Figure shows a block A of mass 6 m 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

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$\sqrt{ga}$

$\sqrt{\frac{2 ga}{7}}$

$\sqrt{\frac{ga}{21}}$

zero

answer is C.

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Detailed Solution

$v_{2} = speed of block A; v_{1} = speed of block B w . r . t . A . By conservation of mementum,$

$6 {mv}_{2} = m (v_{1} - v_{2}) \Rightarrow 7 v_{2} = v_{1} . . . . (i) By conservation of energy mga = \frac{1}{2} \times 6 m \times {v^{2}}_{2} + \frac{1}{2} \times m \times {(v_{1} - v_{2})}^{2} \Rightarrow ga = 3 {v^{2}}_{2} + \frac{1}{2} {(7 v_{2} - v_{2})}^{2} \Rightarrow 2 ga = 6 {v^{2}}_{2} + 36 {v^{2}}_{2} \Rightarrow v_{2} = \sqrt{\frac{ga}{21}}$

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