Question

Two pendulum bobs of mass $m$ and $2m$ collide elastically at the lowest point in their motion. If both the balls are released from height $H$ above the lowest point. The velocity of the bob of mass $m$ just after collision is :

• A. $\sqrt{\frac{2gH}{3}}$
• B. $\frac{5}{3}\sqrt{2gH}$
• C. $\sqrt{2gH}$
• D. None of these

Answer 1

The correct option is B $\frac{5}{3}\sqrt{2gH}$

Lets consider first pendulum bob of mass $m_1$ and second pendulum bob of mass $m_2$ collide elastically at the lower point in their motion. If both are released from the height $H$ above the lower point.

At height $H$, Kinetic energy is equal to potential energy.

$\frac{1}{2}m{u}^2=mgH$

$u=\sqrt{2gH}$

Given ,

$m_1=m$

$m_2=2m$

$u_1=-\sqrt{2gH}$

$u_2=\sqrt{2gH}$

In elastic collision,

The velocity of first $v_1$ after collision at the lower point,

$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2$

By putting the given value of $m_1$, $m_2$, $u_1$, $u_2$ in the above equation,

$v_1=\left(\frac{m-2m}{m+2m}\right)(-\sqrt{2gH})+\left(\frac{4m}{m+2m}\right)\sqrt{2gH}$

$v_1=\frac{5}{3}\sqrt{2gH}$

Thus, the correct option is B.