Q. 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) Zero
B) $\frac{ML^{2}}{K}$
C) $\sqrt{MK}\ L$
D) $\frac{KL^{2}}{2M}$
Explanation
(c) 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
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.