Question

Two identical blocks are placed on a smooth horizontal surface, connected by a light string of length $2 l$. String touches a fixed smooth pulley at its mid-point initially. Which is attached to two smooth vertical walls as shown in figure-$5.127$. Block $A$ is given a speed $V_{0}$ perpendicular to string as shown in diagram. $B$ strikes the pulley and stops.
Speed of block $B$ when it hits the pulley is :

• A. $\frac{V_{0}}{2}$
• B. $V_{0} \frac{\sqrt{3}}{2}$
• C. $V_{0}$
• D. $V_{0} \sqrt{\frac{3}{8}}$

Answer 1

Answer: (D)

Let velocity of block $B$ at the moment of hit be $V$.
So component of velocity of $A$ along string at that instant is also $V$ and component perpendicular to string is $V^{\prime}$, then using conservation of angular momentum about pulley we have
$m V^{\prime} 2 l =m V_{0} I $
$\Rightarrow V^{\prime} =V_{0} / 2$

Using conservation of energy we have
Initial $K E=\operatorname{Final} K E$
$\frac{1}{2} m V^{2}+\frac{1}{2}\left(V^{2}+V^{2}\right)=\frac{1}{2} m V_{0}^{2}=\frac{1}{2} m V^{2}+\frac{1}{2} m\left(V^{2}+V^{2}\right)$
Solving $V=V_{0} \sqrt{\frac{3}{8}}$