Question

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

A. `(ML^(2))/(K)`
B. zero
C. `(KL^(2))/(2M)`
D. `sqrt(MKL)`

Answer 1

Correct Answer - D
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,
`(1)/(2) Mv^(2) = (1)/(2) KL^(2) : 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.