Question

A shell of mass $1\text{kg}$ is projected with velocity $20\text{m/s}$ at an angle ${60}^{\circ }$ with horizontal. It collides inelastically with a ball of mass $1\text{kg}$ which is suspended through a thread of length $1\text{m}$. The other end of the thread is attached to the ceiling of a trolley of mass $\frac{4}{3}\text{kg}$ as shown in figure. Initially the trolley is stationary and it is free to move along horizontal rails without any friction. What is the maximum deflection of the thread with vertical? String does not slack. Take $g=10{\text{m/s}}^2$.

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is D ${60}^{\circ }$

Mass of shell, $m_1=1\text{kg}$
velocity of shell, $u=20\text{m/s}$
Mass of ball, $m_2=1\text{kg}$
Mass of trolley, $m_3=\frac{4}{3}\text{kg}$

Maximum height of the projectile,
$H_{max}=\frac{u^2{\sin }^2\theta }{2g}$

Substituting the values, we get
$H_{max}=\frac{{\left(20\right)}^2\times \frac{3}{4}}{2\times 10}=15\text{m}$

i.e. the shell strikes the ball at highest point of its trajectory.

From the conservation of linear momentum, velocity, $v$ of (ball + shell) just after collision can be calculated,
$m_1{u}_x+m_2{u}_1=m_{1}v+m_{2}v$

Since intially the ball is stationary, so $u_1=0$.
Substituting the values in the above equation,
$1\times 20\cos {60}^{\circ }+1\times 0=1\times v+1\times v$

$\therefore v=\frac{20\cos {60}^o}{2}=5\text{m/s}$


At highest point combined mass is at rest relative to the trolley. Let $v_1$ be the velocity of trolley at this instant.

From conservation of linear momentum we have,
$(m_1+m_2)v+m_3{u}_2=(m_1+m_2+m_3)v_1$

Since initially trolly is at rest, so $u_2=0$.
Substitung the values in the above equation we get,
$(1+1)5+\frac{4}{3}\times 0=(1+1+\frac{4}{3})v_1$

$\Rightarrow 2\times 5=(2+\frac{4}{3})v_1$

$\therefore v_1=3\text{m/s}$

From conservation of energy, we have

Change in kinetic energy of the system= Change in potential energy of the system
$\frac{1}{2}(m_1+m_2)v^2-\frac{1}{2}(m_1+m_2+m_3)v_1^2=(m_1+m_2)gl-(m_1+m_2)gl\cos \theta$

Substituting the values, we get
$\frac{1}{2}(1+1)5^2-\frac{1}{2}(1+1+\frac{4}{3})v_1^2=(1+1)\times 10(1-\cos \theta )$

On solving we get,
$\cos \theta =\frac{1}{2}$

$\therefore \theta ={60}^o$

Hence, option (d) is correct answer.