Work Energy And Power Question 286
Question: 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
Options:
A) $ \frac{kL^{2}}{2M} $
B) $ \sqrt{Mk}L $
C) $ \frac{ML^{2}}{k} $
D) zero
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \frac{1}{2}Mv^{2}=\frac{1}{2}kL^{2} $
$ \Rightarrow v=\sqrt{\frac{k}{M}}L $
Momentum $ =M\times v=M\times \sqrt{\frac{k}{M}}.L $
$ =\sqrt{kM}.L $
