A long block A is at rest on a smooth horizontal surface. A small block B whose mass is half of mass of A is placed on A at one end and is given an initial velocity u as shown in figure. The coefficient of friction between the blocks is `mu`.

Options

A. (a) Finally both move with a common velocity `2u//3`.
B. (b) Acceleration of B relative to A initially is `3mug//2` towards left.
C. (c) Magnitude of total work done by friction is equal to the final kinetic energy of the system.
D. (d) The ratio of initial to final momentum of the system is 1.

Answer

B, D

Explanation

v is the final common velocity.

`(m+2m)v=m uimpliesv=u/3`

`a_1=(mu mg)/(m)=mug, a_2=(mu mg)/(2m)=(mug)/(2)`
Acceleration of B relative to A:
`a_1+a_2=mug+(mug)/(2)=3/2mug` towards left
Work done against friction:
`W_1=K_i-K_f=1/2m u^2-1/23mv^2=(m u^2)/(3)`
Final `KE=1/23mv^2=(m u^2)/(6)`
Hence, c. is not correct.
Option (d) is correct, because momentum is always conserved as there is no external force.