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. Zero
B. `(ML^(2))/(K)`
C. `sqrt(MK)L`
D. `(KL^(2))/(2M)`

Answer

C

Explanation

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 energ
`(1)/(2)Mv^(2)= (1)/(2) KL^(2) therefore v= sqrt((K)/(M))L` where v is the velocity of block by which it collides with spring. So, its maximum momentum
`P= Mv= M sqrt((K)/(M))L= sqrt(MK)L`
After collision the block will rebound with same linear momentum