Question

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

• A. The blocks will reach the 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)

Before the blocks reach a common velocity the acceleration of $A$ relative to $B$ is $\frac{3}{2}\mu g$.

Let $v$ be common velocity of the blocks
Using conservation of linear momentum, since friction is internal for the 2 block system,
$\frac{m}{2}u=(m+\frac{m}{2})v$
$v=\frac{u}{3}$
Initial kinetic energy of the system
$k_i=\frac{1}{2}\left(\frac{m}{2}\right)u^2$
$k_i=\frac{1}{4}m{u}^2$
Using Work energy theorem,
$\frac{m{u}^2}{4}-w_f=\frac{1}{2}\left(\frac{3m}{2}\right)\frac{u^2}{9}$
$\frac{1}{4}m{u}^2-\frac{1}{12}m{u}^2=w_f$
$w_f=\frac{3m{u}^2-m{u}^2}{12}=\frac{m{u}^2}{6}=\frac{2}{3}\times \frac{1}{4}m{u}^2$
$w_f=\frac{2}{3}{k}_i$
Force of friction between blocks
$f_k=\mu \left(\frac{m}{2}\right)g$
Acceleration of A to right.
$a_A=\frac{\mu mg}{2\left(m\right)}=\frac{\mu g}{2}$
Acceleration of B to left,
$a_B=\frac{\mu mg}{2\left(\frac{m}{2}\right)}=\mu g$
Acceleration of A relative to B,
$a_{AB}=a_A-(-a_B)$
$a_{AB}=\frac{\mu g}{2}+\mu g$
$a_{AB}=\frac{3\mu g}{2}$