A shell of mass $1kg$ is projected with velocity $20 \mathrm{m}/\mathrm{s}$ at an angle $60°$with horizontal. It collides inelastically with a ball of mass $1 \mathrm{kg}$which is suspended through a thread of length.$1 \mathrm{m}$. The other end of the thread is attached to the ceiling of a trolley of mass $\frac{4}{3}kg$ 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 $g=10 \mathrm{ms}/{\mathrm{s}}^2$

 $H=\frac{u^2{\sin }^2\theta }{2g}=\frac{(20{)}^2\times \frac{3}{4}}{2\times 10}=15 \mathrm{m}$

i.e, the shell strikes the ball at highest point of is trajectory. Velocity of (ball + shell) just atter collision,

$v=20 \cos {60}^{0 }=10 \mathrm{m}/\mathrm{s}$

Just after impact , the velocity of the combined mass is v' where (1+1)v'=1$\times 10 \Rightarrow v\text{'}=5 m/s$

At highest point combined mass is at rest relative to the trolley. Let v be the velocity of trolley at this instant. From conservation of linear momentum we have,

$$\begin{gathered} 2\times 5=\left(2+\frac{4}{3}\right)v \\ v=3 \mathrm{m}/\mathrm{s} \end{gathered}$$

or From conservation of energy, we have $\frac{1}{2}\times 2\times (5{)}^2-\frac{1}{2}\left(2+\frac{4}{3}\right)(3{)}^2=2\times 10(1-\cos \theta )$ 

Solving we get, 

$\cos \theta =\frac{1}{2}$  $\Rightarrow \theta ={60}^0$