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

• A. Zero
• B. $\frac{M L^{2}}{K}$
• C. $\sqrt{M K} L$
• D. $\frac{K L^{2}}{2 M}$

Answer 1

Answer: (C)

When collision occurs all kinetic energy it possesses get converted into P.E
$\frac{1}{2} M v^{2}=\frac{1}{2} K L^{2}$
$ \Rightarrow M v^{2}=K L^{2}$
Multiplying both side by M. (Mass)
$(M v)^{2}=M K L^{2}$
$M v=\sqrt{M K L^{2}}$
$P=\sqrt{M K} L$ or $\sqrt{M K L^{2}}$