First slide

Simple hormonic motion

Question

A body is hanging from a rigid support by an inextensible string of length $l$ . It is struck by an identical body of mass m with horizontal velocity $u = \sqrt{2 g l}$ . The collision is perfectly inelastic. The increase in tension will be (Just after the collision)

Moderate

Solution

Before Collision, T = mg
Just after Collision $V = \frac{m u}{m + m} = \frac{u}{2}$
$∴$ Just after collision, $T^{'} = 2 m g + \frac{2 m V^{2}}{l}$
$\begin{array}{l} ∴ T^{'} = 2 m g + \frac{2 m . u^{2}}{4 l} = 2 m g + m g = 3 m g \\ Δ T = T^{'} - T = 2 m g \end{array}$

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