A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

Options

A. 0.8
B. 0.25
C. 0.5
D. 0.4

Answer

B

Explanation

If we consider both the blocks as a single syste then due to absence of external force during collision , linear momentum of the system remains conserved . Let v . be the velocity of heavier block after collision .
`mv+4mxx 0 - 4mv +0`
`v . = v/4`
Coefficient of restitution
` e = ("Relaative velocity of seperation ")/("Relative velocity of approach ") = (v/4)/v`
` e = 1/4 = 0.25`.