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' [NEET 2018]

detailed solution

Correct option is B

Since, the collision mentioned is an elastic head on collision. Thus, according to the law of conservation of linearmomentum, we have                                m1u1 + m2u2 = m1v1 + m2v2where, m1 and m2 are the masses of the two blocks respectively, u1 and u2 are their initial velocities and v1 and v2are their final velocities, respectively.Given,               m1=m, m2=4m                         u1=v, u2=0 and v1=0∴  mv + 4m × 0 = 0 + 4mv2⇒               mv = 4mv2    or    v2=v4                            …(i)Now, the coefficient of restitution,                     e= relative velocity of separation  relative velocity of approach                         =-v2-v1u2-u1=-v4-00-v                 [from Eq.(i)]                         = 1/4∴                  e = 0.25