Work Energy And Power Question 148

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

Options:

A) $ \frac{ML^{2}}{k} $

B) zero

C) $ \frac{kL^{2}}{2M} $

D) $ \sqrt{Mk}L $

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Answer:

Correct Answer: D

Solution:

[d] 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, $ \frac{1}{2}Mv^{2}=\frac{1}{2}KL^{2} $
$ \therefore v=\sqrt{\frac{k}{M}L} $

Where v is the velocity of block by which it collides with spring.

So, its maximum momentum. $ P=Mv=M\sqrt{\frac{k}{M}}L=\sqrt{MK}L $

After collision, the block will rebound with same linear momentum.



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