A bob of mass M is suspended by an inextensible string. A particle of mass m, moving with velocity v₀ strike the bob as shown in the figure. If the particle comes to rest just after collision. If $\mathrm{\theta }={30}^{\circ }$, ${\mathrm{v}}_{\mathrm{o}}=4\mathrm{m}/\mathrm{s}\text{ and }\frac{\mathrm{m}}{\mathrm{M}}=10$, then calculate the velocity of the bob (in m/s) just after collision.

Here tension in the string will be impulse in nature. Let us make the impulse diagram of the situation.

Applying impulse moment

       ${\mathrm{mv}}_0-{\mathrm{J}}_1=0\text{ or }{\mathrm{J}}_1={\mathrm{mv}}_0$             …(i)

As the string is inextensible, the bob cannot move in vertical direction.

Hence, ${\mathrm{J}}_{\text{tension }}={\mathrm{J}}_1\cos \mathrm{\theta }\text{ or }{\mathrm{J}}_{\text{tension }}=\mathrm{mvcos}\mathrm{\theta }$

The horizontal component of impulse $\left({\mathrm{J}}_1\sin \mathrm{\theta }\right)$ will be responsible for motion of the bob in horizontal direction, hence 

$\begin{array}{l} {\mathrm{J}}_1\sin \mathrm{\theta }=\mathrm{MV}\text{ or }\mathrm{V}=\frac{{\mathrm{J}}_1\sin \mathrm{\theta }}{\mathrm{M}}\\ \Rightarrow \mathrm{V}=\frac{{\mathrm{mv}}_0\sin \mathrm{\theta }}{\mathrm{M}}=\frac{4\times 0.5\times 10}{1}=20\mathrm{m}/\mathrm{s}\end{array}$

After collision the bob starts moving in the circular path.