Class 11th Physics Chapter 4 Solutions

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

Question 4.1

Give the magnitude and direction of the net force acting on:

(a) A drop of rain falling down with a constant speed

Since the drop is falling with constant speed, its acceleration is zero.
Hence, the net force acting on it is zero.

Answer:

• Magnitude of net force: 0 N

• Direction: None

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(b) A cork of mass 10 g floating on water

The weight of the cork acting downward is balanced by the upthrust (buoyant force) of water acting upward.
Therefore, the resultant force is zero.

Answer:

• Magnitude of net force: 0 N

• Direction: None

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(c) A kite skillfully held stationary in the sky

The forces acting on the kite (weight, lift, tension, air drag) balance each other.
Since the kite is stationary, its acceleration is zero.

Answer:

• Magnitude of net force: 0 N

• Direction: None

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(d) A car moving with a constant velocity of 30 km h⁻¹ on a rough road

Constant velocity means zero acceleration.
The driving force balances the frictional force.

Answer:

• Magnitude of net force: 0 N

• Direction: None

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(e) A high-speed electron in space far from all material objects and free of electric and magnetic fields

No external force acts on the electron.

Answer:

• Magnitude of net force: 0 N

• Direction: None

Question 4.2

A pebble of mass 0.05 kg is thrown vertically upward. Give the direction and magnitude of the net force on the pebble:
(Ignore air resistance)

Weight of the pebble:

$$F=mg=0.05\times 10=0.5\text{ N}$$

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(a) During its upward motion

Only gravitational force acts, directed downward.

Answer:

• Magnitude of net force: 0.5 N

• Direction: Vertically downward

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(b) During its downward motion

Gravity continues to act downward.

Answer:

• Magnitude of net force: 0.5 N

• Direction: Vertically downward

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(c) At the highest point where it is momentarily at rest

Although velocity is zero, gravitational force still acts.

Answer:

• Magnitude of net force: 0.5 N

• Direction: Vertically downward

Question 4.3

Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg:
(Neglect air resistance)

Weight of the stone:

$$F=mg=0.1\times 10=1\text{ N}$$

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(a) Just after it is dropped from the window of a stationary train

Only gravitational force acts.

Answer:

• Magnitude of net force: 1 N

• Direction: Vertically downward

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(b) Just after it is dropped from the window of a train running at a constant velocity of 36 km h⁻¹

Horizontal motion does not affect the force acting on the stone.
Only gravity acts.

Answer:

• Magnitude of net force: 1 N

• Direction: Vertically downward

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(c) Just after it is dropped from the window of a train accelerating at 1 m s⁻²

After release, the stone is no longer influenced by the train’s acceleration.
Only gravity acts.

Answer:

• Magnitude of net force: 1 N

• Direction: Vertically downward

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(d) A stone lying on the floor of a train accelerating at 1 m s⁻², the stone being at rest relative to the train

The stone accelerates along with the train.

$$F=ma=0.1\times 1=0.1\text{ N}$$

Answer:

• Magnitude of net force: 0.1 N

• Direction: In the direction of the train’s acceleration

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Question 4.4

One end of a string of length $l$ is connected to a particle of mass $m$ and the other end 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) $T-{\displaystyle \frac{m{v}^2}{l}}$
(iii) $T+{\displaystyle \frac{m{v}^2}{l}}$
(iv) $0$

($T$ is the tension in the string. Choose the correct alternative.)

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Solution 

The particle is moving in a horizontal circular path on a smooth table, so:

• There is no friction.

• The weight of the particle and the normal reaction act vertically and cancel each other.

• The only horizontal force acting on the particle is the tension $T$ in the string.

This tension provides the centripetal force required for circular motion.

$$\text{Net force towards the centre}=\frac{m{v}^2}{l}$$

But this centripetal force is provided entirely by the tension $T$.

$$\therefore T=\frac{m{v}^2}{l}$$

Hence, the net force acting on the particle towards the centre is equal to the tension $T$.

Question 4.5

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

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Solution 

Given:

• Retarding force, $F=50\text{N}$

• Mass of the body, $m=20\text{kg}$

• Initial velocity, $u=15{\text{m s}}^{-1}$

• Final velocity, $v=0{\text{m s}}^{-1}$ (body stops)

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Step 1: Find the acceleration

Using Newton’s second law,

$$F=ma$$

Since the force is retarding, acceleration will be negative.

$$a=\frac{F}{m}=\frac{50}{20}=2.5{\text{m s}}^{-2}$$
$$\Rightarrow a=-2.5{\text{m s}}^{-2}$$

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Step 2: Use the equation of motion

$$v=u+at$$

Substitute the values:

$$0=15+(-2.5)t$$
$$2.5t=15$$
$$t=\frac{15}{2.5}=6\text{s}$$

Answer:

$$\boxed{{\displaystyle \text{The body takes }6\text{ seconds to stop.}}}$$

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Question 4.6

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?

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Solution 

Given:

• Mass of the body, $m=3.0\text{kg}$

• Initial speed, $u=2.0{\text{m s}}^{-1}$

• Final speed, $v=3.5{\text{m s}}^{-1}$

• Time taken, $t=25\text{s}$

---

Step 1: Find the acceleration

Using the first equation of motion,

$$a=\frac{v-u}{t}$$
$$a=\frac{3.5-2.0}{25}=\frac{1.5}{25}=0.06{\text{m s}}^{-2}$$

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Step 2: Find the force

Using Newton’s second law,

$$F=ma$$
$$F=3.0\times 0.06=0.18\text{N}$$

Direction of the force

Since the speed increases and the direction of motion remains unchanged, the force acts in the direction of motion.

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Question 4.7

A body of mass 5 kg is acted upon by two perpendicular forces of magnitudes 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

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Solution 

Given:

• Mass of the body, $m=5\text{kg}$

• Two perpendicular forces:
  $F_1=8\text{N}$ and $F_2=6\text{N}$

---

Step 1: Find the resultant force

Since the forces are perpendicular, the resultant force is given by:

$$F=\sqrt{F_1^2+F_2^2}$$
$$F=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\text{N}$$

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Step 2: Find the acceleration

Using Newton’s second law,

$$a=\frac{F}{m}$$
$$a=\frac{10}{5}=2{\text{m s}}^{-2}$$

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Step 3: Find the direction of acceleration

The acceleration is in the direction of the resultant force.

Let the direction of the 8 N force be along the x-axis.

$$\tan \theta =\frac{F_2}{F_1}=\frac{6}{8}=\frac{3}{4}$$
$$\theta ={\tan }^{-1}\left(\frac{3}{4}\right)$$

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Question 4.8

The driver of a three-wheeler moving with a speed of 36 km h⁻¹ 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.

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Solution 

Given:

• Speed of the three-wheeler,
  $u=36{\text{km h}}^{-1}=10{\text{m s}}^{-1}$

• Final velocity,
  $v=0{\text{m s}}^{-1}$
  

• Time taken to stop,
  $t=4.0\text{s}$

• Mass of three-wheeler = 400 kg

• Mass of driver = 65 kg

$$\text{Total mass},m=400+65=465\text{kg}$$

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Step 1: Calculate the acceleration

Using the first equation of motion,

$$a=\frac{v-u}{t}$$

$$a=\frac{0-10}{4}=-2.5{\text{m s}}^{-2}$$

(The negative sign indicates retardation.)

Step 2: Calculate the average retarding force

Using Newton’s second law,

$$F=ma$$
$$F=465\times (-2.5)=-1162.5\text{N}$$

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Question 4.9

A rocket with a lift-off mass of 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s⁻². Calculate the initial thrust (force) of the blast.
(Take $g=10{\text{m s}}^{-2}$)

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Solution 

Given:

• Mass of the rocket,
  $m=\mathrm{20,000}\text{kg}$

• Upward acceleration,
  $a=5.0{\text{m s}}^{-2}$
  

• Acceleration due to gravity,
  $g=10{\text{m s}}^{-2}$

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Step 1: Forces acting on the rocket

• Thrust $T$ acts upward

• Weight $mg$ acts downward

The net upward force on the rocket is:

$$T-mg=ma$$

Step 2: Calculate the thrust

$$T=m(a+g)$$
$$T=\mathrm{20,000}\times (5+10)$$
$$T=\mathrm{20,000}\times 15$$
$$T=3.0\times {10}^5\text{N}$$

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Question 4.10

A body of mass 0.40 kg moving initially with a constant speed of 10 m s⁻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 $t=-5$, $25$ and $100$ s.

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Solution 

Given:

• Mass of the body, $m=0.40\text{kg}$

• Initial speed, $u=10{\text{m s}}^{-1}$ (towards north)

• Force applied, $F=8.0\text{N}$ (towards south)

• Time for which force acts = 30 s

• Initial position at $t=0$, $x=0$

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Step 1: Choose sign convention

• Take north as positive direction

• Hence, force towards south is negative

---

Step 2: Find acceleration

Using Newton’s second law:

$$a=\frac{F}{m}=\frac{8.0}{0.40}=20{\text{m s}}^{-2}$$

Since force is towards south:

$$a=-20{\text{m s}}^{-2}$$

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Step 3: Equation of motion

Position as a function of time:

$$x=ut+\frac{1}{2}a{t}^2$$

$$x=10t+\frac{1}{2}(-20)t^2$$

$$x=10t-10t^2$$

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(a) Position at $t=-5$

$$x=10(-5)-10(25)$$

$$x=-50-250=-300\text{m}$$Answer:$$\boxed{{\displaystyle x=-300\text{m}}}(\text{300 m towards south})$$

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(b) Position at $t=25$

$$x=10(25)-10(625)$$

$$x=250-6250=-6000\text{m}$$Answer:$$\boxed{{\displaystyle x=-6000\text{m}}}(\text{6000 m towards south})$$

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(c) Position at $t=100$

$$x=10(100)-10(10000)$$

$$x=1000-100000=-99000\text{m}$$Answer$$\boxed{{\displaystyle x=-99000\text{m}}}(\text{99 km towards south})$$

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Question 4.11

A truck starts from rest and accelerates uniformly at $2.0{\text{m s}}^{-2}$. 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 $t=11$?
(Neglect air resistance; take $g=10{\text{m s}}^{-2}$)

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Solution (NCERT-style)

Given:

• Acceleration of the truck, $a_{\text{truck}}=2.0{\text{m s}}^{-2}$

• Truck starts from rest

• Stone is released at $t=10$

• Time at which quantities are required: $t=11$
  ⇒ Time of motion of stone after release,

  $$\mathrm{\Delta }t=11-10=1\text{s}$$

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Step 1: Velocity of the truck at $t=10$ s

$$v_{\text{truck}}=u+at=0+2.0\times 10=20{\text{m s}}^{-1}$$

When the stone is dropped, it has:

• Horizontal velocity = 20 m s⁻¹

• Vertical velocity = 0

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(a) Velocity of the stone at $t=11$ s

Horizontal component

• No horizontal force acts on the stone after release.

$$v_x=20{\text{m s}}^{-1}$$

Vertical component

$$v_y=u_y+gt=0+10\times 1=10{\text{m s}}^{-1}$$

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Resultant velocity

$$v=\sqrt{v_x^2+v_y^2}$$

$$v=\sqrt{{20}^2+{10}^2}=\sqrt{400+100}=\sqrt{500}$$

$$v=10\sqrt{5}{\text{m s}}^{-1}$$

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Direction of velocity

$$\tan \theta =\frac{v_y}{v_x}=\frac{10}{20}=\frac{1}{2}$$

$$\theta ={\tan }^{-1}\left(\frac{1}{2}\right)$$

The velocity is directed below the horizontal.

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Answer (a):

• Velocity:

$$\boxed{{\displaystyle 10\sqrt{5}{\text{m s}}^{-1}}}$$

• Direction:
  At an angle ${\tan }^{-1}(1\mathrm{/}2)$ below the horizontal

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(b) Acceleration of the stone at $t=11$

After release, the only force acting on the stone is gravity.

$$\text{Acceleration}=g=10{\text{m s}}^{-2}$$

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Answer (b):

• Acceleration:

$$\boxed{{\displaystyle 10{\text{m s}}^{-2}\text{ vertically downward}}}$$

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Question 4.12

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 m s⁻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?

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Solution 

The bob is executing simple pendulum motion. Its motion and hence its trajectory after cutting the string depend on the velocity of the bob at the instant the string is cut.

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(a) String is cut at one of the extreme positions

At the extreme position of oscillation:

• The velocity of the bob is zero.

• Only gravitational force acts on the bob after the string is cut.

Since the bob has no horizontal velocity at that instant, it will simply fall under gravity.

Trajectory:

The bob will move vertically downward in a straight line.

Answer (a):

• Trajectory: Straight vertical line downward

• Reason: Velocity at extreme position is zero

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(b) String is cut at the mean position

At the mean (lowest) position:

• The bob has maximum speed, given as

  $$v=1{\text{m s}}^{-1}$$

• The direction of velocity at the mean position is horizontal.

• After the string is cut, the bob has:

  • Initial horizontal velocity = 1 m s⁻¹

  • Vertical acceleration due to gravity = $g$

Thus, the bob behaves like a horizontally projected projectile.

Trajectory:

The bob will follow a parabolic path.

Answer (b):

• Trajectory: Parabola

• Reason: Horizontal velocity with vertical acceleration due to gravity

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Question 4.13

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 m s⁻¹,
(b) downwards with a uniform acceleration of 5 m s⁻²,
(c) upwards with a uniform acceleration of 5 m s⁻².

What would be the readings on the scale in each case?

(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

(Take $g=10{\text{m s}}^{-2}$)

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Solution 

Given:
Mass of the man, $m=70\text{kg}$
Acceleration due to gravity, $g=10{\text{m s}}^{-2}$

The reading of the weighing scale is the normal reaction $N$ exerted by the scale on the man.

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(a) Lift moving upwards with uniform speed

Uniform speed ⇒ acceleration $a=0$

$$N=mg=70\times 10=700\text{N}$$

Reading on the scale:$$\boxed{{\displaystyle 700\text{N}}}$$

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(b) Lift moving downwards with uniform acceleration $5{\text{m s}}^{-2}$

Acceleration is downward.

$$N=m(g-a)$$

$$N=70(10-5)=70\times 5=350\text{N}$$

Reading on the scale

$$\boxed{{\displaystyle 350\text{N}}}$$

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(c) Lift moving upwards with uniform acceleration $5{\text{m s}}^{-2}$

Acceleration is upward.

$$N=m(g+a)$$

$$N=70(10+5)=70\times 15=1050\text{N}$$

Reading on the scale:

$$\boxed{{\displaystyle 1050\text{N}}}$$

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(d) Lift falls freely under gravity

In free fall:

$$a=g$$

$$N=m(g-g)=0$$

Reading on the scale:$$\boxed{{\displaystyle 0}}$$

(The man feels weightless.)

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Question 4.14

Figure 4.16 shows the position–time graph of a particle of mass 4 kg. What is the
(a) force on the particle for $t<0$, $0<t<4\text{s}$, and $t>4\text{s}$?
(b) impulse at $t=0$ and $t=4\text{s}$?
(Consider one-dimensional motion only.)

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Given / From the position–time graph (Fig. 4.16)

• The graph consists of straight line segments with sharp corners (kinks) at
  $t=0$ and $t=4\text{s}$.

• Slope of a position–time graph = velocity.

From the graph:

• For $t<0$: slope is constant ⇒ velocity is constant

• For $0<t<4\text{s}$: slope is constant (different from earlier)

• For $t>4\text{s}$: slope is constant again

Hence, velocity is constant in each interval, but changes suddenly at
$t=0$ and $t=4\text{s}$.

Mass of particle:

$$m=4\text{kg}$$

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(a) Force on the particle

For $t<0$

Velocity is constant ⇒ acceleration $a=0$

$$F=ma=0$$

---

For $0<t<4\text{s}$

Velocity is again constant ⇒ acceleration $a=0$

$$F=0$$

Force = 0 N

---

For $t>4\text{s}$

Velocity remains constant ⇒ acceleration $a=0$

$$F=0$$

---

Answer (a):

$$\boxed{{\displaystyle \text{Force is zero for }t<0,0<t<4\text{ s and }t>4\text{ s}}}$$

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(b) Impulse at $t=0$ and $t=4\text{s}$

Impulse is produced when velocity changes suddenly.

$$\text{Impulse}=\mathrm{\Delta }p=m(v_2-v_1)$$

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Impulse at $t=0$

From the graph:

• Velocity just before $t=0$: $v_1=2{\text{m s}}^{-1}$

• Velocity just after $t=0$: $v_2=0$

$$I_0=4(0-2)=-8\text{N s}$$

Impulse at $t=0$:$$\boxed{{\displaystyle -8\text{N s}}}$$

(negative sign indicates direction opposite to initial motion)

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Impulse at $t=4\text{s}$

From the graph:

• Velocity just before $t=4$: $v_1=0$

• Velocity just after $t=4$: $v_2=-2{\text{m s}}^{-1}$

$$I_4=4(-2-0)=-8\text{N s}$$

Impulse at $t=4\text{s}$:

$$\boxed{{\displaystyle -8\text{N s}}}$$

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Question 4.15

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 along the direction of the string to
(i) the 10 kg body (A),
(ii) the 20 kg body (B).

Find the tension in the string in each case.

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Solution

Let
Mass of A = 10 kg
Mass of B = 20 kg

Since the surface is smooth, there is no friction.

---

Step 1: Acceleration of the system

Total mass

$$m=10+20=30\text{ kg}$$

$$a=\frac{F}{m}=\frac{600}{30}=20{\text{m s}}^{-2}$$

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(i) Force applied on the 10 kg body (A)

The 20 kg body (B) is pulled only by the tension T.

$$T=m_B\times a=20\times 20=400\text{ N}$$

Tension in the string = $\boxed{{\displaystyle 400\text{ N}}}$

(ii) Force applied on the 20 kg body (B)

The 10 kg body (A) is pulled only by the tension T.

$$T=m_A\times a=10\times 20=200\text{ N}$$

Tension in the string = $\boxed{{\displaystyle 200\text{ N}}}$

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Question 4.16

Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string which passes over a frictionless pulley. Find
(i) the acceleration of the masses, and
(ii) the tension in the string, when the system is released.

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Solution

Let

$$m_1=8\text{ kg},m_2=12\text{ kg}$$

Since $m_2>m_1$, the 12 kg mass moves downward and the 8 kg mass moves upward.

Take acceleration = $a$, tension in string = $T$, and $g=9.8{\text{m s}}^{-2}$.

---

Equations of motion

For 12 kg mass (downward):

$$12g-T=12a\text{(1)}$$

For 8 kg mass (upward):

$$T-8g=8a\text{(2)}$$

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Step 1: Find acceleration

Add equations (1) and (2):

$$12g-8g=12a+8a$$

$$4g=20a$$

$$a=\frac{4g}{20}=\frac{g}{5}$$

$$\boxed{{\displaystyle a=1.96{\text{m s}}^{-2}}}$$

Step 2: Find tension

Substitute $a=\frac{g}{5}$ in equation (2):

$$T-8g=8\left(\frac{g}{5}\right)$$

$$T=8g+\frac{8g}{5}=\frac{48g}{5}$$

$$\boxed{{\displaystyle T=94.08\text{ N}}}$$

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Question 4.17

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.

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Solution

Let the initial nucleus be at rest in the laboratory frame.

$$\text{Initial momentum }=0$$

Suppose the nucleus disintegrates into two smaller nuclei of masses $m_1$ and $m_2$, moving with velocities ${\vec{v}}_1$and ${\vec{v}}_2$respectively.

Using law of conservation of momentum

Since no external force acts on the system,

$$\text{Total momentum before disintegration}=\text{Total momentum after disintegration}$$

$$0=m_1{\vec{v}}_1+m_2{\vec{v}}_2$$

Rearranging,$$m_1{\vec{v}}_1=-m_2{\vec{v}}_2$$Hence proved.

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Question 4.18

Two billiard balls, each of mass 0.05 kg, moving in opposite directions with a speed of
6 m s⁻¹, collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

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Solution

Given:
Mass of each ball$$m=0.05\text{ kg}$$Initial speed$$u=6{\text{ m s}}^{-1}$$

Final speed (after rebound)

$$v=6{\text{ m s}}^{-1}$$

---

Impulse

Impulse $J$ is equal to the change in momentum:

$$J=\mathrm{\Delta }p=m(v-u)$$

---

For one billiard ball

Take the initial direction as positive.

Initial velocity

$$u=+6{\text{ m s}}^{-1}$$

After collision, the ball rebounds, so its velocity becomes

$$v=-6{\text{ m s}}^{-1}$$

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Change in momentum

$$\mathrm{\Delta }p=m(v-u)$$

$$\mathrm{\Delta }p=0.05(-6-6)$$

$$\mathrm{\Delta }p=0.05(-12)=-0.6{\text{ kg m s}}^{-1}$$Impulse

Magnitude of impulse imparted to each ball:

$$\boxed{{\displaystyle J=0.6\text{ N s}}}$$

(The negative sign only indicates change of direction.)

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Question 4.19

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 m s⁻¹, find the recoil speed of the gun.

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Solution

Given:
Mass of shell,

$$m=0.020\text{ kg}$$Mass of gun,$$M=100\text{ kg}$$Velocity of shell,$$v=80{\text{ m s}}^{-1}$$

Let recoil speed of the gun be $V$.

---

Using law of conservation of momentum

Initially, the gun–shell system is at rest, so total momentum is zero.

$$\text{Initial momentum}=0$$

After firing:$$mv-MV=0$$

$$MV=mv$$

$$V=\frac{mv}{M}$$

Calculation

$$V=\frac{0.020\times 80}{100}$$

$$V=\frac{1.6}{100}=0.016{\text{ m s}}^{-1}$$

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Question 4.20

A batsman deflects a ball by an angle of 45° without changing its initial speed, which is
54 km h⁻¹. What is the impulse imparted to the ball?
(Mass of the ball = 0.15 kg)

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Solution

Given:
Mass of ball,$$m=0.15\text{ kg}$$

Initial speed = Final speed$$v=54{\text{ km h}}^{-1}=15{\text{ m s}}^{-1}$$

Angle between initial and final velocities,

$$\theta ={45}^{\circ }$$

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Impulse

Impulse $J$ = change in momentum

$$J=m\mathrm{\mid }\mathrm{\Delta }\vec{v}\mathrm{\mid }$$

Magnitude of change in velocity when direction changes by angle $\theta$:

$$\mathrm{\mid }\mathrm{\Delta }\vec{v}\mathrm{\mid }=\sqrt{v^2+v^2-2v^2\cos \theta }$$

$$\mathrm{\mid }\mathrm{\Delta }\vec{v}\mathrm{\mid }=v\sqrt{2(1-\cos {45}^{\circ })}$$$$\mathrm{\mid }\mathrm{\Delta }\vec{v}\mathrm{\mid }=15\sqrt{2(1-0.707)}$$$$\mathrm{\mid }\mathrm{\Delta }\vec{v}\mathrm{\mid }=15\times 0.765\approx 11.48{\text{ m s}}^{-1}$$

Impulse imparted

$$J=0.15\times 11.48$$

$$\boxed{{\displaystyle J\approx 1.72\text{ N s}}}$$

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Question 4.21

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⁻¹ in a horizontal plane.

(i) What is the tension in the string?
(ii) What is the maximum speed with which the stone can be whirled if the string can withstand a maximum tension of 200 N?

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Solution

Given:

$$m=0.25\text{ kg},r=1.5\text{ m}$$

Speed of rotation:

$$f=40{\text{ rev min}}^{-1}=\frac{40}{60}=\frac{2}{3}{\text{ rev s}}^{-1}$$

Angular speed:

$$\omega =2\pi f=2\pi \times \frac{2}{3}=\frac{4\pi }{3}{\text{ rad s}}^{-1}$$

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(i) Tension in the string

Linear speed:

$$v=\omega r=\frac{4\pi }{3}\times 1.5=6.28{\text{ m s}}^{-1}$$

Tension provides the centripetal force:

$$T=\frac{m{v}^2}{r}$$

$$T=\frac{0.25\times (6.28{)}^2}{1.5}$$

$$T\approx 6.6\text{ N}$$

$$\boxed{{\displaystyle T\approx 6.6\text{ N}}}$$

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(ii) Maximum speed when maximum tension = 200 N

$$T_{\max }=\frac{m{v}_{\max }^{}}{r}$$

$$v_{\max }=\sqrt{\frac{T_{\max }r}{m}}$$

$$v_{\max }=\sqrt{\frac{200\times 1.5}{0.25}}$$

$$v_{\max }=\sqrt{1200}$$

$$\boxed{{\displaystyle v_{\max }\approx 34.6{\text{ m s}}^{-1}}}$$

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Question 4.22

If, in Exercise 4.21, 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.

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Answer and Explanation

The correct option is (b).

Reason

• While the stone is moving in a circle, the tension in the string provides the centripetal force, continuously changing the direction of velocity.

• At the instant the string breaks, this centripetal force vanishes suddenly.

• According to Newton’s first law of motion, the stone will continue to move with the velocity it had at that instant.

• The instantaneous velocity of a particle in circular motion is always along the tangent to the circle.

Conclusion

$$\boxed{{\displaystyle \text{The stone flies off tangentially from the instant the string breaks.}}}$$

Hence, option (b) is correct.

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Question 4.23

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.

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Answer

(a) A horse cannot pull a cart and run in empty space

A horse pulls a cart by pushing the ground backward with its legs. The ground provides an equal and opposite frictional force on the horse, which helps it move forward. In empty space, there is no ground and hence no friction, so the horse cannot get a forward reaction force and cannot pull the cart.

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(b) Passengers are thrown forward when a speeding bus stops suddenly

This happens due to inertia of motion. When the bus stops suddenly, the lower part of the passenger’s body in contact with the seat stops, but the upper part tends to continue moving forward, causing the passenger to be thrown forward.

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(c) It is easier to pull a lawn mower than to push it

When a lawn mower is pulled, the applied force has an upward component which reduces the normal reaction and hence reduces friction. When it is pushed, the downward component of force increases the normal reaction and friction, making it harder to move.

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(d) A cricketer moves his hands backwards while holding a catch

By moving his hands backward, the cricketer increases the time during which the ball’s momentum is brought to zero. Since force is given by

$$F=\frac{\mathrm{\Delta }p}{\mathrm{\Delta }t},$$

increasing the time reduces the force acting on the hands, preventing injury.