The block of mass `M` moving on the frictionless horizontal surface collides with the spring constant `k` and compresses it by length `L` . The maximum momention of the block after collision is

Options

A. `sqrt(mk)L`
B. `(kL^(2))/(2M)`
C. zero
D. `(ML^(2))/k`

Answer

A

Explanation

According to conservation of energy
`1/2 kL^(2)=1/2 M upsilon^(2)`
`rArr kL=((M upsilon)^(2))/M`
`MkL^(2)=p^(2)` `(p=m upsilon)`
`:. P=Lsqrt(MK)`