Question

A particle traveling horizontally with speed $u$ collides and sticks with a particle of equal mass hanging at rest at the end of a light inextensible string of length $2l$. If the string rotates through an angle of ${60}^{\circ }$ before its velocity becomes zero, then $u$ is

• A. $\sqrt{4gl}$
• B. $\sqrt{8gl}$
• C. $\sqrt{2gl}$
• D. $\sqrt{gl}$

Answer 1

The correct option is B $\sqrt{8gl}$

Verticle distance moved by particle is

$r=2l-2l\cos {60}^{\circ }=2l-2l\times \frac{1}{2}$
$r=l$
By conservation of momentum

$uistheinitialvelocity$
$\Rightarrow mu=2mV$ $V$is the velocity just after collosion
$\Rightarrow V=\frac{u}{2}$
By conservation of Energy,
$mgr=\frac{1}{2}m{V}^2$
$\Rightarrow 2gr=V^2$
$\Rightarrow 2gl=\frac{u^2}{4}$
$\Rightarrow u=\sqrt{8gl}$.