Question

A block $B$ rests on a horizontal plane. A small block $A$ of mass that of $B$ is placed at one end of $B$ and given a horizontal velocity $v$. If $\mu$ is the coefficient of friction between the two blocks, then the time elapsed before the two blocks attain a common velocity is:

• A. $\frac{v}{2\mu g}$
• B. $\frac{3v}{2\mu g}$
• C. $\frac{v}{\mu g}$
• D. $\frac{v}{3\mu g}$

Answer 1

The correct option is A $\frac{v}{2\mu g}$

It is given that $m_a=m_b=$ let's assume, $m$.

Let the final, common velocity $=v_f$

Applying 1st equation of motion:

For $\text{A}:v_f=v+(-\frac{\mu mg}{m})t$

For $\text{B}:v_f=0+\left(\frac{\mu mg}{m}\right)t$

Adding the above, we get:

$2v_f=v\Rightarrow v_f=\frac{v}{2}$

Substituting in equation for $B$:

$\frac{v}{2}=\mu gt$

$\therefore t=\frac{v}{2\mu g}\to$ option $\text{(A)}$

$\to QED.$