Question

Figure shows a block $A$ of mass $6m$ having a smooth semicircular groove of radius $a$ placed on a smooth horizontal surface. A block $B$ of mass $m$ is released from rest from horizontal position as shown in the figure. Find the speed of the bigger block $6m$, when the smaller block reaches the bottommost position of the groove. Take $g=10{\text{m/s}}^2$.

• A. $\sqrt{\frac{ga}{14}}$
• B. $\sqrt{\frac{ga}{21}}$
• C. $\sqrt{\frac{ga}{10}}$
• D. $\sqrt{\frac{ga}{36}}$

Answer 1

The correct option is B $\sqrt{\frac{ga}{21}}$

Block $6m$ will move along $-ve$ x-direction,due to component of normal reaction ($N_x$) acting along that direction.


When the smaller block reaches the bottommost position of groove, let it's velocity be $v_1$ relative to the bigger block, and let the mass $6m$ be moving with velocity $v_2$ relative to ground.


Then, $\overrightarrow{v_{Bg}}=\overrightarrow{v_{BA}}+\overrightarrow{v_{Ag}}$
$\therefore$ Velocity of block $B$ w.r.t ground $\overrightarrow{v_{Bg}}=+\left(v_1\right)\hat{i}+(-v_2)\hat{i}=(v_1-v_2)\hat{i}$

Applying conservation of momentum in horizontal direction on system of blocks :
${v_i}_A={v_i}_B=0$ (Initial velocity of both blocks zero)

$P_i=P_f$
[$\because$ External force on system of $(m+6m)$ in horizontal direction$=0$]
$\Rightarrow 0=m\times (v_1-v_2)-(6m\times v_2)$
$\therefore v_2=\frac{v_1}{7}...\left(i\right)$

Applying mechanical energy conservation on system of blocks:
$\text{Loss in Gravitational PE of block B}$
$=\text{Gain in KE of block B+ Gain in KE of block A}$
$\Rightarrow mga=\frac{1}{2}m(v_1-v_2{)}^2+\frac{1}{2}\times 6m\times {v_2}^2$
$\Rightarrow 2ga=6v_2^2+v_1^2+v_2^2-2v_1{v}_2....\left(ii\right)$
From Eq. $\left(i\right)$ and $\left(ii\right)$
$v_2^2=\frac{2ga}{42}$
$\therefore v_2=\sqrt{\frac{ga}{21}}$
will be the velocity of block $6m$.