Question

Figure shows a block A of mass $8m$ having a smooth semi circular groove of radius $a$ placed on a smooth horizontal surface. A block B of mass $m$ is released from a position in groove, where its radius is horizontal. The speed of block A when block B reaches its bottom is

• A. $\sqrt{ga}$
• B. $\sqrt{\frac{2ga}{7}}$
• C. $\sqrt{\frac{ga}{36}}$
• D. $\text{zero}$

Answer 1

The correct option is C $\sqrt{\frac{ga}{36}}$

Let
$v_2=$ speed of block A towards left,
$v_1=$ speed of block B w.r.t. A towards right,
$v_1-v_2=$speed of block B towards right.

By conservation of momentum,
$8m{v}_2=m(v_1-v_2)$
$9v_2=v_1$


By conservation of energy
$mga=\frac{1}{2}\times 8m\times v_2^2+\frac{1}{2}\times m\times (v_1-v_2{)}^2$
$ga=4v_2^2+\frac{1}{2}(9v_2-v_2{)}^2$
$ga=36v_2^2$
$v_2=\sqrt{\frac{ga}{36}}$
$\therefore \left(c\right)$