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Question

# A wedge of mass M has one face making an angle α with horizontal and is resting on a smooth rigid floor. A particle of mass ′m′ hits the inclined face of the wedge with horizontal velocity v0. It is observed that the particle rebounds in vertical direction after impact. Calculate the speed of particle after impact. [Neglect friction between particle and the wedge and take M=2m,v0=10 m/s,tanα=2,g=10 m/s2.]

A
5 m/s
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B
0 m/s
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C
20 m/s
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D
10 m/s
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Solution

## The correct option is B 0 m/sAfter collision, assume speed of wedge is v1 and speed of particle is v2 as demonstrated in the below figure. By momentum conservation along the wedge surface i.e. incline mv0cosα=Mv1cosα+mv2sinα ⇒ mv0cosα=2mv1cosα+mv2sinα (∵ M=2m) (by dividing both sides with cosα, we get) v0=2v1+v2tanα ∵(tanα=2) ⇒ v0=2v1+2v2 ...(1) Now, by momentum conservation perpendicular to the incline, we get mv0sinα=Mv1sinα−mv2cosα ⇒ mv0sinα=2mv1sinα−mv2cosα (∵2m=M) (by dividing both sides with cosα, we get) v0tanα=2v1tanα−v2 ∵ tanα=2 ⇒ 2v0=4v1−v2 ...(II) Now, from eq. (I) and (II) we have v0=2v1 and v2=0 Hence, after the collision speed of particle along vertical direction is zero.

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