First slide

Collisions

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 the velocity of the heavy ball will be

Moderate

A

$\frac{2}{3} \sqrt{gl}$
B

$\frac{4}{3} \sqrt{gl}$
C

$\frac{1}{3} \sqrt{gl}$
D

$\frac{1}{2} \sqrt{gl}$

Solution

Velocity of light ball, when it is descended by height 2l
$v = \sqrt{2 g 2 l} = 2 \sqrt{gl}$
When string will taut then impulsive tension will be generated in the string.
Let velocity all both masses become v' after string taut.
$J = 2 mv' - - - - (1) J = m \times 2 \sqrt{gl} - mv' - - - - (2)$
(1) = (2)
From conservation of linear momentum.
$m 2 \sqrt{gl} = 2 {mv}^{'} + {mv}^{'}$
$2 m \sqrt{gl} = 3 {mv}^{'}$
$\frac{2}{3} \sqrt{gl} = v^{'}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App