Numerical Problems-1 - Chapter 4 Work Energy and Power 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
Calculate the work done by a force of $30 \mathrm{~N}$ in lifting a load of $2 \mathrm{Kg}$ to a height of $10 \mathrm{~m}\left(\mathrm{~g}=10 \mathrm{~ms}^{-1}\right)$

Answer:
Given: $F=30 \mathrm{~N}$, load $(\mathrm{m})=2 \mathrm{~kg}$; height $=10 \mathrm{~m}, \mathrm{~g}=10 \mathrm{~ms}^{-2}$
Gravitational force $\mathrm{F}=\mathrm{mg}=30 \mathrm{~N}$
The distance moved $\mathrm{h}=10 \mathrm{~m}$
Work done on the object $\mathrm{W}=\mathrm{Fh}=30 \times 10=300 \mathrm{~J}$.
Question 2.
A ball with a velocity of $5 \mathrm{~ms}^{-1}$ impinges at angle of $60^{\circ}$ with the vertical on a smooth horizontal plane. If the coefficient of restitution is 0.5 , find the velocity and direction after the impact.
Answer:
Given: Velocity of ball: $5 \mathrm{~ms}^{-1}$
Angle of inclination with vertical: $60^{\circ}$
Coefficient of restitution $=0.5$.
Note: Let the angle reflection is $\theta$ ' and the speed after collision is $v$ '. The floor exerts a force on the ball along the normal during the collision. There is no force
parallel to the surface. Thus, the parallel component of the velocity of the ball remains unchanged. This gives
$\mathrm{v}^{\prime} \sin \theta^{\prime}=\mathrm{v} \sin \theta$
Vertical component with respect to floor $=v^{\prime} \cos \theta^{\prime}$ (velocity of separation)
Velocity of approach $=v \cos \theta$

Coefficient of restitution $e=\frac{\text { velocity of separation }}{\text { velocity of approach }}$
$
e=\frac{v^{\prime} \cos \theta^{\prime}}{v \cos \theta} \quad \therefore \quad v^{\prime} \cos \theta^{\prime}=e v \cos \theta
$
from (i) and (ii)

(ii)
$
\begin{aligned}
v^{\prime} \sqrt{\left(1-\sin ^{\prime 2} \theta\right)} & =e v \cos \theta \\
v^{\prime 2}\left(1-\sin ^{\prime 2} \theta\right) & =e^2 v^2 \cos ^2 \theta \\
v^{\prime 2}-v^{\prime 2} \sin ^{\prime 2} \theta & =e^2 v^2 \cos ^2 \theta \\
v^{\prime 2} & =v^2 \sin ^2 \theta+e^2 v^2 \cos ^2 \theta \\
v^{\prime} & =\sqrt{v^2 \sin ^2 \theta+e^2 v^2 \cos ^2 \theta} \\
v^{\prime} & =v \sqrt{\sin ^2 \theta+e^2 \cos ^2 \theta}
\end{aligned}
$
The speed after collision $v^{\prime}=v \sqrt{\sin ^2 \theta+e^2 \cos ^2 \theta}$
$
\begin{aligned}
& v^{\prime}=5 \sqrt{\sin ^2(60)+(0.5)^2 \cos ^2 60}=5 \sqrt{\frac{3}{4}+0.25 \times \frac{1}{4}} \\
& =\frac{5}{2} \sqrt{3.25}=2.5 \times 1.8=4.5 \mathrm{~ms}^{-1} \\
& \text { Angle of reflection } \theta^{\prime}=\tan ^{-1}\left(\frac{\tan \theta}{e}\right)=\tan ^{-1}\left(\frac{\tan 60^{\circ}}{0.5}\right)=\tan ^{-1}\left(\frac{\sqrt{3}}{0.5}\right) \\
& =\tan ^{-1}(3.464)=73.9^{\circ} \text {. } \\
&
\end{aligned}
$
Question 3.
A bob of mass $\mathrm{m}$ is attached to one end of the rod of negligible mass and length $\mathrm{r}$, the other end of which is pivoted freely at a fixed center $\mathrm{O}$ as shown in the figure.
What initial speed must be given to the object to reach the top of the circle?
(Hint: Use law of conservation of energy). Is this speed. less or greater than speed obtained in the section 4.2.9?

Answer:
To get the vertical speed given to the object to reach the top of the circle, law of conservation of energy can be used at a points (1) and (2)
Total energy at $1=$ Total energy at 2
$\therefore$ Potential energy at point $1=0$

from eqn (i)
$
\begin{aligned}
0+\frac{1}{2} m v_1^2 & =2 m g r+\frac{1}{2} m v_2^2 \\
\frac{1}{2} v_1^2 & =2 g r+\frac{1}{2} v_2^2 \\
v_1^2-v_2^2 & =4 g r
\end{aligned}
$
In this case bob of mass $m$ is connected with a rod of negligible mass, so the velocity of bob at highest point can be equal to zero i.e. $v_2=0$
$\therefore \quad$ eqn. (2) becomes $v_1^2=4 g r ; \quad v_1=\sqrt{4 g r} \mathrm{~ms}^{-1}$ or $v_1=2 \sqrt{g r} \mathrm{~ms}^{-1}$
The speed of bob obtained here is lesser than the speed obtained in section 4.2.9. It is only because of string is replaced by a massless rod here.
Question 4.
Two different unknown masses $A$ and $B$ collide. $A$ is initially at rest when $B$ has a speed $v$. After collision $B$ has a speed $\mathrm{v} / 2$ and moves at right angles to its original direction of motion. Find the direction in which $\mathrm{A}$ moves after collision.

Answer:
Given:
$
\begin{aligned}
& u_2=v, \quad u_1=0 \\
& v_2=\frac{v}{2}, \quad \theta=?
\end{aligned}
$
Component along $x$-axis
$
m_1 v_1 \cos \theta=m_2 v
$
Component along $y$-axis
$
m_1 v_1 \sin \theta=m_2\left(\frac{v}{2}\right)
$
from (1) and (2)
$
\begin{aligned}
\tan \theta & =\frac{\left(\frac{v}{2}\right)}{v}=\frac{1}{2} \\
\theta & =\tan ^{-1}\left(\frac{1}{2}\right)=\tan ^{-1}(0.5) ; \theta=26^{\circ} 33^{\prime}
\end{aligned}
$

Question 5.
A bullet of mass $20 \mathrm{~g}$ strikes a pendulum of mass $5 \mathrm{~kg}$. The centre of mass of pendulum rises a vertical distance of $10 \mathrm{~cm}$. If the bullet gets embedded into the pendulum, calculate its initial speed.
Answer:
Given: $\mathrm{m}_1=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg} ; \mathrm{m}_2=5 \mathrm{~kg} ; \mathrm{s}=10 \times 10^{-2} \mathrm{~m}$.
Let the speed of the bullet be $\mathrm{v}$. The common velocity of bullet and pendulum bob is $\mathrm{V}$. According to law of conservation of linear momentum.
$
\mathrm{V}=\frac{m_1 v}{\left(m_1+m_2\right)}=\frac{20 \times 10^{-3} v}{5+20 \times 10^{-3}}=\frac{0.02}{5.02} v=0.004 v
$
The bob with bullet go up with a deceleration of $g=9.8 \mathrm{~ms}^{-2}$. Bob and bullet come to rest at a height of $10 \times 10^{-2} \mathrm{~m}$
from III rd equation of motion
$
\begin{aligned}
v^2 & =u^2+2 a s \text { here } \\
v^2-2 g s & =0 \\
v^2 & =2 g s \\
(0.004 v)^2 & =2 \times 9.8 \times 10 \times 10^{-2} \\
v^2 & =\frac{2 \times 9.8 \times 10 \times 10^{-2}}{(0.004)^2} \\
v & =350 \mathrm{~ms}^{-1} .
\end{aligned}
$