen already. The weight of the system is 270 N, while the normal force $R^{'}$ is 267.3 N.

EXERCISE QUESTIONS WITH SOLUTIONS

(For simplicity in numerical calculations, take $g = 10 m s^{- 2}$ )

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.

Sol. (a) Zero net force The rain drop is falling with a constant speed. Hence, its 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.

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, (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of $4 5^{\circ}$ with the horizontal direction? Ignore air resistance Sol. 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:

$F = m \times a$

Where, F = Net force $m =$ Mass of the pebble $= 0.05 kg$

$a = g = 10 m / s^{2} F = 0.05 \times 10 = 0.5 N$

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.

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. Sol. (a) 1 N; vertically downward Mass of the stone, $m = 0.1 kg$ Acceleration of the stone, $a = g = 10 m / s^{2}$ As per Newton's second law of motion, the net force acting on the stone, $F = ma = mg$ $= 0.1 \times 10 = 1 N$ 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 $1 m / s^{2}$ . Therefore, the net force acting on the stone,

$F^{'} = ma = 0.1 \times 1 = 0.1 N$

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, $a = 0.1 m / s$ . The net force acting on the stone will be in the direction of motion of the train. Its magnitude is given by:

$F = ma = 0.1 \times 1 = 0.1 N$

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) $T - \frac{mv ^{2}}{ℓ}$ , (iii) $T + \frac{mv ^{2}}{ℓ}$ (iv) 0 T is the tension in the string [Choose the correct alternative. Sol. (i) T, When a particle connected to a string revolves in a circular path around a center, 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.,

$F = T = \frac{mv ^{2}}{ℓ}$

Where F is the net force acting on the particle.

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? Sol. 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, $v = 0$ Using Newton's second law of motion, the acceleration (a) produced in the body can be calculated as:

$F = ma - 50 = 20 \times a ∴ a = \frac{- 50}{20} = - 2.5 m / s^{2}$

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

$v = u + a t ∴ t = \frac{- u}{a} = \frac{- 15}{- 2.5} = 6 s$

A constant force acting on a body of mass 3.0 kg changes its speed from $2.0 ms^{- 1}$ to $3.5 ms^{- 1}$ in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force? Sol. 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:

$v = u + at a = \frac{v - u}{t} = \frac{3.5 - 2}{25} = \frac{1.5}{25} = 0.06 m / s^{2}$

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

$F = ma 3 \times 0.06 = 0.18 N$

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.

A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6N. Give the magnitude and direction of the acceleration of the body. Sol. Mass of the body, $m = 5 kg$ . The given situation can be represented as follows:

The resultant of two forces is given as :

$R = (8)^{2} + (- 6)^{2} = 64 + 36 = 10 N ∴ θ = tan^{- 1} (\frac{- 6}{8}) = - 36.8 7^{\circ}$

The negative sign indicates that $θ$ is in the clockwise direction with respect to the force magnitude 8 N. As per Newton's second law of motion, the acceleration (a) of the body is given as: $F = ma$

$∴ a = \frac{F}{m} = \frac{10}{5} = 2 m / s^{2}$

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 the driver is 65 kg. Sol. Initial speed of the three-wheeler,

$u = 36 km / h = 36 \times \frac{5}{18} = 10 m / s$

Final speed of the three-wheeler, $v = 0$ 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 equation of motion, the acceleration (a) of the three-wheeler can be calculated as:

$v = u + at ∴ a = \frac{v - u}{t} = \frac{0 - 10}{4} = - 2.5 m / s^{2}$

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 calculated as:

$F = Ma = 465 \times (- 2.5) = - 1162.5 N$

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

A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of $5.0 ms^{- 2}$ . Calculate the initial thrust (force) of the blast. Sol. Mass of the rocket, $m = 20, 000 kg$ Initial acceleration, $a = 5 m / s^{2}$ Acceleration due to gravity, $g = 10 m / s^{2}$ Using Newton's second law of motion, the net force (thrust) acting on the rocket is given by the relation.

$F - m g = ma F = m (g + a) = 20000 \times (10 + 5) = 20000 \times 15 = 3 \times 1 0^{5} N$

A body of mass 0.40 kg moving initially with a constant speed of $10 ms^{- 1}$ to the north is subject to a constant force of 8.0 N directed towards the south for 30 s . Take the instant the force is applied to be $t = 0$ , the position of the body at that time to be $x = 0$ and predict its position at $t = - 5 s, 25 s, 100 s$ . Sol. Mass of the body, $m = 0.40 kg$ Initial speed of the body, $u = 10 m / s$ due north Force acting on the body, $F = - 8.0 N$ Acceleration produced in the body,

$a = \frac{F}{m} = \frac{- 8.0}{0.40} = - 20 m / s^{2} At t = - 5 s$

Acceleration, $a^{'} = 0$ and $u = 10 m / s$

$s = u t + \frac{1}{2} a^{'} t^{2} = 10 \times (- 5) = - 50 m$

At $t = 25 s$ Acceleration $a^{''} = - 20 m / s^{2}$ and $u = 10 m / s$

$s^{'} = u t^{'} + \frac{1}{2} a^{''} t^{2} = 10 \times 25 + \frac{1}{2} \times (- 20) \times (25)^{2} = 250 - 6250 = - 6000 m$

At $t = 100 s$ For

$0 \leq t \leq 30 s, a = - 20 m / s^{2} & u = 10 m / s s_{1} = ut + \frac{1}{2} a^{''} t^{2} = 10 \times 30 + \frac{1}{2} \times (- 20) \times (30)^{2} = 300 - 9000 = - 8700 m$

For $3 0^{'} < t \leq 100 s$ As per the first equation of motion, for $t = 30 s$ , final velocity is given by :

$v = u + at = 10 + (- 20) \times 30 = - 590 m / s$

Velocity of the body after $30 s = - 590 m / s$ For motion between 30s to 100s i.e. in 70s, $a^{''} = 0$

$s_{2} = vt + \frac{1}{2} a^{''} t^{2} = - 590 \times 70 = - 41300 m$

$∴$ Total distance, $s^{''} = s_{1} + s_{2}$

$= - 8700 - 41300 = - 50000 m$

A truck starts from rest and accelerates uniformly at $2.0 ms^{- 2}$ . At $t = 10 s$ , a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at $t = 11 s$ ? (Neglect air resistance.) Sol. Initial velocity of the truck, $u = 0$ Acceleration, $a = 2 m / s^{2}$ Time, $t = 10 s$ As per the first equation of motion, final velocity is given as:

$v = u + at = 0 + 2 \times 10 = 20 m / s$

The final velocity of the truck and hence, of the stone is 20 m/s. At $t = 11 s$ , the horizontal component $(v_{x})$ of velocity, in the absence of air resistance, remains unchanged, i.e.,

$v_{x} = 20 m / s$

The vertical component $(v_{y})$ of velocity of the stone is given by the first equation of motion as:

$v_{y} = u + a_{y} δ t$

Where, $δ t = 11 - 10 = 1 s$ and $a = g = 10 m / s^{2}$

$∴ v_{y} = 0 + 10 \times 1 = 10 m / s$

The resultant velocity (v) of the stone is given as :

exercise-ques-11-physics-chap-4-class-11

$v = v_{x}^{2} + v_{y}^{2} = 2 0^{2} + 1 0^{2} = 400 + 100 = 500 = 22.36 m / s$

Let $θ$ be the angle made by the resultant velocity with the horizontal component of velocity, $v_{x}$

$∴ tan θ = (\frac{v _{y}}{v _{x}}) θ = tan^{- 1} (\frac{10}{20}) = tan^{- 1} (0.5) = 26.5 7^{\circ}$

When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is $10 m / s^{2}$ and it acts vertically downward.

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? Sol. (a) Vertically down ward Parabolic path. At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then he 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.
A man of mass 70 kg stands on a weighing scale in a lift which is moving (a) upwards with a uniform speed of $10 ms^{- 1}$ , (b) downwards with a uniform acceleration of $5 ms^{- 2}$ , (c) upwards with a uniform acceleration of $5 ms^{- 2}$ . What would be the readings on the scale in each case? What would be the reading if the lift mechanism failed and it hurtled down freely under gravity? Sol. (a) Mass of the man, $m = 70 kg$ . Acceleration, $a = 0$ Using Newton's second law of motion, we can write the equation of motion as: $R - mg = ma$ Where, ma is the net force acting on the man. As the lift is moving at a uniform speed, acceleration $a = 0$

$∴ R = m g = 70 \times 10 = 700 N$

$∴$ Reading on the weighing scale

$= \frac{700}{g} = \frac{700}{10} = 70 kg$

(b) Mass of the man, $m = 70 kg$ Acceleration, $a = 5 m / s^{2}$ downward Using Newton's second law of motion, we can write the equation of motion as:

$mg - R = ma \Rightarrow R = m (g - a) = 70 (10 - 5) = 70 \times 5 = 350 N$

$∴$ Reading on the weighing scale

$= \frac{350}{g} = \frac{350}{10} = 35 kg$

Acceleration, $a = 5 m / s^{2}$ upward Using Newton's second law of motion, we can write the equation of motion as : (c) Mass of the man, $m = 70 kg$ Acceleration, $a = 5 m / s^{2}$ upward Using Newton's second law of motion, we can write the equation of motion as :

$R - mg = ma \Rightarrow R = m (g + a) = 70 (10 + 5) = 70 \times 15 = 1050 N$

Reading on the weighing scale $= \frac{1050}{g}$

$= \frac{1050}{10} = 105 kg$

Figure shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for $t < 0, t > 4 s, 0 < t < 4 s$ ? (b) impulse at $t = 0$ and $t = 4 s$ ? (Consider one dimensional motion only).

exercise-ques-14-chap-4-class-11-physics

Sol. For $t < 0$ It can be observed from the given graph that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero. Hence, the force acting on the particle is zero. For $t > 4s$ It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of 3 m from the origin. Hence, no force is acting on the particle. For $0 < t < 4$ It can be observed that the given position-time graph has a constant slope. Hence, the acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero. At $t = 0$ Impulse = Change in momentum = mv-mu Mass of the particle, $m = 4 kg$ Initial velocity of the particle, $u = 0$ Final velocity of the particle, $v = \frac{3}{4} m / s$

$∴ Impulse = 4 (\frac{3}{4} - 0) = 3 kg m / s$

At $t = 4s$ Initial velocity of the particle, $u = \frac{3}{4} m / s$ Final velocity of the particle, $v = 0$

$∴ Impulse = 4 (0 - \frac{3}{4}) = - 3 kg m / s$

Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force $F = 600 N$ is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case? Sol. Horizontal force, $F = 600 N$ Mass of body A, $m = 10 kg$ Mass of body $B, m_{2} = 20 kg$ Total mass of the system,

$m = m_{1} + m_{2} = 30 kg$

Using Newton's second law of motion, the acceleration (a) produced in the system can be calculated as:

$F = ma ∴ a = \frac{F}{m} = \frac{600}{30} = 20 m / s^{2}$

(i) When force F is applied on body A:

The equation of motion can be written as:

$F - T = m_{1} a T = F - m_{1} a = 600 - 10 \times 20 = 400 N$

(ii) When for F is applied on body B

physics-exercise-ques-15-(iii)-chap-4-class-11

The equation of motion can be written as

$F - T = m_{2} a T = F - m_{2} a ∴ T = 600 - 20 \times 20 = 200 N$

Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released. Sol. The given system of two masses and a pulley can be represented as shown in the following figure:

physics-chap-4-class-11-exercise-ques-16

Smaller mass, $m_{1} = 8 kg$ Larger mass, $m_{2} = 12 kg$ Tension in the string $= T$ Mass $m_{2}$ , owing to its weight, moves downward with acceleration a, and mass $m_{1}$ moves upward. Applying Newton's second law of motion to the system of each mass: For mass $m_{1}$ : The equation of motion can be written as:

$T - m_{1} g = ma$

For mass $m_{2}$ : The equation of motion can be written as: $m_{2} g$

$- T = m_{2} a$

Adding equations (i) and (ii), we get:

$(m_{2} - m_{1}) g = (m_{1} + m_{2}) a ∴ a = (\frac{m _{2} - m _{1}}{m _{1} + m _{2}}) g \dots (iii) = (\frac{12 - 8}{12 + 8}) \times 10 = \frac{4}{20} \times 10 = 2 m / s^{2}$

Therefore, the acceleration of the masses is $2 m / s^{2}$ . Substituting the value of a in equation (ii), we get

$m_{2} g - T = m_{2} a \Rightarrow T = m_{2} (g - a) = 12 (10 - 2) = 96 N$

Therefore, the tension in the string is 96N

A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions. Sol. Let $m, m_{1}$ and $m_{2}$ be the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest. Initial momentum of the system (parent nucleus) $= 0$ Let $v_{1}$ and $v_{2}$ be the respective velocities of the daughter nuclei having masses $m_{1}$ and $m_{2}$ . Total linear momentum of the system after disintegration $= m_{1} v_{1} + m_{2} v_{2}$ . According to the law of conservation of momentum: Total initial momentum = Total final momentum.

$0 = m_{1} v_{1} + m_{2} v_{2} v_{1} = \frac{- m _{2} v _{2}}{m _{1}}$

Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

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

$= P_{f} - P_{i} = - 0.3 - 0.3 = - 0.6 kg m / s$

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

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 ? Sol. 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

$mv - MV = 0 ∴ V = \frac{mv}{M} = \frac{0.020 \times 80}{100 \times 1000} = 0.016 m / s$

A batsman deflects a ball by an angle of $4 5^{\circ}$ without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.) Sol. The given situation can be represented as shown in the following figure.

class-11-chap-4-exercise-ques-20-physics

Where, AO = Incident path of the ball OB = Path followed by the ball after deflection $∠ AOB =$ Angle between the incident and deflected paths of the ball $=45^{\circ}$$\angle \mathrm{AOP}=\angle \mathrm{BOP}=22.5^{\circ}=\theta$ Initial and final velocities of the ball = v Horizontal component of the initial velocity $= v cos θ$ , along RO Vertical component of the initial velocity $= v sin θ$ , along PO Horizontal component of the final velocity $= v cos θ$ , along OS Vertical component of the final velocity $= v sin θ$ , along OP The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions. ∴ Impulse imparted to the ball = Change in the linear momentum of the ball $= mv cos θ - (- mv cos θ) = 2 mv cos θ$ Mass of ball, $m = 0.15 kg$ Velocity of the ball, $v = 54 km / h = 15 m / s$ ∴ Impulse $= 2 \times 0.15 \times 15 cos 22. 5^{\circ}$ $= 4.16 kg m / s$

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N? Sol. Mass of the stone, $m = 0.25 kg$ Radius of the circle, $r = 1.5 m$ Number of revolution per second,

$n = \frac{40}{60} = \frac{2}{3} rps$

Angular velocity, $ω = \frac{v}{r} = 2 π n$ The centripetal force for the stone is provided by the tension T, in the string

$i.e., T = F_{Centripetal} = \frac{m v ^{2}}{r} = m r ω^{2} = m r (2 πn)^{2} = 0.25 \times 1.5 \times (2 \times 3.14 \times \frac{2}{3})^{2} = 6.57 N$

maximum tension in the string, $T_{max} = 200 N$

$T_{m a x} = \frac{mv _{m a x}^{2}}{r} ∴ v_{m a x} = \frac{T _{m a x} \times r}{m} = \frac{200 \times 1.5}{0.25} = 1200 = 34.64 m / s$

Therefore, the maximum speed of the stone is 34.64 m/s

If, in Q.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, (c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ? Sol. Option (b) is correct. 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.
Explain why (a) a horse cannot pull a cart and run in empty space, (b) passengers are thrown forward from their seats when a speeding bus stops suddenly, (c) it is easier to pull a lawn mower than to push it, (d) a cricketer moves his hands backwards while holding a catch.

Sol. (a) In order to pull a cart, a horse pushes the ground backward with some force. The ground in turn exerts an equal and opposite reaction force upon the feet of the horse. This reaction force causes the horse to move forward. An empty space is devoid of any such reaction force. Therefore, a horse cannot pull a cart and run in empty space. (b) When a speeding bus stops suddenly, the lower portion of a passenger's body, which is in contact with the seat, suddenly comes to rest. However, the upper portion tends to remain in motion (as per the first law of motion). As a result, the passenger's upper body is thrown forward in the direction in which the bus was moving. (c) While pulling a lawn mower, a force at an angle $θ$ is applied on it, as shown in the following figure.

class-11-exercise-ques-23-(i)-physics-chap-4

The vertical component of this applied force acts upward. This reduces the effective weight of the mower.

On the other hand, while pushing a lawn mower, a force at an angle $θ$ is applied on it, as shown in the following figure.

class-11-exercise-ques-23-(ii)-chap-4-physics

In this case, the vertical component of the applied force acts in the direction of the weight of the mower. This increases the effective weight of the mower.

Since the effective weight of the lawn mower is lesser in the first case, pulling the lawn mower is easier than pushing it.

(d) According to Newton's second law of motion, we have the equation of motion:

$F = ma = m \frac{Δ v}{Δ t}$

Where, $F =$ Stopping force experienced by the cricketer as he catches the ball $m =$ Mass of the ball $Δ t =$ Time of impact of the ball with the hand. It can be inferred from equation (i) that the impact force is inversely proportional to the impact time,

$i.e. F \propto \frac{1}{Δ t}$

Equation (ii) shows that the force experienced by the cricketer decreases if the time of impact increases and vice versa. While taking a catch, a cricketer moves his hand backward so as to increase the time of impact $(Δ t)$ . This is turn results in the decrease in the stopping force, thereby preventing the hands of the cricketer from getting hurt.

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

NCERT Solutions for Class 11 Physics chapter-wise are a clear explanation for all textbook questions. Understand the fundamental ideas, learn the right steps to solve questions and build a strong base of Physics.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 2	Motion in a Straight Line
Chapter 3	Motion in a Plane
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 11	Thermodynamics
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 4

Free Body Diagram Understanding: Learn how to draw Free Body Diagrams FBDs and identify forces such as tension, normal reaction, friction, and gravitational force.
Constraint Relations in Connected Bodies: Understand how strings, pulleys, and connected bodies move together using constraint relations and step-by-step equations.
Clear Understanding of Friction: Learn the cause, direction, and limiting value of friction and understand how friction depends on the tendency of relative motion between surfaces.
Concept-Based Questions: Understand Newton’s laws through practical examples, such as why pulling a lawn mower is easier than pushing it.
Prepared by ALLEN Subject Experts: Get accurate and detailed NCERT Solutions for Class 11 Physics Chapter 4, prepared according to the latest NCERT syllabus.
Simple and Step-by-Step Explanations: Understand Newton’s Laws of Motion, friction, tension, normal force, and connected body problems through clear explanations and systematic methods.
Complete NCERT Exercise Coverage: Get detailed solutions to NCERT Class 11 Physics Chapter 4 examples and exercise questions, covering the important concepts of Newton’s laws of motion for CBSE, JEE, and NEET preparation.

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