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 $
Show Answer
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.
