Question

# A small particle of mass m=2 kg moving with constant horizontal velocity u=10 ms−1 strikes a wedge shaped block of mass M=4 kg placed on the smooth surface as shown in figure. After collision, the particle starts moving up the inclined plane. Find the velocity of the wedge immediately after collision-

A
1 ms1
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B
3 ms1
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C
2 ms1
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D
4 ms1
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Solution

## The correct option is C 2 ms−1 As given, after the collision particle starts moving up the inclined plane, it means particle velocity with respect to the wedge is along the incline i.e., Vrel. Force on the wedge at the time of the collision will be perpendicular to the inclined plane. Hence, velocity will not change along the inclined plane. ucosα=Vrel+Vcosα ⇒u√2=Vrel+V√2 [∵α=45∘] Vrel=u−V√2 ......(1) Horizontal component of velocity of particle after collision, Vx=Vrelcos45∘+V=Vrel√2+V Vertical component of velocity after collision, Vy=Vrelsin45∘=Vrel√2 At the time of collision, there will be an external impulse force from the ground act on the wedge in vertical direction only. So, along horizontal direction, momentum will be conserved. mu=mVx+MV mu=m(Vrel√2+V)+MV ........(2) From (1) and (2) we get, mu=m(u−V2+V)+MV ⇒2×10=2(10−V2+V)+4V ⇒20=10+5V ∴V=2 ms−1 <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (C) is the correct answer. Why this question ? Caution: Since there will be impulsive force from the ground to wedge in vertical direction. Thus, momentum is not conserved in the vertical direction. Only horizontal direction momentum will be conserved.

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