Question

In the figure shown, the heavy ball of mass $2m$ rests on the horizontal surface and the lighter ball of mass $m$ is dropped from a height $h>2l$. At the instant, the string gets taut, the upward velocity $\left(\text{in}\text{m/s}\right)$ of the the heavy ball? $\left(Given:l=3.6\text{m}\right)$.

• A. $\frac{2}{3}\sqrt{gl}$
• B. $\frac{4}{3}\sqrt{gl}$
• C. $\frac{1}{3}\sqrt{gl}$
• D. $\frac{1}{2}\sqrt{gl}$

Answer 1

The correct option is A $\frac{2}{3}\sqrt{gl}$

Velocity of light ball, when it is descended by height $2l$

$v=\sqrt{2g2l}=2\sqrt{gl}$

When string will tout then impulsive tension will be generated in the string.

Let Velocity all both masses become v' after string tout.

$J=2m{v}^{\prime }$...........(1)

$J=m\times 2\sqrt{gl}-m{v}^{\prime }$ ...........(2)

On solving above equation

$m2\sqrt{gl}=2m{v}^{\prime }+m{v}^{\prime }$

$2m\sqrt{gl}=3m{v}^{\prime }$

$\therefore \frac{2}{3}\sqrt{gl}=v^{\prime }$

$\therefore V^{\prime }=\frac{2}{3}\sqrt{gl}$