Question

A long block A is at rest on a smooth horizontal surface. A small block B, whose mass is half of A, is placed on A at one end and moved along A with some velocity $u$. The coefficient of friction between the blocks is $\mu$ :

• A. the blocks will reach a final common velocity $\frac{u}{3}$
• B. the work done against friction is two-thirds of the initial kinetic energy of B
• C. before the blocks reach a common velocity, the acceleration of A relative to B is $\frac{2}{3}\mu g$
• D. before the blocks reach a common velocity the acceleration of A relative to B is $\frac{3}{2}\mu g$

Answer 1

Answer: (A), (B), (D)

the work done against friction is two-thirds of the initial kinetic energy of B
the blocks will reach a final common velocity $\frac{u}{3}$
before the blocks reach a common velocity the acceleration of A relative to B is $\frac{3}{2}\mu g$

As there are no external forces acting on the 'A + B' system, its total momentum is conserved. If the masses of A and B are 2m and 2 respectively, and v is the final common velocity, $mu=(m+2m)v$ or $v=u/3$

Work done against friction $=$ loss in KE $=\frac{1}{2}m{u}^2-\frac{1}{2}\left(3m\right)v^2$

$=\frac{1}{2}m{u}^2-\frac{1}{2}\left(3m\right)\frac{u^2}{9}=\frac{1}{2}m{u}^2[1-\frac{1}{3}]=\frac{2}{3}\times \frac{1}{2}m{u}^2$

The force of friction between the blocks is $\mu$mg.

Acceleration of A (to the right) $=a_1=\frac{\mu mg}{2m}=\frac{\mu g}{2}$

Acceleration of B (to the left) $=a_2=\frac{\mu mg}{m}=\mu g$

Acceleration of A relative to B $=a_1-(-a_2)=\frac{3}{2}\mu g$.