First slide

Work

Question

The block of mass M moving on a frictionless horizontal surface collides with a spring of spring constant k and compresses it by length L. The maximum momentum of the block after collision is

Moderate

A

$\frac{M L^{2}}{k}$
B

$z e r o$
C

$\frac{k L^{2}}{2 M}$
D

$\sqrt{M k} L$

Solution

When block of mass M collides with the spring, its kinetic energy gets converted into elastic potential energy of the spring. From the law of conservation of energy, conservation of energy KE = energy stored in a spring
$\frac{1}{2} M v^{2} = \frac{1}{2} k L^{2} \Rightarrow ∴ v = \sqrt{\frac{k}{M}} L$ here M is mass, length of compression in spring =L
Where v is the velocity of block by which it collides with spring. So its maximum momentum,
$m o m e n t u m = p = M v = M \sqrt{\frac{k}{M}} L = \sqrt{M k} L$
After collision, the block will rebound with same linear momentum.

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