Additional Questions - Chapter 4 Work Energy and Power 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions Solved
Multiple Choice Questions
Question 1.
Thrust and linear momentum
(a) Thrust and linear momentum
(b) Work and energy
(c) Work and power
(d) Power and energy
Answer:
(b) Work and energy
Question 2.
The rate of work done is called as
(a) energy
(b) power
(c) force
(d) mechanical energy
Answer:
(b) power
Question 3.
Unit of work done
(a) $\mathrm{Nm}$
(b) joule
(c) either $a$ or $b$
(d) none
Answer:
(c) either a or b

Question 4.
Dimensional formula for work done is
(a) $\mathrm{MLT}^{-1}$
(b) $\mathrm{ML}^2 \mathrm{~T}^2$
(c) $\mathrm{M}^{-1} \mathrm{~L}^{-1} \mathrm{~T}^2$
(d) $\mathrm{ML}^2 \mathrm{~T}^{-2}$
Answer:
(d) $\mathrm{ML}^2 \mathrm{~T}^{-2}$
Question 5 .
When a body moves on a horizontal direction, the amount of work done by the gravitational force is
(a) positive
(b) negative
(c) zero
(d) infinity
Answer:
(c) zero
Question 6.
The amount of work done by centripetal force on the object moving in a circular path is
(a) zero
(b) infinity
(c) positive
(d) negative
Answer:
(a) zero
Question 7.
The work done by the goal keeper catches the ball coming towards him by applying a force is
(a) positive
(b) negative
(c) zero
(d) infinity
Answer:
(b) negative
Question 8 .
If the angle between force and displacement is acute then the work done is
(a) positive
(b) negative
(c) zero
(d) maximum
Answer:
(a) positive

Question 9.
If the force and displacement are perpendicular to each other, then the work done is
(a) positive
(b) negative
(c) zero
(d) maximum
Answer:
(c) zero
Question 10.
If the angle between force and displacement is obtuse, then the work done is
(a) positive
(b) negative
(c) zero
(d) minimum
Answer:
(b) negative
Question 11.
The area covered under force and displacement graph is
(a) work done
(b) acceleration
(c) power
(d) kinetic energy
Answer:
(a) work done
Question 12.
The capacity to do work is
(a) force
(b) energy
(c) work done
(d) power
Answer:
(b) energy

Question 13.
The energy possessed by a body due to its motion is called as
(a) potential energy
(b) kinetic energy
(c) mechanical energy
(d) none
Answer:
(b) kinetic energy
Question 14 .
The energy possessed by the body by virtue of its position is called as
(a) potential energy
(b) kinetic energy
(c) mechanical energy
(d) none
Answer:
(a) potential energy
Question 15 .
1 erg is equivalent to
(a) $10^{-7} \mathrm{~J}$
(b) $1.6 \times 10^{-19} \mathrm{~J}$
(c) $4.186 \mathrm{~J}$
(d) $3.6 \times 10^{-6} \mathrm{~J}$
Answer:
(a) $10^{-7} \mathrm{~J}$
Question 16.
1 electron volt is equivalent to
(a) $10^{-7} \mathrm{~J}$
(b) $1.6 \times 10^{-19} \mathrm{~J}$
(c) $4.186 \mathrm{~J}$
(d) $3.6 \times 10^{-6} \mathrm{~J}$
Answer:

(b) $1.6 \times 10^{-19} \mathrm{~J}$
Question 17.
1 kilowatt hour is equivalent to
(a) $10^{-7} \mathrm{~J}$
(b) $1.6 \times 10^{-19} \mathrm{~J}$
(c) $4.186 \mathrm{~J}$
(d) $3.6 \times 10^{-6} \mathrm{~J}$
Answer:
(d) $3.6 \times 10^{-6} \mathrm{~J}$
Question 18.
1 calorie is equivalent to
(a) $10^{-7} \mathrm{~J}$
(b) $1.6 \times 10^{-19} \mathrm{~J}$
(c) $4.186 \mathrm{~J}$
(d) $3.6 \times 10^6 \mathrm{~J}$
Answer:
(c) $4.186 \mathrm{~J}$
Question 19.
The amount of work done by a moving body depends on the
(a) mass of the body
(b) velocity
(c) both (a) and (b)
(d) time
Answer:
(c) both (a) and (b)
Question 20.
The kinetic energy of a body is given by
(a) $\frac{1}{2} m v^2$
(b) $m a$
(c) Fs
(d) $\left(v^2-u^2\right) m$

Answer:
(a) $\frac{1}{2} m v^2$
Question 21.
Kinetic energy of the body is always
(a) zero
(b) infinity
(c) negative
(d) positive
Answer:
(d) positive
Question 22.
If the work done by the force on the body is positive then its kinetic energy
(a) increases
(b) decreases
(c) zero
(d) either increases or decreases
Answer:
(a) increases
Question 23.
If $p$ is the momentum of the particle then its kinetic energy is
(a) $\sqrt{2 \mathrm{Mp}}$
(b) $\frac{p}{2 \mathrm{M}}$
(c) $\frac{p^2}{2 m}$.
(d) $\frac{2 m}{p}$
Answer:
(c) $\frac{\mathrm{p}^2}{2 \mathrm{~m}}$
Question 24 .
If two objects of masses $\mathrm{m}_1$ and $\mathrm{m}_2\left(\mathrm{~m}_1>\mathrm{m}_2\right)$ are moving with the same momentum then the kinetic energy will be greater for
(a) $\mathrm{m}_1$
(b) $\mathrm{m}_2$
(c) $\mathrm{m}_1$ or $\mathrm{m}_2$
(d) both will have equal kinetic energy
Answer:
(b) $\mathrm{m}_2$
Question 25.
For a given momentum, the kinetic energy is proportional to
(a) $m$
(b) $\frac{1}{m}$
(c) $m^2$
(d) $\sqrt{m}$

Answer:
(b) $\frac{1}{\mathrm{~m}}$
Question 26.
Elastic potential energy possessed by a spring is
(a) $\frac{1}{2} m v^2$
(b) $m g h$
(c) $\frac{1}{2} k x^2$
(d) $k x^2$
Answer:
(c) $\frac{1}{2} \mathrm{kx}^2$
Question 27.
Potential energy stored in the spring depends on
(a) spring constant
(b) mass
(c) gravity
(d) length
Answer:
(b) mass
Question 28.
Two springs of spring constants $\mathrm{k}_1$ and $\mathrm{k}_2\left(\mathrm{k}_1>\mathrm{k}_2\right)$. If they are stretched by the same force then $\left(\mathrm{u}_1, \mathrm{u}_2\right.$ are potential energy of the springs) is
(a) $u_1>u_2$
(b) $\mathrm{u}_2>\mathrm{u}_1$
(c) $u_1=u_2$
(d) $u_1 \geq u_2$
Answer:
(b) $\mathrm{u}_2>\mathrm{u}_1$

Question 29.
Conservative force is
(a) electrostatic force
(b) magnetic force
(c) gravitational force
(d) all the above
Answer:
(d) all the above
Question 30 .
Non conservative force is
(a) frictional force
(b) viscous force
(c) air resistance
(d) all the above
Answer:
(d) all the above
Question 31.
If the work done is completely recoverable, then the force is
(a) conservative
(b) non-conservative
(c) both (a) and (b)
(d) frictional in nature
Answer:
(b) non-conservative
Question 32 .
The work done by the conservative forces in a cycle is
(a) zero
(b) one
(c) infinity
(d) having negative value
Answer:
(a) zero

Question 33.
Negative gradient of potential energy gives
(a) conservative force
(b) non conservative force
(c) kinetic energy
(d) frictional force
Answer:
(a) conservative force
Question 34.
When a particle moving in a vertical circle, the variable is/are
(a) velocity of the particle
(b) tension of the string
(c) both (a) and (b)
(d) mass of the particle
Answer:
(c) both (a) and (b)
Question 35 .
Which of the following is zero at the highest point in vertical circular motion?
(a) velocity of the particle
(b) tension of the spring
(c) potential energy
(d) none
Answer:
(a) velocity of the particle
Question 36.
The body must have a speed at highest point in vertical circular motion to stay in the circular path
(a) $\geq \sqrt{g r}$
(b) $\geq \sqrt{2 g r}$
(c) $\geq \sqrt{5 g r}$
(d) $\geq 5 \mathrm{gr}$
Answer:
(a) $\geq \sqrt{\mathrm{gr}}$
Question 37.
The body must have a minimum speed of lowermost point in vertical circular motion to complete the circle

(a) $\geq \sqrt{g r}$
(b) $\geq \sqrt{2 g r}$
(c) $\geq \sqrt{5 g r}$
(d) $\geq 5 g r$
Answer:
(c) $\geq \sqrt{5 \mathrm{gr}}$
Question 38 .
The rate of work done is
(a) energy
(b) force
(c) power
(d) energy flow
Answer:
(c) power
Question 39.
The unit of power is
(a) $\mathrm{J}$
(b) $\mathrm{W}$
(c) $\mathrm{J} \mathrm{s}^{-1}$
(d) both (b) and (c)
Answer:
(d) both (b) and (c)
Question 40 .
One horse power (1 $\mathrm{hp})$ is
(a) $476 \mathrm{~W}$
(b) $674 \mathrm{~W}$
(c) $746 \mathrm{~W}$
(d) $764 \mathrm{~W}$
Answer:
(c) $746 \mathrm{~W}$
Question 41.
The dimension of power is
(a) $\mathrm{ML}^2 \mathrm{~T}^{-2}$
(b) $\mathrm{ML}^2 \mathrm{~T}^{-3}$
(c) $\mathrm{ML}^{-2} \mathrm{~T}^2$
(d) $\mathrm{ML}^{-2} \mathrm{~T}^3$
Answer:
(b) $\mathrm{ML}^2 \mathrm{~T}^{-3}$
Question 42.
$\mathrm{kWh}$ is the practical unit of
(a) energy
(b) power
(c) electrical energy

(d) none
Answer:
(a) energy
Question 43.
If a force $F$ is applied on a body and the body moves with velocity $\mathrm{v}$, the power will be
(a) F.V
(b) $\mathrm{F} / \mathrm{V}$
(c) $\mathrm{FV}^2$
(d) $\mathrm{FW}^2$
Answer:
(a) F.V
Question 44 .
A body of mass $m$ is thrown vertically upward with a velocity $v$. The height at which the kinetic energy of the body is one third of its initial value is given by
(a) $\frac{v^2}{2}$
(b) $\frac{v^2}{4}$
(c) $\frac{v^2}{6 g}$
(d) $\frac{v^2}{3}$
Answer:
(c) $\frac{v^2}{6 g}$
Solution:
Initial
$
\mathrm{K} . \mathrm{E}=\frac{1}{2} m v^2
$
$
\begin{aligned}
\therefore m g h=\frac{1}{3}\left(\frac{1}{2} m v^2\right) \Rightarrow m g h & =\frac{1}{6} m v^2 \\
h & =\frac{v^2}{6 g}
\end{aligned}
$

Question 45 .
A body of mass $5 \mathrm{~kg}$ is initially at rest. By applying a force of $20 \mathrm{~N}$ at an angle of $60^{\circ}$ with horizontal the body is moved to a distance of $4 \mathrm{~m}$. The kinetic energy acquired by the body is
(a) $80 \mathrm{~J}$
(b) $60 \mathrm{~J}$
(c) $40 \mathrm{~J}$
(d) $17.2 \mathrm{~J}$
Answer:
(c) $40 \mathrm{~J}$
Solution:
The work done is equal to its kinetic energy
$\therefore \mathrm{K}$.E gained $=$ Fs $\cos \theta=20 \times 4 \cos 60^{\circ}=40 \mathrm{~J}$
Question 46.
A bullet is fired normally on an immovable wooden plank of thickness $2 \mathrm{~m}$. It loses $20 \%$ of its kinetic energy in penetrating a thickness $0.2 \mathrm{~m}$ of the plank. The distance penetrated by the bullet inside the wooden plank is
(a) $0.2 \mathrm{~m}$
(b) $0.8 \mathrm{~m}$
(c) $1 \mathrm{~m}$
(d) $1.5 \mathrm{~m}$
Answer:
(c) $1 \mathrm{~m}$
Solution:
The wood offers a constant retardation. If the bullet loses $20 \%$ of its kinetic energy by penetrating $0.2 \mathrm{~m}$. it can penetrate further into $4 \times 0.2=0.8 \mathrm{~m}$ with the remaining kinetic energy. So the total distance penetrated by the bullet is $0.2+0.8=1 \mathrm{~m}$.
Question 47.
Which of the following quantity is conserved in all collision process?
(a) kinetic energy
(b) linear momentum
(c) both (a) and (b)
(d) none.
Answer:
(b) linear momentum

Question 48.
The kinetic energy is conserved in
(a) elastic collision
(b) inelastic collision
(c) both (a) and (b)
(d) none
Answer:
(a) Elastic collision
Question 49.
The kinetic energy is not conserved in
(a) Elastic collision
(b) In elastic collision
(c) both (a) and (b)
(d) none
Answer:
(b) In elastic collision
Question 50 .
In inelastic collision, which is conserved
(a) linear momentum
(b) total energy
(c) both (a) and (b)
(d) none
Answer:
(c) both (a) and (b)
Question 51.
If the two colliding bodies stick together after collision such collisions are
(a) elastic collision
(b) inelastic collision
(c) perfectly inelastic collision
(d) head on collision
Answer:
(c) perfectly inelastic collision

Question 52.
When bubblegum is thrown on a moving vehicle, it sticks is an example for
(a) elastic collision
(b) inelastic collision
(c) perfectly inelastic collision
(d) none
Answer:
(c) perfectly inelastic collision
Question 53.
Elastic collision is due to
(a) conservative force
(b) non conservative force
(c) gravitational force
(d) electrostatic force
Answer:
(b) non conservative force
Question 54.
Inelastic collision is due to
(a) conservative force
(b) non conservative force
(c) gravitational force
(d) electrostatic force
Answer:
(b) non conservative force
Question 55.
If the velocity of separation is equal to the velocity of approach, then the collision is
(a) conservative force
(b) non conservative force
(c) gravitational force
(d) electrostatic force
Answer:
(a) conservative force
Question 56.
For elastic collision, coefficient of restitution is
(a) 0
(b) 1
(c) $0<\mathrm{e}<1$
(d) $\infty$
Answer:
(b) 1
Question 57.
For inelastic collision co-efficient of restitution is
(a) 0

(b) 1
(c) $0<$ e $<1$
(d) $\infty$
Answer:
(c) $0<\mathrm{e}<1$
Question 58 .
For perfectly inelastic collision, coefficient of restitution is
(a) 0
(b) 1
(c) $0<\mathrm{e}<1$
(d) $\infty$
Answer:
(a) 0
Question 59.
The ratio of velocities of equal masses in an inelastic collision with one of the masses is stationary is 60. A box is dragged across a surface by a rope which makes an angle $45^{\circ}$ with the horizontal. The tension in the rope is $100 \mathrm{~N}$ when the box is dragged $10 \mathrm{~m}$. The work done is
(a) $\frac{1-e}{1+e}$
(b) $\frac{1+e}{e-1}$
$(c)(1+e)(1-e)$
(d) $\frac{e-1}{e+1}$
Answer:
(a) $\frac{1-e}{1+e}$
Question 60 .
A box is dragged across a surface by a rope which makes an angle $45^{\circ}$ with the horizontal. The tension in the rope is $100 \mathrm{~N}$ when the box is dragged $10 \mathrm{~m}$. The work done is
(a) $707.1 \mathrm{~J}$
(b) $607.1 \mathrm{~J}$
(c) $1414.2 \mathrm{~J}$
(d) $900 \mathrm{~J}$
Answer:
(a) $707.1 \mathrm{~J}$

The component of force acting along the surface is $T \cos \theta$

$\begin{aligned}
& \therefore \text { Work done }=\mathrm{T} \cos \theta \times \mathrm{x} \\
& =10 \circ \cos 45^{\circ} \times 10 \\
& =707.1 \mathrm{~J}
\end{aligned}$
Question 61.
A position dependent force $\mathrm{F}=\left(7-2 \mathrm{x}+3 \mathrm{x}^2\right) \mathrm{N}$ acts on a small body of mass $2 \mathrm{~kg}$ and displaces it from $\mathrm{x}=0$ to $\mathrm{x}=5 \mathrm{~m}$. Work done is
(a) $35 \mathrm{~J}$
(b) $70 \mathrm{~J}$
(c) $135 \mathrm{~J}$
(d) $270 \mathrm{~J}$
Answer:
(c) $135 \mathrm{~J}$
Solution:
$
\mathrm{W}=\int_{x_1}^{x_2} \mathrm{~F} \cdot d x=\int_0^5\left(7-2 x 3 x^2\right) d x=\left[7 x-x^2+x^3\right]_0^5=135 \mathrm{~J}
$
Question 62 .
In gravitational field, the work done in moving a body from one point into another depends on
(a) initial and final positions
(b) distance between them
(c) actual distance covered
(d) velocity of motion
Answer:
(c) initial and final positions

Question 63.
A particle of mass " $m$ " moving with velocity $\mathrm{v}$ strikes a particle of mass " $2 \mathrm{~m}$ " at rest and sticks to it. The speed of the combined mass is
(a) $\frac{v}{2}$
(b) $2 v$
(c) $\frac{v}{3}$
(d) $\frac{v}{4}$
Answer:
(c) $\frac{v}{3}$
Solution:
According to conservation of linear momentum
$
\begin{aligned}
& m v=(m+2 m) v_1 \\
& v_1=\frac{m v}{3 m}=\frac{v}{3}
\end{aligned}
$
Question 64.
A force of $(10 \hat{i}-3 \hat{j}+6 \hat{k}) \mathrm{N}$ acts on a body of $5 \mathrm{~kg}$ and displaces it from $(6 \hat{i}+5 \hat{j}-3 \hat{k})$ to ( $10 \hat{i}-2 \hat{j}+7 k) \mathrm{m}$. The work done is
(a) $100 \mathrm{~J}$
(b) 0
(c) $121 \mathrm{~J}$
(d) none of these
Answer:
(c) $121 \mathrm{~J}$

Solution:
Here
$
\mathrm{F}=(10 \hat{i}-3 \hat{j}+6 \hat{k}) \mathrm{N}
$
$
\begin{aligned}
\text { Displacement vector } & =\bar{x}=\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k} \\
\bar{x} & =(10-6) \hat{i}+(-2-5) \hat{j}+(7+3) \hat{k}=4 \hat{i}-7 \hat{j}+10 \hat{k}
\end{aligned}
$
$
\begin{aligned}
\text { We know that work done } & =w=\overline{\mathrm{F}} \cdot \bar{s}=(10 \hat{i}-3 \hat{j}+6 \hat{k}) \cdot(4 \hat{i}-7 \hat{j}+10 k) \\
& =40+21+60=121 \mathrm{~J} .
\end{aligned}
$
Question 65.
A $9 \mathrm{~kg}$ mass and $4 \mathrm{~kg}$ mass are moving with equal kinetic energies. The ratio of their momentum is
(a) $1: 1$
(b) $3: 2$
(c) $2: 3$
(d) $9: 4$
Answer:
(b) $3: 2$
Solution:
Given that K.E are equal
$
\therefore \quad \frac{\mathrm{P}_1^2}{2 m_1}=\frac{\mathrm{P}_2^2}{2 m_2} \quad \therefore \quad \frac{\mathrm{P}_1}{\mathrm{P}_2}=\sqrt{\frac{m_1}{m_2}}=\sqrt{\frac{9}{4}}=\frac{3}{2}
$
Question 66.
If momentum of a body increases by $25 \%$ its kinetic energy will increase by
(a) $25 \%$
(b) $50 \%$
(c) $125 \%$
(d) $56.25 \%$
Answer:
(d) $56.25 \%$
Solution:

Let momentum of $\mathrm{p}_1=100 \%$ momentum of $\mathrm{p}_2=125 \%$.
$
\begin{aligned}
& \therefore \\
& \frac{p_2}{p_1}=\frac{5}{4} \quad \Rightarrow \quad p_2=\frac{5}{4} p_1 \quad \therefore \quad v_2=\frac{5}{4} v_1 \\
& \text { Kinetic energy }=\frac{1}{2} m v^2 \\
& \mathrm{E} \propto v^2 \quad \mathrm{E}_2=\frac{25}{16} \mathrm{E}_1 \\
& \text { Increase in } \mathrm{K} . \mathrm{E}=\frac{E_2-E_1}{E_1} \times 100=\frac{\frac{25}{16} E_1-E_1}{E_1} \times 100 \\
& =\frac{\frac{25-16}{16} \mathrm{E}_1}{\mathrm{E}_1} \times 100=\frac{9}{16} \times 100=56.25 \% \\
&
\end{aligned}
$
Question 67.
A missile fired from a launcher explodes in mid air, its total
(a) kinetic energy increases
(b) momentum increases
(c) kinetic energy decreases
(d) momentum decreases
Answer:
(a) kinetic energy increases
Question 68 .
A bullet hits and gets embedded in a wooden block resting on a horizontal friction less surface. Which of the following is conserved?
(a) momentum alone
(b) kinetic energy alone
(c) both momentum and kinetic energy
(d) no quantity is conserved
Answer:
(a) momentum alone
Question 69.
Two balls of equal masses moving with velocities $10 \mathrm{~m} / \mathrm{s}$ and $-7 \mathrm{~m} / \mathrm{s}$ respectively collide elastically.
Their velocities after collision will be
(a) $3 \mathrm{~ms}^{-1}$ and $17 \mathrm{~ms}^{-1}$
(b) $-7 \mathrm{~ms}^{-1}$ and $10 \mathrm{~ms}^{-1}$
(c) $10 \mathrm{~ms}^{-1}$ and $-7 \mathrm{~ms}^{-1}$

(d) $3 \mathrm{~ms}^{-1}$ and $-70 \mathrm{~ms}^{-1}$
Answer:
(b) $-7 \mathrm{~ms}^{-1}$ and $10 \mathrm{~ms}^{-1}$
Question 70 .
A spring of negligible mass having a force constant of $10 \mathrm{Nm}^{-1}$ is compressed by a force to a distance of $4 \mathrm{~cm}$. A block of mass $900 \mathrm{~g}$ is free to leave the top of the spring. If the spring is released, the speed of the block is
(a) $11.3 \mathrm{~ms}^{-1}$
(b) $13.3 \times 101 \mathrm{~ms}^{-1}$
(c) $13.3 \times 10^{-2} \mathrm{~ms}^{-1}$
(d) $13.3 \times 10^{-3} \mathrm{~ms}^{-1}$
Answer:
(c) $13.3 \times 10^{-2} \mathrm{~ms}^{-1}$
Solution:
We know that, the potential energy of the spring $=\frac{1}{2} \mathrm{kx}^2$. Here the potential energy of the spring is converted into kinetic energy of the block.
$
\begin{aligned}
& \therefore \quad \frac{1}{2} k x^2=\frac{1}{2} m v^2 \\
& v^2=\frac{k}{m} x^2 \Rightarrow v=\sqrt{\frac{k}{m}} x \\
& v=\sqrt{\frac{20}{90}} \times 4 \times 10^{-2}=\sqrt{\frac{100}{9}} \times 4 \times 10^{-2}=13.3 \times 10^{-2} \mathrm{~ms}^{-1} \text {. } \\
&
\end{aligned}
$
Question 71.
A particle falls from a height ftona fixed horizontal plate and rebounds. If e is the coefficient " of restitution, the total distance travelled by the particle on rebounding when it stops is
(a) $\frac{h(1+e)^2}{(1-e)^2}$
(b) $\frac{h(1+e)}{(1-e)}$
(c) $\frac{h\left(1+e^2\right)}{1-e^2}$
(d) $\frac{h(1-e)^2}{(1+e)^2}$

Answer:
(c) $\frac{h\left(1+e^2\right)}{1-e^2}$
$
\begin{aligned}
& S=h+2 e^2 h+2 e^4 h+2 e^6 h+\ldots \ldots \\
& S=h+2 h\left(e^2+e^4+e^6+\ldots\right)
\end{aligned}
$
By using binomixal expansion we can write it as
$
\begin{aligned}
& \mathrm{S}=h+2 h\left(\frac{e^2}{1-e^2}\right)=h\left(1+\frac{2 e^2}{1-e^2}\right)=h\left(\frac{1-e^2+2 e^2}{1-e^2}\right) \\
& \mathrm{S}=h \frac{1+e^2}{1-e^2}
\end{aligned}
$

Question 72.
If the force $F$ acting on a body as a function of $x$ then the work done in moving a body from $x=1 \mathrm{~m}$ to $\mathrm{x}$ $=3 \mathrm{~m}$ is

(a) $6 \mathrm{~J}$
(b) $4 \mathrm{~J}$
(c) $2.5 \mathrm{~J}$
(d) $1 \mathrm{~J}$
Answer:
(b) $4 \mathrm{~J}$
Question 73 .
A boy "A" of mass $50 \mathrm{~kg}$ climbs up a staircase in $10 \mathrm{~s}$. Another boy "B" of mass $60 \mathrm{~kg}$ climbs up a Same staircase in $15 \mathrm{~s}$. The ratio of the power developed by the boys " $\mathrm{A}$ " and " $\mathrm{B}$ " is
(a) $\frac{5}{4}$
(b) $\frac{3}{2}$
(c) $\frac{4}{5}$
(d) $\frac{2}{3}$
Answer:
(a) $\frac{5}{4}$
Solution:
Work done by the boy 'A' $\left(\mathrm{W}_1\right)=m_1 g h=50 \mathrm{gh}$
Work done by the boy ' $\mathrm{B}$ ' $\left(\mathrm{W}_2\right)=m_2 g h=60 \mathrm{gh}$
Power developed by $\quad \mathrm{A}=\mathrm{P}_1=\frac{w_1}{t_1}=\frac{50 \mathrm{gh}}{10}$
Power developed by $\mathrm{B}=$ is $\quad \mathrm{P}_2=60 \mathrm{gh} / 15$. then, $p_1 / p_2=\frac{50 \mathrm{gh} / 10}{60 \mathrm{gh} / 15}=\frac{5}{4}$
Short Answer Questions
Question 1.
Define work, energy, power.
Answer:
Work: Work is said to be done by the force when the force applied on a body displaces it.
Energy: Energy is defined as the ability to do work.
Power: The rate of work done is called power.
Question 2.
Discuss the possibilities of work done to be zero.

Answer:
Work done is zero in the following cases.
(i) When the force is zero $(\mathrm{F}=0)$. For example, , a body moving on a horizontal smooth frictionless surface will continue to do so as no force (not even friction) is acting along the plane. (This is an ideal situation.)
(ii) When the displacement is zero $(\mathrm{dr}=0$ ). For example, when force is applied on a rigid wall it does not produce any displacement. Hence, the work done is zero as shown in figure.
(iii) When the force and displacement are perpendicular $\left(0=90^{\circ}\right)$ to each other, when a body moves on a horizontal direction, the gravitational force $(\mathrm{mg}$ ) does not work on the body, since it acts at right angles to the displacement as shown in Figure (b). In circular motion the centripetal force does not do work on the object moving on a circle as it is always perpendicular to the displacement as shown in Figure (c).

Question 3.
Derive the relation between momentum and kinetic energy.
Answer:
Consider an object of mass $m$ moving with a velocity $\mathrm{v}$. Then its linear momentum is
$\vec{p}=m \vec{v}$ and its kinetic energy, $\mathrm{KE}=\frac{1}{2} m v^2$.
$
\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2} m(\vec{v} \cdot \vec{v})
$
Multiplying both the numerator and denominator of equation (i) by mass $\mathrm{m}$
$
\begin{aligned}
& \mathrm{KE}=\frac{1}{2} \frac{m^2(\vec{v} \cdot \vec{v})}{m}=\frac{1}{2} \frac{(m \vec{v}) \cdot(m \vec{v})}{m}=\frac{1}{2} \frac{\vec{p} \cdot \vec{p}}{m}=\frac{p^2}{2 m} \quad[\because \vec{p}=m \vec{v}] \\
& \mathrm{KE}=\frac{p^2}{2 m}
\end{aligned}
$
where $|\vec{p}|$ is the magnitude of the momentum. The magnitude of the linear momentum can be obtained by
$
|\vec{p}|=p=\sqrt{2 m(\mathrm{KE})}
$
Note that if kinetic energy and mass are given, only the magnitude of the momentum can be calculated but not the direction of momentum. It is because the kinetic energy and mass are scalars.
Question 4.
How can an object move with zero acceleration (constant velocity) when the external force is acting on the object?
Answer:
It is possible when there is another force which acts exactly opposite to the external applied force. They both cancel each other and the resulting net force becomes zero, hence the object moves with zero acceleration.
Question 5.
Why should the object be moved at constant velocity when we define potential energy?
Answer:

If the object does not move at constant velocity, then it will have different velocities at the initial and final locations. According to work-kinetic energy theorem, the external force will impart some extra kinetic energy. But we associate potential energy to the forces like gravitational force, spring force and coulomb force. So the external agency should not impart any kinetic energy when the object is taken from initial to final location.
Question 6.
Derive an expression for potential energy near the surface of the earth.
Answer:
The gravitational potential energy $(\mathrm{U})$ at some height $\mathrm{h}$ is equal to the amount of work required to take the object from ground to that height $\mathrm{h}$ with constant velocity. Let us consider a body of mass $m$ being moved from ground to the height $\mathrm{h}$ against the gravitational force as shown.

The gravitational force $\overrightarrow{\mathrm{F}}_g$ acting on the body is, $\overrightarrow{\mathrm{F}}_g=-m g \hat{j}$ (as Gravitational potential energy the force is in y direction, unit vector $\hat{j}$ is used). Here, negative sign implies that the force is acting vertically downwards. In order to move the body without acceleration (or with constant velocity), an external applied force $\overrightarrow{\mathrm{F}}_a$, equal in magnitude but opposite to that of gravitational force $\overrightarrow{\mathrm{F}}_g$ has to be applied on the body i.e., $\overrightarrow{\mathrm{F}}_a=-\overrightarrow{\mathrm{F}}_g$.
This implies that $\overrightarrow{\mathrm{F}}_a=+m g \hat{j}$. The positive sign implies that the applied force is in vertically upward direction. Hence, when the body is lifted up its velocity remains unchanged and thus its kinetic energy also remains constant.
The gravitational potential energy (U) at some height $\mathrm{h}$ is equal to the amount of work required to take the object from the ground to that height $h$.
$
\mathrm{U}=\int \overrightarrow{\mathrm{F}}_a \cdot d \vec{r}=\int_0^h\left|\overrightarrow{\mathrm{F}}_a\right||d \vec{r}| \cos \theta
$
Since the displacement and the applied force are in the same upward direction, the angle between them, $\theta$ $=0^{\circ}$. Hence, $\cos 0^{\circ}=1$ and $\left|\overrightarrow{\mathrm{F}}_a\right|=\mathrm{mg}$ and $|d \vec{r}|=\mathrm{dr}$
$
\begin{aligned}
& \mathrm{U}=m g \int_0^h d r \\
& \mathrm{U}=m g[r]_0^h=m g h
\end{aligned}
$
Question 7.
Explain force displacement graph for a spring.
Answer:

Since the restoring spring force and displacement are linearly related as $\mathrm{F}=-\mathrm{kx}$, and are opposite in direction, the graph between $F$ and $x$ is a straight line with dwelling only in the second and fourth quadrant as shown in Figure. The elastic potential energy can be easily calculated by drawing a $F-x$ graph. The shaded area (triangle) is the work done by the spring force.
$
\text { Area } \left.=\frac{1}{2} \text { (base) (height }\right)=\frac{1}{2} \times(x) \times(k x)=\frac{1}{2} k x^2
$
Question 8.
Explain the potential energy - displacement graph for a spring.
Answer:
A compressed or extended spring will transfer its stored potential energy into kinetic energy of the mass attached to the spring. The potential energy-displacement graph is shown in Figure.

In a frictionless environment, the energy gets transferred from kinetic to potential and potential to kinetic repeatedly such that the total energy of the system remains constant. At the mean position, $\Delta \mathrm{KE}=\Delta \mathrm{U}$
Question 9.
Define unit of power.
Answer:
The unit of power is watt. One watt is defined as the power when one joule of work is done in one second.
Question 10.
Define average power and instantaneous power.
Answer:
The average power is defined as the ratio of the total work done to the total time taken. $\mathrm{P}_{\mathrm{av}}=$ total work done/total time taken The instantaneous power is defined as the power delivered at an instant $\mathrm{p}_{\text {inst }}=\mathrm{dw} / \mathrm{dt}$
Question 11.
Define elastic and inelastic collision.
Answer:
In any collision, if the total kinetic energy of the bodies before collision is equal to the total final kinetic energy of the bodies after collision then it is called as elastic collision.
In a collision the total initial kinetic energy of the bodies before collision is not equal to the .total final kinetic energy of the bodies after collision. Then it is called as inelastic collision.
Question 12.
What will happen to the potential energy of the system.
If (i) Two same charged particles are brought towards each other
(ii) Two oppositely charged particles are brought towards each other.
Answer:
(i) When the same charged particles are brought towards each other, the potential energy of the system will increase. Because work has to be done against the force of repulsion. This work done only stored as potential energy.
(ii) When two oppositely charged particles are brought towards each other, the potential energy of the system will decrease. Because work is done by the force of attraction between the charged particles.

Question 13.
Define the conservative and non-conservative forces. Give examples of each.
Answer:
Conservative force : e.g., Gravitational force, electrostatic force.
Non-Conservative force : e.g., forces of friction, viscosity.
Question 14.
A light body and a heavy body have same linear momentum. Which one has greater K.E ?
Answer:
Lighter body has more K.E. as K.E. $=\frac{p^2}{2 m}$ and for constant $p, K . E . \propto \frac{1}{m}$
Question 15.
The momentum of the body is doubled, what $\%$ does its $\mathrm{K} . \mathrm{E}$ change?
Answer:
K.E. $=\frac{p^2}{2 m}$ when $p$ is doubled K.E. becomes 4 times.
$
\therefore \% \text { Increase in K.E. }=\frac{\Delta \text { K.E. }}{\text { K.E. }} \times 100=\frac{4 \text { K.E. }- \text { K.E. }}{\text { K.E. }} \times 100=3 \times 100=300 \%
$
Question 16.
A body is moving along a circular path. How much work is done by the centripetal force?
Answer:
$
\mathrm{W}=\mathrm{FS} \cos 90^{\circ}=0
$
Question 17.
Which spring has greater value of spring constant - a hard spring or a delicate spring?
Answer:
Hard spring.
Question 18.
Two bodies stick together after collision. What type of collision is in between these two bodies? .
Answer:
Inelastic collision.
Question 19.
State the two conditions under which a force does not work.
Answer:
1. Displacement is zero or it is perpendicular to force.
2. Conservative force moves a body over a closed path.
Question 20.
How will the momentum of a body changes if its $\mathrm{K} . \mathrm{E}$. is doubled?
Answer:
Momentum becomes $\sqrt{2}$ times.

Question 21 .
K.E. of a body is increased by $300 \%$. Find the $\%$ increase in its momentum.
Answer:
$
\begin{aligned}
& \text { K.E. }=\frac{p^2}{2 m} \text { so } p=\sqrt{2 m k} \\
& \text { Increase in K.E. }=300 \% \text { of } k=3 k \\
& \text { Final K.E., } k^{\prime}=k+3 k=4 k \\
& \text { Final momentum, } p^{\prime}=\sqrt{2 m k^{\prime}}=\sqrt{2 m \times 4 k}=2 \sqrt{2 m k}=2 p \\
& \% \text { Increase in momentum }=\frac{p^{\prime}-p}{p} \times 100=100 \%
\end{aligned}
$
Question 22 .
A light and a heavy body have same K.E., which of the two have more momentum and why?
Answer:
Heavier body.
Question 23.
Does the P.E. of a spring decreases or increases when it is compressed or stretched?
Answer:
Increases because W.D. on it when it increases is compressed or stretched.
Question 24 .
Name a process in which momentum changes but $\mathrm{K}$.E. does not.
Answer:
Uniform circular motion.
Question 25.
What happens to the P.E. of a bubble when it rises in water?
Answer:
Decreases.

Question 26.
A body is moving at constant speed over a frictionless surface. What is the work done by the weight of the body?
Answer:
$
\mathrm{W}=0 \text {. }
$
Question 27.
Define spring constant of a spring.
Answer:
It is the restoring force set up in a string per unit extension.
Short Answer Questions 2 Marks
Question 28.
How much work is done by a coolie walking on a horizontal platform with a load on his head? Explain.

Answer:
$\mathrm{W}=0$ as his displacement is along the horizontal direction and in order to balance the load on his head,
he applies a force on it in the upward direction equal to its weight. Thus angle between force and displacement is zero.
Question 29.
Mountain roads rarely go straight up the slope, but wind up gradually. Why?
Answer:
If roads go straight up then angle of slope 0 would be large so frictional force $f=\mu \mathrm{mg}$ cos $\theta$ would be less and the vehicles may slip. Also greater power would be required.
Question 30 .
A truck and a car moving with the same K.E. on a straight road. Their engines are simultaneously switched off which one will stop at a lesser distance?
Answer:
By Work - Energy Theorem,
Loss in K.E. = W.D. against the force $\times$ distance of friction
or
$
\mathrm{K} . \mathrm{E} .=\mu m g \mathrm{~S}
$

For constant K.E., $\mathrm{S} \propto \frac{1}{m}$
$\therefore$ Truck will stop in a lesser distance because of greater mass.
Question 31.
Is it necessary that work done in the motion of a body over a closed loop is zero for every force in nature? Why?
Answer:
No. W.D. is zero only in case of a conservative force.
Question 32 .
How high must a body be lifted to gain an amount of P.E. equal to the K.E. it has when moving at speed $20 \mathrm{~ms}^{-1}$. (The value of acceleration due to gravity at a place is $9.8 \mathrm{~ms}^{-2}$ ).
Answer:
$
\begin{aligned}
& m g h=\frac{1}{2} m v^2 \\
& \text { so } h=20.2 \mathrm{~m}
\end{aligned}
$
Question 33.
Give an example in which a force does work on a body but fails to change its K.E.
Answer:
When a body is pulled on a rough, horizontal surface with constant velocity. Work is done on the body but K.E. remains unchanged.
Question 34.
A bob is pulled sideway so that string becomes parallel to horizontal and released. Length of the pendulum is $2 \mathrm{~m}$. If due to air resistance loss of energy is $10 \%$, what is the speed with which the bob arrived at the lowest point.

Answer:
$
\begin{aligned}
& \frac{1}{2} m v^2=90 \% \text { of } \mathrm{mgh} \\
& \therefore v=6 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
Question 35 .
Two springs $A$ and $B$ are identical except that $A$ is harder than $B\left(K_A>K_B\right)$ if these are stretched by the equal force. In which spring will more work be done?
Answer:
$
\mathrm{F}=\mathrm{K} \cdot x \text { so } x=\frac{\mathrm{F}}{\mathrm{K}}
$
For same $\mathrm{F}, \quad \mathrm{W}_{\mathrm{A}}=\frac{1}{2} \mathrm{~K}_{\mathrm{A}} x^2=\frac{1}{2} \frac{\mathrm{F}^2}{\mathrm{~K}_{\mathrm{A}}}$ and
$
\mathrm{W}_{\mathrm{B}}=\frac{\mathrm{F}^2}{2 \mathrm{~K}_{\mathrm{B}}}
$
$
\therefore \quad \quad \frac{\mathrm{W}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{B}}}=\frac{\mathrm{K}_{\mathrm{B}}}{\mathrm{K}_{\mathrm{A}}}
$
As $\mathrm{K}_{\mathrm{A}}>\mathrm{K}_{\mathrm{B}}$ so $\mathrm{W}_{\mathrm{A}}<\mathrm{W}_{\mathrm{B}}$
Question 36.
Find the work done if a particle moves from position $\mathrm{r}_1=$ to a position $(3 \hat{i}+2 \hat{j}-6 \hat{k})$ to a position $\vec{r}_2=(14 \hat{i}+13 \hat{j}-9 \hat{k})$ under the effect of force $\overrightarrow{\mathrm{F}}=(4 \hat{i}+\hat{j}+3 \hat{k}) \mathrm{N}$
Answer:
$
\begin{aligned}
& \vec{r}=\vec{r}_2-\vec{r}_1=11 \hat{i}+11 \hat{j}-3 \hat{k} \\
& \overrightarrow{\mathrm{F}}=(4 \hat{i}+\hat{j}+3 \hat{k}) \mathrm{N} \\
\therefore \quad & \mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \vec{r}=46 \mathrm{~J} .
\end{aligned}
$
Question 37.
Spring $A$ and $B$ are identical except that $A$ is stiffer than $B$, i.e., force constant $k_A>k_B$. In which spring is more work expended if they are stretched by the same amount?

Answer:
$
\begin{aligned}
& \mathrm{W}=\frac{1}{2} \mathrm{Kx}^2 \\
& \therefore \quad \frac{\mathrm{W}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{B}}}=\frac{\mathrm{K}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{B}}}, \text { for same } x \\
& \text { As } \mathrm{K}_{\mathrm{A}}>\mathrm{K}_{\mathrm{B}} \text { so } \mathrm{W}_{\mathrm{A}}>\mathrm{W}_{\mathrm{B}}
\end{aligned}
$
Question 38 .
A ball at rest is dropped from a height of $12 \mathrm{~m}$. It loses $25 \%$ of its kinetic energy in striking the ground, find the height to which it bounces. How do you account for the loss in kinetic energy?
Answer:
If ball bounces to height $\mathrm{h}^{\text {', then }}$
$\mathrm{mgh}=75 \%$ of $\mathrm{mgh}$
$\therefore \mathrm{h}^{\prime}=0.75 \mathrm{~h}=9 \mathrm{~m}$
Question 39.
Which of the two kilowatt hour or electron volt is a bigger unit of energy and by what factor?
Answer:
kwh is a bigger unit of energy.
$
\frac{1 k w h}{1 \mathrm{eV}}=\frac{3.6 \times 10^6 \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J}}
$
Question 40 .
A spring of force constant $\mathrm{K}$ is cut into two equal pieces. Calculate force constant of each part.
Answer:
Force constant of each half becomes twice the force constant of the original spring.
Short Answer Questions 3 Marks
Question 41
A car of mass $2000 \mathrm{~kg}$ is lifted up a distance of $30 \mathrm{~m}$ by a crane in $1 \mathrm{~min}$. A second crane does the same job in $2 \mathrm{~min}$. Do the cranes consume the same or different amounts of fuel? What is the power supplied by each crane? Neglect Power dissipation against friction.
Answer:
$
\begin{aligned}
& \mathrm{t}_1=1 \min =60 \mathrm{~s}, \mathrm{t}_2=2 \min =120 \mathrm{~s} \\
& \mathrm{~W}=\mathrm{Fs}=\mathrm{mgs}=5.88 \times 105 \mathrm{~J}
\end{aligned}
$
As both cranes do same amount of work so both consume same amount of fuel.
$
\begin{aligned}
& \mathrm{P}_1=\frac{\mathrm{W}}{t_1} \text { and } \mathrm{P}_2=\frac{\mathrm{W}}{t_2} \\
& \therefore \quad \mathrm{P}_1=9800 \mathrm{~W} \text { and } \mathrm{P}_2=4900 \mathrm{~W} .
\end{aligned}
$

Question 42 .
$20 \mathrm{~J}$ work is required to stretch a spring through $0.1 \mathrm{~m}$. Find the force constant of the spring. If the spring is further stretched through $0.1 \mathrm{~m}$, calculate work done.
Answer:
P.E. of spring when stretched through a distance $01 \mathrm{~m}$,
$
\mathrm{U}=\mathrm{W} . \mathrm{D} .=\frac{1}{2} \mathrm{~K} \cdot x^2=20 \mathrm{~J}
$
or
$
\mathrm{K}=4000 \mathrm{~N} / \mathrm{m}
$
when spring is further stretched through $01 \mathrm{~m}$, then P.E. will be :
$
\begin{array}{ll} 
& U^{\prime}=\frac{1}{2} \mathrm{~K}(0.2)^2=80 \mathrm{~J} \\
\therefore \quad \text { W.D. }=\mathrm{U}^{\prime}-\mathrm{U}=80-20=60 \mathrm{~J} .
\end{array}
$
Question 43.
A pump on the ground floor of a building can pump up water to fill a tank of volume $30 \mathrm{~m}^3$ in $15 \mathrm{~min}$. If the tank is $40 \mathrm{~m}$ above the ground, how much electric power is consumed by the pump. The efficiency of the pump is $30 \%$.
Answer:
$
\begin{aligned}
30 \% \text { of Power } & =\frac{\mathrm{W}}{t}=\frac{m g h}{t}=\frac{V \rho g h}{t} \\
\frac{30}{100} \times \mathrm{P} & =\frac{V \rho g h}{t} \\
\therefore \quad \mathrm{P} & =43.6 \mathrm{KW} .
\end{aligned}
$
Question 44 .
A ball bounces to $80 \%$ of its original height. Calculate the mechanical energy lost in each bounce.
Answer:
Let Initial P.E. = $\mathrm{mgh}$
P.E. after first bounce $=\mathrm{mg} \times 80 \%$ of $\mathrm{h}=0.80 \mathrm{mgh}$
P.E. lost in each bounce $=0.20 \mathrm{mgh}$

$\therefore$ Fraction of P.E. lost in each bounce $=\frac{0.20 m g h}{m g h}=0 \cdot 20$
Long Answer Questions
Question 1.
Obtain an expression for the critical vertical of a body revolving in a vertical circle
Answer:
Imagine that a body of mass $(\mathrm{m})$ attached to one end of a massless and inextensible string executes circular motion in a vertical plane with the other end of the string fixed. The length of the string becomes the radius ( $\mathrm{r}$ ) of the circular path (See figure).

Let us discuss the motion of the body by taking the free body diagram (FBD) at a position where the position vector $(\vec{r})$ makes an angle $\theta$ with the vertically downward direction and the instantaneous velocity is as shown in Figure.
There are two forces acting on the mass.
1. Gravitational force which acts downward
2. Tension along the string.
Applying Newton's second law on the mass, in the tangential direction,
$m g \sin \theta=m a_t$.
$m g \sin \theta=-m\left(\frac{d v}{d t}\right)$
where, $a_t=-\frac{d v}{d t}$ is tangential retardation
In the radial direction,
$
\begin{aligned}
& \mathrm{T}-m g \cos \theta=m a_r \\
& \mathrm{~T}-m g \cos \theta=\frac{m v^2}{r}
\end{aligned}
$
where, $a_r=\frac{v^2}{r}$ is the centripetal acceleration.
The circle can be divided into four sections A, B, C, D for better understanding of the motion. The four important facts to be understood from the two equations are as follows:
(i) The mass is having tangential acceleration $(g \sin \theta)$ for all values of $\theta$ (except $\theta=0^{\circ}$ ), it is clear that this vertical circular motion is not a uniform circular motion.
(ii) From the equations (ii) and (i) it is understood that as the magnitude of velocity is not a constant in the course of motion, the tension in the string is also not constant.

(iii) The equation (ii), $\mathrm{T}=m g \cos \theta \frac{m v^2}{r}$ highlights that in sections $\mathrm{A}$ and $\mathrm{D}$ of the circlt $\left(\right.$ for $-\frac{\pi}{2}<\theta<\frac{\pi}{2} ; \cos \theta$ is positive $)$, the term $m g \cos \theta$ is always greater than zero. Henc the tension cannot vanish even when the velocity vanishes.
(iv) The equation (ii), $\frac{m v^2}{r}=\mathrm{T}-m g \cos \theta$; further highlights that in sections $\mathrm{B}$ and $\mathrm{C}$ of th circle, $\left(\right.$ for $\frac{\pi}{2}<\theta<\frac{3 \pi}{2} ; \cos \theta$ is negative $)$, the second term is always greater than zerc
Hence velocity cannot vanish, even when the tension vanishes.
These points are to be kept in mind while solving problems related to motion in vertical circle. To start with let us consider only two positions, say the lowest point 1 and the highest point 2 as shown in Figure for further analysis. Let the velocity of the body at the lowest point 1 be $\vec{v}_1$, at the highest point 2 be $\vec{v}_2$ and $\vec{v}$ at any other point. The direction of velocity is tangential to the circular path at all points.
Let $\vec{T}_1$ be the tension in the string at the lowest point and $\vec{T}_2$ be , the tension at the highest point and $\overrightarrow{\mathrm{T}}$ be the tension at any other point. Tension at each point acts towards the center. The tensions and velocities at these two points can be found by applying the law of conservation of energy.

For the lowest point (1)
When the body is at the lowest point 1 , the gravitational force $m \vec{g}$ which acts on the body (vertically downwards) and another one is the tension $\overrightarrow{\mathrm{T}}_1$, acting vertically upwards, i.e. towards the center. From the equation (ii), we get
$
\begin{aligned}
\mathrm{T}_1-m g & =\frac{m v_1^2}{r} \\
\mathrm{~T}_1 & =\frac{m v_1^2}{r}+m g
\end{aligned}
$
For the highest point (2)
At the highest point 2 , both the gravitational force $\mathrm{mg}$ on the body and the tension $\mathrm{T}_2$ act downwards, i.e. towards the center again.
$
\begin{aligned}
\mathrm{T}_2+m g & =\frac{m v_2^2}{r_2} \\
\mathrm{~T}_2 & =\frac{m v^2}{r_2}-m g
\end{aligned}
$
From equations (iv) and (ii), it is understood that $T_1>T_2$. The difference in tension $T_1-T_2$ is obtained by subtracting equation (iv) from equation (ii).
$
\begin{aligned}
& \mathrm{T}_1-\mathrm{T}_2=\frac{m v_1^2}{r}+m g-\left(\frac{m v_2^2}{r}-m g\right)=\frac{m v_1^2}{r}+m g-\frac{m v_2^2}{r}+m g \\
& \mathrm{~T}_1-\mathrm{T}_2=\frac{m}{r}\left[v_1^2-v_2^2\right]+2 m g
\end{aligned}
$

The term $\left[v_1^2-v_2^2\right]$ can be found easily by applying law of conservation of energy at point 1 and also at point 2 .
Note: The tension will not do any work on the mass as the tension and the direction of motion is always perpendicular.
The gravitational force is doing work on the mass, as it is a conservative force the total energy of the mass is conserved throughout the motion.
Total energy at point $1\left(E_1\right)$ is same as the total energy at a point $2\left(E_2\right)$ $\mathrm{E}_1=\mathrm{E}_2$
Potential energy at point $1, U_1=0$ (by taking reference as point 1 )
Kinetic energy at point $1, \mathrm{KE}_1=\frac{1}{2} m v_1^2$
Total energy at point $1, \mathrm{E}_1=\mathrm{U}_1+\mathrm{KE}_1=0+\frac{1}{2} m v_1^2=\frac{1}{2} m v_1^2$
Similarly, Potential energy at point $2, \mathrm{U}_2=\mathrm{mg}(2 \mathrm{r})$ ( $\mathrm{h}$ is $2 \mathrm{r}$ from point 1$)$

Kinetic Energy at point 2, $\mathrm{KE}_2=\frac{1}{2} m v_2^2$
Total energy at point $2, \mathrm{E}_2=\mathrm{U}_2+\mathrm{KE}_2=2 m g r+\frac{1}{2} m v_2^2$
From the law of conservation of energy given in equation (vi), we get
$
\frac{1}{2} m v_1^2=2 m g r+\frac{1}{2} m v_2^2
$
After rearranging, $\frac{1}{2} m\left(v_1^2-v_2^2\right)=2 m g r$
$
v_1^2-v_2^2=4 g r
$
Substituting equation (vii) in equation (iv) we get,
$
\mathrm{T}_1-\mathrm{T}_2=\frac{m}{r}[4 g r]+2 m g
$
Therefore, the difference in tension is
$
\mathrm{T}_1-\mathrm{T}_2=6 \mathrm{mg} \ldots \text { (viii) }
$
Minimum speed at the highest point (2)
The body must have a minimum speed at point 2 otherwise, the string will slack before reaching point 2 and the body will not loop the circle. To find this minimum speed let us take the tension $T_2=0$ in equation (iv).
$
\begin{aligned}
0 & =\frac{m v_2^2}{r}-m g \\
\frac{m v_2^2}{r} & =m g \\
v_2^2 & =r g \\
v_2 & =\sqrt{g r}
\end{aligned}
$
The body must have a speed at point $2, v_2 \geq \sqrt{g r}$ to stay in the circular path.
Maximum speed at the lowest point 1
To have this minimum speed $\left(v_2=\sqrt{g r}\right)$ at point 2 , the body must have minimum speed also at point 1 . By making use of equation (vii) we can find the minimum speed at point 1.
$
v_1^2-v_2^2=4 g r
$
Substituting equation (ix) in (vii),

$
\begin{aligned}
v_1^2-g r & =4 g r \\
v_1^2 & =5 g r \\
v_1 & =\sqrt{5 g r}
\end{aligned}
$
The body must have a speed at point $1, v_1 \geq \sqrt{5 g r}$ to stay in the circular path. From equations (ix) and (x), it is clear that the minimum speed at the lowest point 1 should be $\mathrm{v} 5$ times more than the minimum speed at the highest point 2 , so that the body loops without leaving the circle.
Question 2.
Obtain the expressions for the velocities of the two bodies after collision in the case of one dimensional elastic collision and discuss the special cases.
Answer:
Consider two elastic bodies of masses $\mathrm{m}_1$ and $\mathrm{m}_2$ moving in a straight line (along positive $\mathrm{x}$ direction) on a frictionless horizontal surface as shown in figure

In order to have collision, we assume that the mass $m_1$ moves faster than mass $m_2$ i.e., $u_1>u_2$. For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same

From the law of conservation of linear momentum,
Total momentum before collision $\left(\mathrm{p}_{\mathrm{i}}\right)=$ Total momentum after collision $\left(\mathrm{p}_f\right)$
$
m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_1
$
Or $\quad m_1\left(u_1-v_1\right)=m_2\left(v_2-u_2\right)$
Further,

For elastic collision,
Total kinetic energy before collision $\mathrm{KE}_{\mathrm{i}}$ = Total kinetic energy after collision $\mathrm{KF}_{\mathrm{f}}$
$
\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2
$
After simplifying and rearranging the terms,
$
m_1\left(u_1^2-v_1^2\right)=m_2\left(v_2^2-u_2^2\right)
$
Using the formula $a^2-b^2=(a+b)(a-b)$, we can rewrite the above equation as $\mathrm{m}_1\left(\mathrm{u}_1+\mathrm{v}_1\right)\left(\mathrm{u}_1-\mathrm{v}_1\right)=\mathrm{m}_2\left(\mathrm{v}_2+\mathrm{u}_2\right)\left(\mathrm{v}_2-\mathrm{u}_2\right) \ldots$ (iv)
Dividing equation (iv) by (ii) gives,
$
\begin{aligned}
\frac{m_1\left(u_1+v_1\right)\left(u_1-v_1\right)}{m_1\left(u_1-v_1\right)} & =\frac{m_2\left(v_2+u_2\right)\left(v_2-u_2\right)}{m_2\left(v_2-u_2\right)} \\
u_1+v_1 & =v_2+u_1 \\
u_1-u_2 & =v_2-v_1
\end{aligned}
$
Rearranging, (v)
Equation (v) can be rewritten as
$
u_1-u_2=-\left(v_1-v_2\right)
$
This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for $\mathrm{V}_1$ and $\mathrm{v}_2$,
$
\mathrm{v}_1=\mathrm{v}_2+\mathrm{u}_2-\mathrm{u}_1
$
Or $\mathrm{v}_2=\mathrm{u}_1+\mathrm{v}_1-\mathrm{u}_1$
To find the final velocities $v_1$ and $v_2$ :
Substituting equation (vii) in equation (ii) gives the velocity of $\mathrm{m}_1$ as

$
\begin{aligned}
& \mathrm{m}_1\left(\mathrm{u}_1-\mathrm{v}_1\right)=\mathrm{m}_2\left(\mathrm{u}_1+\mathrm{v}_1-\mathrm{u}_2-\mathrm{u}_2\right) \\
& \mathrm{m}_1 \mathrm{u}_1-\mathrm{m}_1 \mathrm{v}_1=\mathrm{m}_2\left(\mathrm{u}_1+\mathrm{v}_1-2 \mathrm{u}_2\right) \\
& \mathrm{m}_1 \mathrm{u}_1+2 \mathrm{~m}_2 \mathrm{u}_2=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_1 \\
& \text { Or } \quad v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right) u_1+\left(\frac{2 m_2}{m_1+m_2}\right) u_2
\end{aligned}
$
Similarly, by substituting (vi) in equation (ii) or substituting equation (viii) in equation (vii), we get the final velocity of $\mathrm{m}_2$ as
$
v_2=\left(\frac{2 m_1}{m_1+m_2}\right) u_1+\left(\frac{m_2-m_1}{m_1+m_2}\right) u_2
$
Case 1: When bodies has the same mass i.e., $\mathrm{m}_1=\mathrm{m}_2$,
$
\begin{aligned}
\text { equation (viii) } \Rightarrow \quad v_1 & =(0) u_1+\left(\frac{2 m_2}{2 m_2}\right) u_2 \\
v_1 & =u_2 \\
\text { equation (ix) } \Rightarrow \quad v_2 & =\left(\frac{2 m_1}{2 m_1}\right) u_1+(0) u_2 \\
v_2 & =u_1
\end{aligned}
$
The equations (x) and (xi) show that in one dimensional elastic collision, when two bodies of equal mass collide after the collision their velocities are exchanged.
Case 2: When bodies have the same mass i.e., $\mathrm{m}_1=\mathrm{m}_2$ and second body (usually called target) is at rest $\left(u_2=0\right)$
By substituting $\mathrm{m}_1 \mathrm{~m}_2=$ and $\mathrm{u}_2=0$ in equations (viii) and equations (ix) we get, from equation (viii) $\Rightarrow V_1=0$...(xii)
from equation (ix) $\Rightarrow \mathrm{v}_2=\mathrm{u}_1 \ldots$ (xiii)
Equations (xii) and (xiii) show that when the first body comes to rest the second body moves with the initial velocity of the first body.
Case 3: The first body is very much lighter than the second body

$\left(m_1 \ll m_2, \frac{m_1}{m_2} \ll 1\right)$ then the ratio $\frac{m_1}{m_2} \approx 0$ and also if the target is at rest $\left(u_2=0\right)$
Dịviding numerator and denominator of equation (viii) by $m_2$, we get
$
\begin{aligned}
& v_1=\left(\frac{\frac{m_1}{m_2}-1}{\frac{m_1}{m_2}+1}\right) u_1+\left(\frac{2}{\frac{m_1}{m_2}+1}\right)(0) \\
& v_1=\left(\frac{0-1}{0+1}\right) u_1 \\
& v_1=-u_1 \\
&
\end{aligned}
$
Similarly, Dividing numerator and denominator of equation (ix) by $\mathrm{m}_2$, we get
$
v_2=\left(\frac{2 \frac{m_1}{m_2}}{\frac{m_1}{m_2}+1}\right) u_1+\left(\frac{1-\frac{m_1}{m_2}}{\frac{m_1}{m_2}+1}\right)
$

$
\begin{aligned}
& v_2=(0) u_1+\left(\frac{1-\frac{m_1}{m_2}}{\frac{m_1}{m_2}+1}\right)(0) \\
& \mathrm{v}_2=0 \\
&
\end{aligned}
$
The equation (xiv) implies that the first body which is lighter returns back (rebounds) in the opposite direction with the same initial velocity as it has a negative sign. The equation (xv) implies that the second body which is heavier in mass continues to remain at rest even after collision. For example, if a ball is thrown at a fixed wall, the ball will bounce back from the wall with the same velocity with which it was thrown but in opposite direction.

Case 4: The second body is very much lighter than the first body $\left(m_2 \ll m_1, \frac{m_2}{m_1} \ll 1\right)$ ther the ratio $\frac{m_2}{m_1} \approx 0$ and also if the target is at rest $\left(u_2=0\right)$
Dividing numerator and denominator of equation (4.53) by $m_1$, we get
$
\begin{aligned}
v_1 & =\left(\frac{1-\frac{m_2}{m_1}}{1+\frac{m_2}{m_1}}\right) u_1+\left(\frac{2 \frac{m_2}{m_1}}{1+\frac{m_2}{m_1}}\right)(0) \\
v_1 & =\left(\frac{1-0}{1+0}\right) u_1+\left(\frac{0}{1+0}\right)(0) \\
v_1 & =u_1
\end{aligned}
$
Similarly,
Dividing numerator and denominator of equation (xiii) by $\mathrm{m}_1$, we get
$
v_2=\left(\frac{2}{1+\frac{m_2}{m_1}}\right) u_1+\left(\frac{\frac{m_2}{m_1}-1}{1+\frac{m_2}{m_1}}\right)(0)
$

$
\begin{aligned}
& v_2=\left(\frac{2}{1+0}\right) u_1 \\
& v_2=2 u_1
\end{aligned}
$
The equation (xvi) implies that the first body which is heavier continues to move with the same initial velocity. The equation (xvii) suggests that the second body which is lighter will move with twice the initial velocity of the first body. It means that the lighter body is thrown away from the point of collision.