question_answer

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

A) \[\frac{k{{L}^{2}}}{2M}\]      

B) \[\sqrt{Mk}L\]

C) \[\frac{M{{L}^{2}}}{k}\]

D) zero

Correct Answer: B

Solution :

[b] \[\frac{1}{2}M{{v}^{2}}=\frac{1}{2}k\,{{L}^{2}}\] \[\Rightarrow v=\sqrt{\frac{k}{\operatorname{M}}}\,L\] Momentum\[=M\times v=M\times \sqrt{\frac{k}{M}}.L\] \[=\sqrt{kM}.L\]