NCERT Solutions for Class 11 Physics Chapter 5

1. Give the magnitude and direction of the net force acting on 

1. a drop of rain falling down with a constant speed, 

Ans: The net force is zero.

As the speed of the rain drop falling down is constant, its acceleration is zero.

Therefore, from Newton’s second law of motion, the net force acting on the rain drop is zero.

2. a cork of mass $10g$ floating on water, 

Ans: The net force is zero.

It is known that the weight of a cork floating on water acts downward. 

The weight of the cork is balanced by buoyant force exerted by the water in the upward direction.

Therefore, no net force acts on the floating cork.

3. a kite skilfully held stationary in the sky, 

Ans: The net force is zero.

The kite is stationary in the sky indicates that it is not moving.

Therefore, from Newton’s first law of motion, the net force acting on the kite is zero.

4. a car moving with a constant velocity of $30{km}/{h}\;$ on a rough road, 

Ans: The net force is zero.

As the car is moving with constant velocity, its acceleration is zero.

Therefore, from Newton’s second law of motion, net force acting on the car is equal to zero.

5. a high-speed electron in space far from all material objects, and free of electric and magnetic fields. 

Ans: The net force is zero.

As the high speed electron is free from the influence of all the fields, no net force acts on the electron.

2. A pebble of mass \[\mathbf{0}.\mathbf{05kg}\] is thrown vertically upwards. Ignore air resistance and give the direction and magnitude of the net force on the pebble,

1. during its upward motion, 

Ans: It is known that,

Acceleration due to gravity always acts downward irrespective of the direction of motion of an object. The only force that acts on the pebble thrown vertically upward during its upward motion is the gravitational force.

From Newton’s second law of motion: $F=m\times a$ 

Where,

$F$ is the net force 

$m$ is the mass of the pebble, $m=0.05kg$ 

a is the acceleration due to gravity, $a=g=10{m}/{{{s}^{-2}}}\;$ 

$\Rightarrow F=0.05\times 10=0.5N$ 

Therefore, the net force on the pebble is $0.5N$ and this force acts in the downward direction. 

2. during its downward motion, 

Ans: The only force that acts on the pebble during its downward motion is the gravitational force.

Therefore, the net force on the pebble in its downward direction is same as in upward direction i.e., $0.5N$ and this force acts in the downward direction. 

3. at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of ${{45}^{\circ }}$ with the horizontal direction? 

Ans: When the pebble is thrown at an angle of ${{45}^{\circ }}$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. 

Therefore, the net force on the pebble is $0.5N$.

3. Neglect air resistance throughout and give the magnitude and direction of the net force acting on a stone of mass $0.1kg$ ,

1. just after it is dropped from the window of a stationary train, 

Ans: It is given that,

Mass of the stone, \[\mathbf{m}=\mathbf{0}.\mathbf{1kg}\] 

Acceleration of the stone, \[\mathbf{a}=g=10{m}/{{{s}^{2}}}\;\] 

From Newton’s second law of motion, 

The net force acting on the stone is $F=ma=mg$ 

$\Rightarrow F=0.1\times 10=1N$ 

It is known that acceleration due to gravity always acts in the downward direction. 

Therefore, the magnitude of force is $1N$ and its direction is vertically downward.

2. just after it is dropped from the window of a train running at a constant velocity of $36{km}/{h}\;$, 

Ans: It is given that,

The train is moving with a constant velocity. 

Therefore, its acceleration is zero in the direction of its motion, i.e. in the horizontal direction. 

Thus, 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. 

Therefore, the magnitude of force is $1N$ and its direction is vertically downward.

3. just after it is dropped from the window of a train accelerating with $1m{{s}^{-2}}$ 

Ans: It is given that, 

The train is accelerating at the rate of $1m{{s}^{-2}}$.

Therefore, the net force acting on the stone is $F'=ma=0.1\times 1=1N$ 

This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force 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 i.e., $F=mg=1N$.

Therefore, the magnitude of force is $1N$ and its direction is vertically downward.

4. lying on the floor of a train which is accelerating with $1m{{s}^{-2}}$, the stone being at rest relative to the train. 

Ans: It is known that,

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, $a=1m{{s}^{-2}}$

The net force acting on the stone will be in the direction of motion of the train. 

Magnitude: $F=ma=0.1\times 1=0.1N$ 

Therefore, the magnitude of force is $0.1N$ and its direction is in the direction of motion of the train.

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. $T-\frac{m{{v}^{2}}}{l}$ ,

iii. $T+\frac{m{{v}^{2}}}{l}$,

iv. $0$ 

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

Ans: (i) $T$ 

The centripetal force of a particle connected to a string revolving in a circular path around a centre is provided by the tension produced in the string.

Therefore, the net force on the particle is the tension $T$ , i.e., 

$F=T=\frac{m{{v}^{2}}}{l}$ 

Where $F$ is the net force acting on the particle. 

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

Ans: It is given that,

Retarding force, \[F=50N\] 

Mass of the body, $m=20kg$ 

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

Final velocity of the body, $v=0$ 

From Newton’s second law of motion,

The acceleration $(a)$ produced in the body: $F=ma$ 

$\Rightarrow -50=20\times a$ 

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

From the first equation of motion, 

The time $(t)$ taken by the body to come to rest: $v=u+at$ 

$\Rightarrow 0=15+(-2.5)t$ 

$\Rightarrow t=\frac{-15}{-2.5}=6s$ 

Therefore, the time taken by the body to stop is $6s$.

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

Ans: It is given that,

Mass of the body, $m=3kg$ 

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

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

Time, $t=25s$ 

From the first equation of motion, 

The acceleration $(a)$ produced in the body: $v=u+at$

$\Rightarrow a=\frac{v-u}{t}$ 

$\Rightarrow a=\frac{3.5-2}{25}=\frac{1.5}{25}=0.6m{{s}^{-2}}$ 

From Newton’s second law of motion, 

Force, $F=ma$ 

\[\Rightarrow F=3\times 0.06=0.18N\] 

As 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. 

Therefore, the magnitude of force is $0.18N$ and direction is along the direction of motion.

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

Ans: It is given that,

Mass of the body, $m=5kg$ 

Representation of given data:

(Image will be uploaded soon)

Resultant of two forces $8N$ and $6N$, $R=\sqrt{{{\left( 8 \right)}^{2}}+{{\left( -6 \right)}^{2}}}$ 

$\Rightarrow R=\sqrt{64+36}$ 

$\Rightarrow R=10N$ 

Angle made by $R$ with the force of $8N$ 

$\theta ={{\tan }^{-1}}\left( \frac{-6}{8} \right)=-{{36.87}^{\circ }}$ 

The negative sign indicates that $\theta $ is in the clockwise direction with respect to the force of magnitude $8N$. 

From Newton’s second law of motion,

The acceleration $(a)$ produced in the body: $F=ma$ 

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

Therefore, the magnitude of acceleration is $2m{{s}^{-2}}$ and direction is ${{37}^{\circ }}$ with a force of $8N$.

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 $4.0s$  just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is $400kg$ and the mass of the driver is $65kg$.

Ans: It is given that,

Initial speed of the three-wheeler, \[u=36\text{ }km/h\] 

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

Time, $t=4s$ 

Mass of the three-wheeler, $m=400kg$ 

Mass of the driver, $m'=65kg$ 

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

From the first law of motion, 

The acceleration $(a)$ of the three-wheeler can be calculated from: $v=u+at$ 

$\Rightarrow a=\frac{v-u}{t}=\frac{0-10}{4}$ 

$\Rightarrow a=-2.5{m}/{{{s}^{2}}}\;$ 

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

From Newton’s second law of motion, 

The net force acting on the three-wheeler can be calculated as: $F=ma$ 

$\Rightarrow F=465\times (-2.5)$ 

$\Rightarrow F=-1162.5N$ 

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

Therefore, the average retarding force on the vehicle is $-1162.5N$.

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

Ans: It is known that,

Mass of the rocket, $m=20,000kg$ 

Initial acceleration, $a=5m{{s}^{-2}}$ 

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

By using Newton’s second law of motion, 

The net force (thrust) acting on the rocket can be written as:

$F-mg=ma$ 

$\Rightarrow F=m(g+a)$ 

$\Rightarrow F=20000(10+5)=20000\times 15$ 

$\Rightarrow F=3\times {{10}^{5}}N$  

Therefore, the initial thrust(force) of the blast is $3\times {{10}^{5}}N$.

10. A body of mass \[\mathbf{0}.\mathbf{40kg}\] moving initially with a constant speed of $10m{{s}^{-1}}$ subject to a constant force of $8.0N$ directed towards the south for $30s$. Take the instant the force is applied to be $t=0$ , the position of the body at that time to be predict its position at \[\mathbf{t}=-\mathbf{5}\text{ }\mathbf{s},\text{ }\mathbf{25}\text{ }\mathbf{s},\text{ }\mathbf{100}\text{ }\mathbf{s}\].

Ans: It is given that, 

Mass of the body, $m=0.40kg$ 

Initial speed of the body, \[u=10m/s\] due north 

Force acting on the body, $F=-8.0N$ 

Acceleration produced in the body, $a=\frac{F}{m}$

At $t=0$ 

$\Rightarrow a=\frac{-8}{0.4}=-20m{{s}^{-2}}$ 

At $t=-5s$ 

Acceleration, $a'=0$ and $u=10m/s$ 

$s=ut+\frac{1}{2}a'{{t}^{2}}$ 

$\Rightarrow s=10\times (-5)=-50m$ 

At $t=25s$

Acceleration, $a=-20m{{s}^{-2}}$ and $u=10m/s$ 

$s'=ut'+\frac{1}{2}a{{(t')}^{2}}$ 

$\Rightarrow s'=10\times (25)+\frac{1}{2}\times (-20){{(25)}^{2}}$ 

$\Rightarrow s'=250+(-6250)$ 

$\Rightarrow s'=-6000m$ 

At $t=100s$

For $0\le t\le 30s$, $a=-20m{{s}^{-2}}$ and $u=10m/s$

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

$\Rightarrow {{s}_{1}}=10\times 30+\frac{1}{2}\times (-20)\times {{(30)}^{2}}$

$\Rightarrow {{s}_{1}}=300-9000$

$\Rightarrow {{s}_{1}}=8700m$

For $30s\le t\le 100s$

First equation of motion: $v=u+at$ 

$\Rightarrow v=10+(-20)\times 30$ 

$\Rightarrow v=-590m{{s}^{-1}}$ 

Velocity of body after $30s=-590m/s$ 

For motion between $30s$ to $100s$ i.e., in $70s$ 

${{s}_{2}}=vt+\frac{1}{2}a'{{t}^{2}}$

${{s}_{2}}=-590\times 70=-41300m$

Total distance, $s''={{s}_{1}}+{{s}_{2}}$ 

$\Rightarrow s''=-8700+(-41300)=-50000m$ 

Therefore, the position of the body at $t=-5s$ is $-50m$ at $t=25s$ is  $-6000m$ and at $t=100s$ is$-50,000m$.

11. A truck starts from rest and accelerates uniformly at $2.0m{{s}^{-2}}$. At $t=10s$, a stone is dropped by a person standing on the top of the truck ($6m$ high from the ground). Neglect air resistance. What are the

1. velocity, and 

Ans: It is given that,

Initial velocity of the truck, $u=0$ (Initially at rest)

Acceleration, $a=2m{{s}^{-2}}$ 

Time, $t=10s$ 

From first equation of motion: $v=u+at$ 

$\Rightarrow v=0+2\times 10=20{m}/{s}\;$ 

Therefore, the final velocity of the truck and the stone is$20{m}/{s}\;$.

At $t=11s$:

The horizontal component $({{v}_{x}})$  of velocity, in the absence of air resistance, remains unchanged, i.e. \[{{v}_{x}}=20m/s\].

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

${{v}_{y}}=u+{{a}_{y}}\delta t$ 

Where, 

$\delta t=11-10=1s$ 

$a=g=10m/{{s}^{2}}$ 

$\Rightarrow {{v}_{y}}=0+10\times 1=10{m}/{{{s}^{2}}}\;$ 

The resultant velocity $(v)$  of the stone is:

(Image will be uploaded soon)

$\Rightarrow v=\sqrt{v_{x}^{2}+v_{y}^{2}}$ 

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

$\Rightarrow v=\sqrt{500}=22.36m/s$

Consider $\theta $  as the angle made by the resultant velocity with the horizontal component of velocity, ${{v}_{x}}$.

$\Rightarrow \tan \theta =\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)$ 

$\Rightarrow \theta ={{\tan }^{-1}}\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)$ 

$\Rightarrow \theta ={{\tan }^{-1}}\left( \frac{10}{20} \right)$

$\Rightarrow \theta ={{\tan }^{-1}}\left( 0.5 \right)$

$\Rightarrow \theta ={{26.57}^{\circ }}$

Therefore, the magnitude of resultant velocity is $22.36{m}/{s}\;$ making an angle of ${{26.57}^{\circ }}$ with the horizontal component of velocity.

2. acceleration of the stone at $t=11s$?

Ans: 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. 

Therefore, the acceleration of the stone is $10m{{s}^{-2}}$  and it acts vertically downward. 

12. A bob of mass \[\mathbf{0}.\mathbf{1kg}\] hung from the ceiling of a room by a string $2m$ long is set into oscillation. The speed of the bob at its mean position is $1m{{s}^{-2}}$. What is the trajectory of the bob if the string is cut when the bob is

1. at one of its extreme positions, 

Ans: If the string is cut when the bob is at one of its extremes then the bob will fall vertically on the ground.

Therefore, at the extreme position, the velocity of the bob becomes zero.

2. at its mean position. 

Ans: If the string is cut when the bob is at its mean position then the bob will trace a projectile path having the horizontal components of velocity only.

The direction of this velocity is tangential to the arc formed by the oscillating bob. At the mean position, the velocity of the bob is $1m/s$. 

Therefore, it will follow a parabolic path. 

13. What would be the readings on the scale of a man of mass \[\mathbf{70kg}\] stands on a weighing scale in a lift which is moving

1. upwards with a uniform speed of $10m{{s}^{-1}}$,

Ans: It is given that,

Mass of the man, $m=70kg$ 

Acceleration, $a=0$(uniform speed)

From Newton’s second law: $R-mg=ma$ 

Where,

$ma$ is the net force acting on the man.

$\Rightarrow R-70\times 10=0$ 

$\Rightarrow R=700N$ 

Reading on the weighing scale$=\frac{700}{g}=\frac{700}{10}=70kg$ 

Therefore, the mass of the man, $m=70kg$ 

2. downwards with a uniform acceleration of $5m{{s}^{-2}}$,

Ans: Acceleration,$a=5{m}/{{{s}^{2}}}\;$ downward

From Newton’s second law: $R=m(g-a)$

$\Rightarrow R=70\left( 10-5 \right)=70\times 5$ 

$\Rightarrow R=350N$ 

Reading on the weighing scale$=\frac{350}{g}=\frac{350}{10}=35kg$ 

Therefore, the mass of the man, $m=35kg$ 

3. upwards with a uniform acceleration of $5m{{s}^{-2}}$.

Ans: Acceleration,$a=5{m}/{{{s}^{2}}}\;$ upward

From Newton’s second law: $R=m(g+a)$

$\Rightarrow R=70\left( 10+5 \right)=70\times 15$ 

$\Rightarrow R=1050N$ 

Reading on the weighing scale$=\frac{1050}{g}=\frac{1050}{10}=105kg$ 

Therefore, the mass of the man, $m=105kg$ 

4. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity? 

Ans: When the lift moves freely under gravity,

Acceleration, $a=g=10m{{s}^{-2}}$ 

From Newton’s second law: $R=m(g-a)$

$\Rightarrow R=70\left( 10-10 \right)=0$ 

Reading on the weighing scale$=\frac{0}{g}=\frac{0}{10}=0kg$ 

Therefore, the man will be in a state of weightlessness.

14. Figure shows the position-time graph of a particle of mass $4kg$ . Consider one-dimensional motion only and find the 

(Image will be uploaded soon)

1. force on the particle for $t<0,t>4s$,$0<t<4s$ ?

Ans:  For $t<0$ :

From the given graph, it is observed 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. 

Therefore, the force acting on the particle is zero. 

For $t>4s$ :

From the given graph, it is observed that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of $3m$ from the origin.

Therefore, no force acts on the particle. 

For $0<t<4$ :

From the given position-time graph, it is observed that it has a constant slope. Thus, the acceleration produced in the particle is zero. 

Therefore, the force acting on the particle is zero. 

2. impulse at $t=0$ and $t=4s$ ? 

Ans: At $t=0$:

\[Impulse=Change\text{ }in\text{ }momentum=mv-mu\] 

Mass of the particle, $m=4kg$

Initial velocity of the particle, $u=0$ 

Final velocity of the particle, $v=\frac{3}{4}m/s$ 

Impulse $=4\left( \frac{3}{4}-0 \right)=3kgm{{s}^{-1}}$ 

At $t=4s$:

Initial velocity of the particle, $u=\frac{3}{4}m/s$ 

Final velocity of the particle, $v=0$ 

Impulse\[=4\left( 0-\frac{3}{4} \right)=-3kgm{{s}^{-1}}\]

Therefore, the impulse at $t=0$ is $3kgm{{s}^{-1}}$ and at $t=4s$is $-3kgm{{s}^{-1}}$.

15. Two bodies of masses $10kg$  and $20kg$ respectively kept on a smooth, horizontal surface are tied to the ends of a light string. What is the tension in the string if a horizontal force $F=600N$ is applied along the direction of string to a) A, b) B along the direction of the string. What is the tension in the string in each case?

1. A, 

Ans: It is given that,

Horizontal force, $F=600N$ 

Mass of body A, ${{m}_{1}}=10kg$ 

Mass of body B, ${{m}_{2}}=20kg$

Total mass of the system, $m={{m}_{1}}+{{m}_{2}}=30kg$

From Newton’s second law of motion, 

The acceleration $(a)$ produced in the system is: $F=ma$ 

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

When force $F$ is applied on body A: 

(Image will be uploaded soon)

The equation of motion can be written as: $F-T=ma$

$\Rightarrow T=F-{{m}_{1}}a$ 

\[\Rightarrow T=600-10\times 20=400N\]

Therefore, the tension in the string is $400N$.

2. B along the direction of the string

Ans: When force $F$  is applied on body B: 

(Image will be uploaded soon)

The equation of motion can be written as:  $F-T={{m}_{2}}a$ 

$\Rightarrow T=F-{{m}_{2}}a$ 

\[\Rightarrow T=600-20\times 20=200N\]

Therefore, the tension in the string is $200N$.

16. Two masses $8kg$ and $12kg$ 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.

Ans: The given system of two masses and a pulley are represented in the following figure: 

(Image will be uploaded soon)

Smaller mass, ${{m}_{1}}=8kg$ 

Larger mass, ${{m}_{2}}=12kg$

Tension in the string $=T$ 

Mass \[{{m}_{2}}\], owing to its weight, moves downward with acceleration $a$, and mass moves ${{m}_{1}}$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: \[T-{{m}_{1}}g=ma\]    .......$(1)$ 

For mass ${{m}_{2}}$:

The equation of motion can be written as: \[{{m}_{2}}g-T={{m}_{2}}a\]   ........$(2)$ 

Adding equations $(1)$ and $(2)$, we get:

$({{m}_{2}}-{{m}_{1}})g=({{m}_{1}}+{{m}_{2}})a$ 

$\Rightarrow a=\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{2}}+{{m}_{1}}} \right)g$        ........$(3)$ 

$\Rightarrow a=\left( \frac{12-8}{12+8} \right)\times 10=\frac{4}{20}\times 10$

$\Rightarrow a=2m{{s}^{-2}}$ 

Thus, the acceleration of the masses is $2m{{s}^{-2}}$ . Substituting the value of  $a$ in equation $(2)$:

\[\Rightarrow {{m}_{2}}g-T={{m}_{2}}\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{2}}+{{m}_{1}}} \right)g\]

\[\Rightarrow T=\left( {{m}_{2}}-\frac{m_{2}^{2}-{{m}_{1}}{{m}_{2}}}{{{m}_{2}}+{{m}_{1}}} \right)g\]

\[\Rightarrow T=\left( \frac{2{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)g\]

\[\Rightarrow T=\left( \frac{2\times 12\times 8}{12+8} \right)\times 10\]

\[\Rightarrow T=96N\]

Thus, the tension in the string is $96N$.

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. 

Ans: Consider $m$, ${{m}_{1}}$ and ${{m}_{2}}$ as 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$={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$ 

From the law of conservation of momentum: 

\[\mathbf{Total}\text{ }\mathbf{initial}\text{ }\mathbf{momentum}=\mathbf{Total}\text{ }\mathbf{final}\text{ }\mathbf{momentum}\] 

 $\Rightarrow 0={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$ 

$\Rightarrow {{m}_{1}}{{v}_{1}}=-{{m}_{2}}{{v}_{2}}$

$\Rightarrow {{v}_{1}}=\frac{-{{m}_{2}}{{v}_{2}}}{{{m}_{1}}}$

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

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

Ans: It is given that,

Mass of each ball$=0.05kg$ 

Initial velocity of each ball$=6m/s$ 

Magnitude of the initial momentum of each ball, ${{p}_{i}}=0.3kgm{{s}^{-1}}$ 

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

magnitudes of their velocity. 

Final momentum of each ball, ${{p}_{f}}=-0.3kgm{{s}^{-1}}$

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

$\Rightarrow \operatorname{Im}pulse={{p}_{f}}-{{p}_{i}}$ 

$\Rightarrow \operatorname{Im}pulse=-0.3-0.3=-0.6kgm{{s}^{-1}}$

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

19. A shell of mass \[\mathbf{0}.\mathbf{020kg}\] is fired by a gun of mass \[\mathbf{100kg}\]. If the muzzle speed of the shell is $80m{{s}^{-1}}$ what is the recoil speed of the gun?

Ans: It is given that,

Mass of the gun, $M=100kg$ 

Mass of the shell, $m=0.020kg$ 

Muzzle speed of the shell, $v=80m/s$ 

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. 

From the law of conservation of momentum:

\[Final\text{ }momentum=Initial\text{ }momentum\] 

$mv-MV=0$

$\Rightarrow V=\frac{mv}{M}$ 

$\Rightarrow V=\frac{0.020\times 80}{100\times 1000}=0.016m/s$ 

Therefore, the recoil speed of the gun is $0.016m/s$.

20. A batsman deflects a ball by an angle of ${{45}^{\circ }}$ 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 $0.15kg$) 

Ans: The given situation can be represented as:

(Image will be uploaded soon)

Where, 

$AO=$ Incident path of the ball 

$OB=$ Path followed by the ball after a deflection 

$\angle AOB=$ Angle between the incident and deflected paths of the ball$={{45}^{\circ }}$ 

$\angle AOB=\angle BOP={{22.5}^{\circ }}=\theta $

Initial and final velocities of the ball$=v$ 

The horizontal component of the initial velocity$=v\cos \theta $ along $RO$ 

Vertical component of the initial velocity$=v\sin \theta $ along $PO$ 

The horizontal component of the final velocity$=v\cos \theta $ along $OS$ 

The vertical component of the final velocity $=v\sin \theta $ along $OP$ 

The horizontal components of velocities suffer no change. The vertical components of velocities are in opposite directions. 

It is known that Impulse imparted to the ball$=$ Change in the linear momentum of the ball.

$\operatorname{Im}pulse=mv\cos \theta -(-mv\cos \theta )=2mv\cos \theta $ 

It is given that,

Mass of the ball, $m=0.15kg$ 

Velocity of the ball, $v=54km/h=54\times \frac{5}{18}=15m/s$ 

Impulse$=2\times 0.15\times 15\cos {{22.5}^{\circ }}=4.16kgm{{s}^{-1}}$ 

Therefore, impulse imparted to the ball is $4.16kgm{{s}^{-1}}$.

21. A stone of mass $0.25kg$  tied to the end of a string is whirled round in a circle of radius $1.5m$ with a speed of $40rev./\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 $200N$?

Ans: It is given that,

Mass of the stone, $m=0.25kg$ 

Radius of the circle, $r=1.5m$ 

Number of revolution per second, $n=\frac{40}{60}=\frac{2}{3}rps$ 

Angular velocity, $\omega =\frac{v}{r}=2\pi n$     ........$(1)$ 

The centripetal force for the stone is provided by the tension $T$ , in the string, i.e., $T={{F}_{Centripetal}}$

$\Rightarrow \frac{m{{v}^{2}}}{r}=mr{{\omega }^{2}}=mr{{\left( 2\pi n \right)}^{2}}$ 

$\Rightarrow {{F}_{Centripetal}}=0.25\times 1.5\times {{\left( 2\times 3.14\times \frac{2}{3} \right)}^{2}}$

$\Rightarrow {{F}_{Centripetal}}=6.57N$

Maximum tension in the string, ${{T}_{\max }}=200N$ 

${{T}_{\max }}=\frac{mv_{\max }^{2}}{r}$

$\Rightarrow {{v}_{\max }}=\sqrt{\frac{{{T}_{\max }}\times r}{m}}$

$\Rightarrow {{v}_{\max }}=\sqrt{\frac{200\times 1.5}{0.25}}$

\[\Rightarrow {{v}_{\max }}=\sqrt{1200}=34.64m/s\]

Thus, the maximum speed of the stone is \[34.64m/s\].

22. If, in Exercise 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: 

1. the stone moves radially out wards, 

2. the stone flies off tangentially from the instant the string breaks, 

3. the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle? 

Ans: $(ii)$ 

From the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. So, if the string breaks, the stone will move in the direction of the velocity at that instant. 

Therefore, the stone will fly off tangentially from the instant the string breaks. 

23. Explain why

1. a horse cannot pull a cart and run in empty space, 

Ans: 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. 

Hence, a horse cannot pull a cart and run in empty space.

2. passengers are thrown forward from their seats when a speeding bus stops suddenly, 

Ans: If 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).

So, the passenger’s upper body is thrown forward in the direction in which the bus was moving. 

3. it is easier to pull a lawn mower than to push it, 

Ans: While pulling a lawn mower, a force at an angle $\theta $  is applied on it, as shown in the following figure:

 (Image will be uploaded soon)

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

While pushing a lawn mower, a force at an angle $\theta $ is applied on it, as shown in the following figure:

(Image will be uploaded soon)

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. 

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

From Newton’s second law of motion: $F=ma=m\frac{\Delta v}{\Delta t}$  ........$(1)$ 

Where, 

$F=$ Stopping force experienced by the cricketer as he catches the ball 

$m=$ Mass of the ball 

$t=$ Time of impact of the ball with the hand 

From equation $(1)$it can be observed that the impact force is inversely proportional to the impact time, i.e., $F<\frac{1}{\Delta t}$       ........$(2)$ 

Equation $(2)$  shows that the force experienced by the cricketer decreases if the time of impact increases and vice versa. 

Therefore, it is easier to pull a lawn mower than to push it.

4. a cricketer moves his hands backwards while holding a catch.

Ans: While taking a catch, a cricketer moves his hand backward so as to increase the time of impact $\Delta t$. This in turn results in the decrease in the stopping force, thereby preventing the hands of the cricketer from getting hurt.

Therefore, a cricketer moves his hands backwards while holding a catch.

Additional Exercise

24. Figure shows the position-time graph of a body of mass $0.04kg$ . Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse? 

(Image will be uploaded soon)

Ans: In the given case, a ball is rebounding between two walls located between at $x=0$ and $x=2cm$ ; after every $2s$, the ball receives an impulse of magnitude $0.08\times {{10}^{-2}}kgm{{s}^{-1}}$ from the walls.

The given graph shows that a body changes its direction of motion after every $2s$.

This situation can be visualized as a ball rebounding to and fro between two stationary walls situated between positions $x=0$ and $x=2cm$.

As the slope of the x-t graph reverses after every $2s$ , the ball collides with a wall after every $2s$. Therefore, ball receives an impulse after every $2s$.

Mass of the ball, $m=0.04kg$ 

The slope of the graph gives the velocity of the ball. 

From the graph, initial velocity is $u=\frac{(2-0)\times {{10}^{-2}}}{(2-0)}={{10}^{-2}}m/s$ 

Velocity of the ball before collision, $u={{10}^{-2}}m/s$

Velocity of the ball after collision, $v=-{{10}^{-2}}m/s$

(The negative sign arises as the ball reverses its direction of motion.) 

Magnitude of impulse = Change in momentum 

$\Rightarrow \left| mv-mu \right|=\left| 0.04(-{{10}^{-2}}-{{10}^{-2}}) \right|$ 

$\Rightarrow 0.08\times {{10}^{-2}}kgm{{s}^{-1}}$ 

Therefore, the magnitude of impulse is $0.08\times {{10}^{-2}}kgm{{s}^{-1}}$.

25. Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with $1m{{s}^{-2}}$. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is $0.2$, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man $=65kg$)

(Image will be uploaded soon)

Ans: It is given that,

Mass of the man, $m=65kg$ 

Acceleration of the belt, $a=1m{{s}^{-2}}$ 

Coefficient of static friction, $\mu =0.2$ 

The net force $F$, acting on the man is given by Newton’s second law of motion as: ${{F}_{net}}=ma$

$\Rightarrow {{F}_{net}}=65\times 1=65N$ 

The man will continue to be stationary with respect to the convey or belt until the net force on the man is less than or equal to the frictional force ${{f}_{s}}$ exerted by the belt, i.e., $F{{'}_{net}}={{f}_{s}}$ 

$\Rightarrow ma'=\mu mg$ 

$\Rightarrow a'=0.2\times 10=2m{{s}^{-2}}$ 

Therefore, the maximum acceleration of the belt up to which the man can stand stationary is $a=2m{{s}^{-2}}$.

26. A stone of mass $m$ tied to the end of a string revolves in a vertical circle of radius $R$. The net forces at the lowest and highest points of the circle directed vertically downwards are: (Choose the correct alternative)

${{T}_{1}}$ and ${{v}_{1}}$ denote the tension and speed at the lowest point. ${{T}_{2}}$ and ${{v}_{2}}$ denote the tension and speed at the highest point.

Ans: (a)

The free body diagram of the stone at the lowest point is:

(Image will be uploaded soon)

From Newton’s second law of motion, 

The net force acting on the stone at this point is equal to the centripetal force:${{F}_{net}}=T-mg=\frac{mv_{1}^{2}}{R}..........(1)$ 

Where, ${{v}_{1}}$ is the velocity at the lowest point.

The free body diagram of the stone at the highest point is:

(Image will be uploaded soon)

From Newton’s second law of motion: $T+mg=\frac{mv_{2}^{2}}{R}.........(2)$ 

Where, ${{v}_{2}}$ is the velocity at the highest point.

From equations $(1)$  and $(2)$ the net force acting at the lowest and the highest points are $(T-mg)$ and $(T+mg)$ respectively.

27. A helicopter of mass $1000kg$rises with a vertical acceleration of $15m{{s}^{-2}}$. The crew and the passengers weigh $300kg$. Give the magnitude and direction of the 

1. force on the floor by the crew and passengers, 

Ans: It is given that,

Mass of the helicopter, ${{m}_{h}}=1000kg$ 

Mass of the crew and passengers, ${{m}_{p}}=300kg$ 

Total mass of the system, $m=1300kg$ 

Acceleration of the helicopter, $a=15m{{s}^{-2}}$ 

From Newton’s second law of motion, 

The reaction force R, on the system by the floor is $R-{{m}_{p}}g=ma$ 

$\Rightarrow R={{m}_{p}}(g+a)$ 

$\Rightarrow R=1300(10+15)=1300\times 25$

$\Rightarrow R=7500N$

As the helicopter is accelerating vertically upward, the reaction force will also be directed upward. 

Hence, from Newton’s third law of motion, the force on the floor by the crew and passengers is $7500N$, directed downward. 

2. action of the rotor of the helicopter on the surrounding air, 

Ans: From Newton’s second law of motion, 

The reaction force $R'$ experienced by the helicopter is: $R'-mg=ma$ 

$\Rightarrow R'=m(g+a)$ 

$\Rightarrow R'=1300(10+15)$

$\Rightarrow R'=1300\times 25$

$\Rightarrow R'=32500N$

Hence, from Newton’s third law of motion, the action of the rotor on the surrounding air will be $32500N$, directed downward. 

3. force on the helicopter due to the surrounding air. 

Ans: The action of the rotor on the surrounding air is $32500N$, directed downward. Therefore, the reaction force experienced by the helicopter from the surrounding air is acting upward. 

Thus, the force on the helicopter due to the surrounding air is$32500N$, directed upward. 

28. A stream of water flowing horizontally with a speed of $15m{{s}^{-1}}$ gushes out of a tube of cross-sectional area ${{10}^{-2}}{{m}^{2}}$ and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound? 

Ans: It is given that,

Speed of the water stream, $v=15m/s$ 

Cross-sectional area of the tube, $A={{10}^{-2}}{{m}^{2}}$ 

Volume of water coming out from the pipe per second, $V=Av=15\times {{10}^{-2}}{{m}^{3}}/s$ 

Density of water, $\rho ={{10}^{3}}kg/{{m}^{3}}$ 

Mass of water flowing out through the pipe per second$=\rho \times V=150kg/s$ 

The water strikes the wall and does not rebound.

Therefore, the force exerted by the water on the wall is given by Newton’s second law of motion as: $F=\frac{\Delta P}{\Delta t}$ 

Where, $F=$ Rate of change of momentum per second

$\Rightarrow F=\frac{mv}{t}=\frac{150\times 15}{1}$ 

$\Rightarrow F=2250N$

Therefore, the force exerted on the wall is $2250N$.

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 

1. the force on the ${{7}^{th}}$ coin (counted from the bottom) due to all the coins on its top, 

Ans: It is given that,

Weight of one coin$=mg$ 

Weight of three coins$=3mg$ 

Force on the seventh coin is exerted by the weight of the three coins on its top. 

Therefore, the force exerted on the ${{7}^{th}}$coin by the three coins on its top is $3mg$ . This force acts vertically downward. 

2. the force on the ${{7}^{th}}$ coin by the eighth coin, 

Ans: 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$=3mg$

Therefore, the force exerted on the ${{7}^{th}}$ coin by the eighth coin is $3mg$. This force acts vertically downward. 

3. the reaction of the ${{6}^{th}}$ coin on the ${{7}^{th}}$ coin. 

Ans: The ${{6}^{th}}$ 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. 

Thus, the total downward force experienced by the ${{6}^{th}}$ coin is $4mg$. 

From Newton’s third law of motion, 

The ${{6}^{th}}$ coin will produce an equal reaction force on the ${{7}^{th}}$ coin, but in the opposite direction. 

Therefore, the reaction force of the ${{6}^{th}}$ coin on the ${{7}^{th}}$coin is of magnitude $4mg$. This force acts in the upward direction. 

30. An aircraft executes a horizontal loop at a speed of $720km/h$ with its wings banked at ${{15}^{\circ }}$. What is the radius of the loop?

Ans: It is given that

Speed of the aircraft, $v=720km/h$ 

$\Rightarrow v=720\times \frac{5}{18}=200m/s$ 

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

Angle of banking, $\theta ={{15}^{\circ }}$ 

Let $r$ be the radius of the loop.

It is known that the angle of banking is related to the radius $(r)$  and speed $(v)$ by the relation: $\tan \theta =\frac{{{v}^{2}}}{rg}$

 $\Rightarrow r=\frac{{{v}^{2}}}{g\tan \theta }$ 

$\Rightarrow r=\frac{{{(200)}^{2}}}{10\times \tan {{15}^{\circ }}}=\frac{4000}{0.2679}$

$\Rightarrow r=14,928.2m$

Therefore, the radius of the loop is $14,928.2m$.

31. A train runs along an unbanked circular track of radius $30m$ at a speed of \[\mathbf{54km}/\mathbf{h}\]. The mass of the train is ${{10}^{6}}kg$ . 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? 

Ans: It is given that,

Radius of the circular track, $r=30m$ 

Speed of the train, \[\mathbf{v}=\mathbf{54}\text{ }\mathbf{km}/\mathbf{h}=54\times \frac{5}{18}=15\mathbf{m}/\mathbf{s}\] 

Mass of the train, $m={{10}^{6}}kg$ 

The centripetal force is provided by the lateral thrust of the rail on the wheel. 

Therefore, from 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.

It is known that the angle of banking is related to the radius $(r)$  and speed $(v)$ by the relation: $\tan \theta =\frac{{{v}^{2}}}{rg}$

$\Rightarrow \tan \theta =\frac{{{(15)}^{2}}}{30\times 10}=\frac{225}{300}$

$\Rightarrow \theta ={{\tan }^{-1}}(0.75)={{36.87}^{\circ }}$ 

Therefore, the angle of banking required to prevent wearing out of the rail is ${{36.87}^{\circ }}$.

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

(Image will be uploaded soon)

Ans: It is given that,

Mass of the man, $M=50kg$ 

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

Force applied on the block, \[\mathbf{F}=\mathbf{25}\times \mathbf{10}=\mathbf{250N}\] 

Weight of the man, \[\mathbf{W}=\mathbf{50}\times \mathbf{10}=\mathbf{500N}\] 

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 \[=\mathbf{250}+\mathbf{500}=\mathbf{750N}\] 

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\[=\mathbf{500}\mathbf{250}=\mathbf{250N}\] 

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

Therefore, case $(b)$ is adopted.

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

(Image will be uploaded soon)

1. climbs up with an acceleration of $6m{{s}^{-2}}$ 

Ans: It is given that,

Mass of the monkey, $m=40kg$ m

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

Maximum tension that the rope can bear, ${{T}_{\max }}=600N$ 

Acceleration of the monkey, $a=6{m}/{{{s}^{2}}}\;$ upward

From Newton’s second law of motion: 

It can be written: \[Tmg=ma\] 

\[\Rightarrow T=m(a+g)\]

\[\Rightarrow T=40(6+10)\]

\[\Rightarrow T=40\times 16=640N\]

Therefore, as $T>{{T}_{\max }}$ the rope will break.

2. climbs down with an acceleration of $4m{{s}^{-2}}$

Ans: Acceleration of the monkey, $a=4{m}/{{{s}^{2}}}\;$ downward 

From Newton’s second law of motion: $mg-T=ma$ 

$\Rightarrow T=m(g-a)$ 

$\Rightarrow T=40(10-4)=240N$

Therefore, as $T<{{T}_{\max }}$ the rope will not break.

3. climbs up with a uniform speed of $5m{{s}^{-1}}$

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

From Newton’s second law of motion, $T-mg=ma$ 

$\Rightarrow T=mg$ 

$\Rightarrow T=40\times 10=400N$ 

Therefore, as $T<{{T}_{\max }}$ the rope will not break.

4. falls down the rope nearly freely under gravity? 

Ans: When the monkey falls freely under gravity, its acceleration will become equal to the acceleration due to gravity, $a=g$ 

From Newton’s second law of motion: $mg-T=mg$ 

$\Rightarrow T=m(g-g)=0$ 

Therefore, as $T<{{T}_{\max }}$ the rope will not break.

34. Two bodies A and B of masses $5kg$ and $10kg$ in contact with each other rest on a table against a rigid wall (figure). The coefficient of friction between the bodies and the table is $0.15$. Ignore the difference between ${{\mu }_{s}}$ and ${{\mu }_{k}}$.A force of $200N$is applied horizontally to A. What are

(Image will be uploaded soon)

1. the reaction of the partition

Ans: It is given that,

Mass of body A, ${{m}_{A}}=5kg$ 

Mass of body B, ${{m}_{B}}=10kg$

Applied force, $F=200N$ 

Coefficient of friction, ${{\mu }_{s}}=0.15$ 

The force of friction is: ${{f}_{s}}=\mu ({{m}_{A}}+{{m}_{B}})g$ 

$\Rightarrow {{f}_{s}}=0.15(5+10)\times 10$

$\Rightarrow {{f}_{s}}=1.5\times 15=22.5N$ leftward 

Net force acting on the partition \[=20022.5=177.5N\] rightward 

From Newton’s third law of motion, 

The reaction force of the partition will be in the direction opposite to the net applied force. 

Therefore, the reaction of the partition will be $177.5N$, in the leftward direction.

2. the action-reaction forces between A and B? What happens when the wall is removed? Does the answer to $(b)$ change, when the bodies are in motion? 

Ans: Force of friction on mass $A$ : ${{f}_{A}}=\mu {{m}_{A}}g$ 

$\Rightarrow {{f}_{A}}=0.15\times 5\times 10=7.5N$ leftward 

Net force exerted by mass $A$ on mass $B$  \[=2007.5=192.5N\] rightward 

From Newton’s third law of motion,

An equal amount of reaction force will be exerted by mass $B$ on mass $A$, i.e., \[192.5N\] acting leftward. 

When the wall is removed, the two bodies will move in the direction of the applied force. 

Net force acting on the moving system$=177.5N$ 

The equation of motion for the system of acceleration $a$,$Net\text{ }force=\left( {{m}_{A}}+{{m}_{B}} \right)a$ 

$\Rightarrow a=\frac{Net\text{ }force}{\left( {{m}_{A}}+{{m}_{B}} \right)}$

$\Rightarrow a=\frac{177.5}{\left( 5+10 \right)}=\frac{177.5}{15}=11.83m{{s}^{-2}}$

Net force causing mass $A$ to move: ${{F}_{A}}={{m}_{A}}a$ 

$\Rightarrow {{F}_{A}}=5\times 11.83=59.15N$

Net force exerted by mass $A$ on mass $B$\[=192.559.15=133.35N\] 

This force will act in the direction of motion. 

From Newton’s third law of motion, an equal amount of force will be exerted by mass $B$ on mass $A$ , i.e., $133.3N$ , acting opposite to the direction of motion. 

35. A block of mass $15kg$ 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 $0.5m{{s}^{-2}}$ for $20s$ and then moves with uniform velocity. Discuss the motion of the block as viewed by 

1. a stationary observer on the ground, 

Ans: It is given that,

Mass of the block, $m=15kg$ 

Coefficient of static friction, $\mu =0.18$ 

Acceleration of the trolley, $a=0.5m{{s}^{-2}}$ 

From Newton’s second law of motion, 

The force $(F)$ on the block caused by the motion of the trolley is: $F=ma$ 

$\Rightarrow F=15\times 0.5=7.5N$ 

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

Force of static friction between the block and the trolley

$f=\mu mg$ 

$f=0.18\times 15\times 10=27N$

The force of static friction between the block and the trolley is greater than the applied external force. 

Therefore, 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.

2. an observer moving with the trolley. 

Ans: 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. 

Therefore, the trolley will appear to be at rest for the observer moving with the trolley. 

36. The rear side of a truck is open and a box of $40kg$ mass is placed $5m$ away from the open end as shown in figure. The coefficient of friction between the box and the surface below it is $0.15$. On a straight road, the truck starts from rest and accelerates with $2m{{s}^{-2}}$. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

(Image will be uploaded soon)

Ans: It is given that,

Mass of the box, $m=40kg$ 

Coefficient of friction, $\mu =0.15$ 

Initial velocity, $u=0$ 

Acceleration, $a=2m/{{s}^{2}}$ 

Distance of the box from the end of the truck, $s'=5m$ 

From Newton’s second law of motion, 

The force on the box caused by the accelerated motion of the truck is: $F=ma=2\times 40=80N$ 

From Newton’s third law of motion, a reaction force of $80N$  is acting on the box in the backward direction.

The backward motion of the box is opposed by the force of friction $f$, acting between the box and the floor of the truck. 

$f=\mu mg$ 

$\Rightarrow f=0.15\times 40\times 10=60N$ 

Net force acting on the block: ${{F}_{net}}=80-60=20N$ backward

The backward acceleration produced in the box is given by: 

${{a}_{back}}=\frac{{{F}_{net}}}{m}=\frac{20}{40}=0.5m{{s}^{-2}}$ 

From the second equation of motion, time $t$ can be calculated as:$s'=ut+\frac{1}{2}{{a}_{back}}{{t}^{2}}$ 

$\Rightarrow 5=0+\frac{1}{2}\times 0.5\times {{t}^{2}}$ 

$\Rightarrow t=\sqrt{20}s$ 

Therefore, the box will fall from the truck after $\sqrt{20}s$ from start.

The distance $s$, travelled by the truck in $\sqrt{20}s$ is: $s=ut+\frac{1}{2}a{{t}^{2}}$ 

$\Rightarrow s=0+\frac{1}{2}\times 2\times {{\left( \sqrt{2} \right)}^{2}}$ 

$\Rightarrow s=20m$ 

Therefore, at a distance of $20m$ from the starting point the box falls off the truck.

37. A disc revolves with a speed of $33\frac{1}{3}{rev}/{\min }\;$, and has a radius of $15cm$. Two coins are placed at $4cm$ and $14cm$ away from the centre of the record. If the co-efficient of friction between the coins and the record is $0.15$, which of the coins will revolve with the record?

Ans: It is given that,

Radius of the disc, \[r=15cm=0.15m\] 

Frequency of revolution, $\nu =33\frac{1}{3}rev/\min $

$\nu =\frac{100}{3\times 60}=\frac{5}{9}{rev}/{s}\;$ 

Coefficient of friction, $\mu =0.15$

Let, mass of each coin be $m$. 

In the given case, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc.

If this is not the case, then the coin will slip from the disc. 

When coin is placed at $4cm$: 

Radius of revolution, $r'=4cm=0.04m$ 

Angular frequency, $\omega =2\pi \nu =2\times \frac{22}{7}\times \frac{5}{9}=3.49{{s}^{-1}}$ 

Frictional force, $f=\mu mg=0.15\times m\times 10=1.5mN$

Centripetal force on the coin: ${{F}_{cent}}=mr'{{\omega }^{2}}$ 

$\Rightarrow {{F}_{cent}}=m\times 0.04\times {{(3.49)}^{2}}$

$\Rightarrow {{F}_{cent}}=0.49mN$

Therefore, as $f>{{F}_{cent}}$the coin will revolve along with the record.

When coin is placed at $14cm$: 

Radius of revolution, $r''=14cm=0.14m$ 

Angular frequency, $\omega =2\pi \nu =2\times \frac{22}{7}\times \frac{5}{9}=3.49{{s}^{-1}}$ 

Frictional force, $f'=\mu mg=0.15\times m\times 10=1.5mN$

Centripetal force on the coin: ${{F}_{cent}}=mr''{{\omega }^{2}}$ 

$\Rightarrow {{F}_{cent}}=m\times 0.14\times {{(3.49)}^{2}}$

$\Rightarrow {{F}_{cent}}=1.7mN$

Therefore, as $f<{{F}_{cent}}$ the coin will slip from the surface of the record.

Thus, the coin placed at $4cm$ from the centre will revolve with the record.

38. You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is $25m$? 

Ans: In a death-well, a motorcyclist does not fall at the top point of a vertical loop because both the force of normal reaction and the weight of the motorcyclist act downward and are balanced by the centripetal force. This situation is shown in the following figure. 

(Image will be uploaded soon)

The net force acting on the motorcyclist is the sum of the normal force and the force due to gravity $({{F}_{g}}=mg)$.

The equation of motion for the centripetal acceleration ${{a}_{c}}$, can be written as: 

$F=m{{a}_{c}}$

$\Rightarrow {{F}_{N}}+{{F}_{g}}=m{{a}_{c}}$ 

$\Rightarrow {{F}_{N}}+mg=\frac{m{{v}^{2}}}{r}$

For ${{v}_{\min }}$ , ${{F}_{N}}=0$ 

$\Rightarrow 0+mg=\frac{m{{v}^{2}}}{r}$

$\Rightarrow {{v}_{\min }}=\sqrt{gr}$ 

It is given that,$r=25m$ 

$\Rightarrow {{v}_{\min }}=\sqrt{25\times 10}=15.8m/s$

Therefore, the minimum speed required is $15.8m/s$.

39. A $70kg$  man stands in contact against the inner wall of a hollow cylindrical drum of radius $3m$ rotating about its vertical axis with $200rev/\min $. The coefficient of friction between the wall and his clothing is $0.15$. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

Ans: It is given that,

Mass of the man, $m=70kg$ 

Radius of the drum, $r=3m$ 

Coefficient of friction, $\mu =0.15$ 

Frequency of rotation, $\nu =200{rev}/{\min }\;=\frac{200}{60}=\frac{10}{3}{rev}/{s}\;$

The necessary centripetal force required for the rotation of the man is provided by the normal force $({{N}_{F}})$.

When the floor revolves, the man sticks to the wall of the drum.

Therefore, the weight of the man $(mg)$ acting downward is balanced by the frictional force $(f=\mu {{F}_{N}})$ acting upward.

Thus, the man will not fall until:$mg<f$ 

$\Rightarrow mg<\mu {{F}_{N}}=\mu mr{{\omega }^{2}}$ 

$\Rightarrow g<\mu r{{\omega }^{2}}$ 

$\Rightarrow \omega >\sqrt{\frac{g}{\mu r}}$ 

The minimum angular speed is given as: 

${{\omega }_{\min }}>\sqrt{\frac{g}{\mu r}}$

$\Rightarrow {{\omega }_{\min }}=\sqrt{\frac{10}{0.15\times 3}}=4.71rad{{s}^{-1}}$ 

Therefore, the minimum rotational speed of the cylinder is $4.71rad{{s}^{-1}}$.

40. A thin circular loop of radius $R$ rotates about its vertical diameter with an angular frequency $\omega $. Show that a small bead on the wire loop remains at its lowermost point for $\omega \le \sqrt{\frac{g}{R}}$. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for $\omega =\sqrt{\frac{2g}{R}}$ ? Neglect friction. 

Ans:

(Image will be uploaded soon)

Let the radius vector joining the bead with the centre make an angle $\theta $, with the vertical downward direction. 

$OP=R=$ Radius of the circle 

$N=$ Normal reaction 

Respective vertical and horizontal equations of forces can be written as:

$mg=N\cos \theta $    .........$(1)$ 

$ml{{\omega }^{2}}=N\sin \theta $    .........$(2)$

From $\Delta OPQ$:$\sin \theta =\frac{l}{R}$

$\Rightarrow l=R\sin \theta $      .........$(3)$

Substituting equation $(3)$ in equation $(2)$:

$\Rightarrow m(R\sin \theta ){{\omega }^{2}}=N\sin \theta $ 

$\Rightarrow mR{{\omega }^{2}}=N$    .........$(4)$

Substituting equation $(4)$ in equation $(1)$:

$\Rightarrow mg=mR{{\omega }^{2}}\cos \theta $ 

$\Rightarrow \cos \theta =\frac{g}{R{{\omega }^{2}}}$       .........$(5)$

It is known that, $\cos \theta \le 1$,

The bead will remain at its lowermost point for $\frac{g}{R{{\omega }^{2}}}\le 1$ 

$\Rightarrow \omega \le \sqrt{\frac{g}{R}}$ 

For $\omega =\sqrt{\frac{2g}{R}}$ or ${{\omega }^{2}}=\frac{2g}{R}$     .........$(6)$

Equating equation $(5)$ and equation $(6)$ 

$\Rightarrow \frac{2g}{R}=\frac{g}{R\cos \theta }$ 

$\Rightarrow \cos \theta =\frac{1}{2}$ 

$\Rightarrow \theta ={{\cos }^{-1}}(0.5)={{60}^{\circ }}$ 

Therefore, the angle made by the radius vector joining the centre to the bead with the vertical is ${{60}^{\circ }}$.

NCERT Exercise Class 11 Chapter 5 Laws of Motion – All Questions

Significance of Laws of Motion

Newton's Law of Motion Class 11 is fundamental because they tie into nearly everything we see in daily life. These laws show us exactly how bodies move or sit still, like why you don't glide out of bed or fall through the floor of your home. Newton's laws guide how cars operate, how water runs, how buildings don't collapse and primarily how everything around us works. It isn't always apparent how vital these laws are, because to utilise them in complicated circumstances, you require to know a lot of things. Laws of Motion Class 11 NCERT Solutions aims at explaining these concepts to you. Newton's laws address very generally all forces, but to employ them for any specific problem, you have to understand all the forces associated, like gravity, tension and friction.

The Three Laws of Motion Proposed By Newton

Newton's first law asserts that if a body is at rest or moving at a uniform speed in a straight line, it will remain at rest or keep moving in a straight line at uniform speed unless it is acted upon by an external force. Newton's second law asserts that the time rate of change of the momentum of a body is equal in both direction and magnitude to the force inflicted on it. The third law of Newton asserts that when two bodies associate, they apply forces to one another that are opposite in direction and equal in magnitude. Newton's third law is additionally known as the law of action and reaction. Laws of Motion Class 11 NCERT Solutions makes it easier to understand the laws in detail. 

Basic Ideas That NCERT Solutions of Class 11 Physics Chapter 5 Explain

Newton’s Laws of Motion indicate the associating/linking an object’s motion and the forces working on it. Laws of motion Class 11 NCERT Solutions talks about how reactions are a two-way mechanism. Class 11 Physics Chapter 5 NCERT Solutions also explores the three basic Laws of Motion that were proposed by Newton. It also tries to relate these laws to everyday situations. Since it might get complicated to understand the theories working behind the laws, therefore it is always advisable to have NCERT Solutions that CoolGyan provides.

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We, at CoolGyan, offer Class 11 Physics Chapter 5 NCERT Solutions in simplistic language to help students build a grip on the chapter. The study material, along with several shortcut methods and step-by-step solution helps students crack their assessment with ease. CoolGyan provides NCERT Solution for all subjects with free PDF download option without any charges. We are also one of the most suitable platforms for online studies and provide solutions for multiple subjects, penned by our expert teachers. We have also introduced one-on-one interaction with subject teachers which can aid students further in their academic endeavours. So to approach your exams more confidently, check out our NCERT Solutions of Class 11 Physics Chapter 5 without delay!

Solved Examples

1. Conservation of Momentum in a Clash Between Particles Can Be Explained With

(a) Conservation of energy

(b) Newton's second law only

(c) Newton's first law only

(d) Both Newton's second and third law

Answer: (d) Both Newton's second and third law

2. An Object of Mass 10 Kg Is Acted Upon by Two Vertical Forces, 6n and 8n. the Resultant Acceleration of the Object Is

(a) 1 m s-2 at the angle of tan-1 (4 / 3) w.r.t. 6N force

(b) 0.2 m s-2 at the angle of tan-1 (3 / 4) w.r.t. 8N force

(c) 0.2 m s-2 at the angle of tan-1 (4 / 3) w.r.t. 6N force

(d) 1 m s-2 at the angle of tan-1 (3 / 4) w.r.t. 8N force.

Answer: (a) and (d)