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Q.

A particle of mass $m_{1}$ is released from the top of a smooth wedge of mass $m_{2}$ which rests on a smooth horizontal floor. If $m_{2} = 2 m_{1}, θ = 45 °$ , the distance moved by the wedge till the mass $m_{1}$ reaches the floor is

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a

$h$

b

$\frac{h}{2}$

c

$\frac{h}{3}$

d

$\frac{h}{4}$

answer is C.

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$S_{12 x} = h S_{1 x} - S_{2 x} = h m_{1} S_{1 x} + m_{2} S_{2 x} = 0 S_{1 x} = - 2 S_{2 x} - 3 S_{2 x} = h S_{2 x} = \frac{h}{3}$

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