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Q.

A ball of mass m is released from A inside a smooth wedge of mass m as shown in fig. What is the speed of the wedge when the ball reaches point B?

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a

gR3212

b

2gR

c

5gR2312

d

32gR

answer is A.

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Detailed Solution

Loss of PE=Gain in KEmgR cosθ=12mv2+12mv1 cosθ-v2+12mv1 sinθ2---1 From conservation of linear momentum mv1 cos45o-v=mv-------(2) Here v1 is the velocity of ball w.r.t wedge solve to get speed of wedge (v)
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