# 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…

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. Then, maximum momentum of the block after collision is

(a) $\displaystyle L \sqrt{M k}$

(b) $\displaystyle \frac{k L^2}{2 M}$

(c) zero

(d) $\displaystyle \frac{M L^2}{k}$

Ans: (a)
Sol: Applying Conservation of Energy ,

$\displaystyle \frac{1}{2}Mv^2 = \frac{1}{2}k L^2$

$\displaystyle Mv^2 = k L^2$

$\displaystyle v^2 = \frac{k}{M} L^2$

$\displaystyle v = \sqrt{\frac{k}{M}} L$

Maximum Momentum $\displaystyle = M v = M \sqrt{\frac{k}{M}} L$

$\displaystyle = L \sqrt{M k}$