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Q.

A particle of mass $m = 0.1 k g$ is released from rest from a point $A$ of a wedge of mass $M = 2.4 k g$ free to slide on a frictionless horizontal plane. The particle slides down the smooth face $A B$ of the wedge. When the velocity of the wedge is $0.2 m / s$ the velocity of the particle in $m / s$ relative to the wedge is:

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a

$4.8$

b

$5$

c

$7.5$

d

$\frac{10}{\sqrt{3}}$

answer is D.

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Detailed Solution

Apply conservation of linear momentum along x-axis

${Mv}_{w} = m (v_{bw} cosα - v_{w})$

Substituting given values we get

$v_{bw} = \frac{10}{\sqrt{3}} m / s$

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A particle of mass m=0.1kg is released from rest from a point A of a wedge of mass M = 2.4 kg free to slide on a frictionless horizontal plane. The particle slides down the smooth face AB of the wedge. When the velocity of the wedge is 0.2 m/s the velocity of the particle in m/s relative to the wedge is: