Q2 / 105 · NTA Abhyas (JEE Main) 2020 · Physics · Center of Mass Momentum and Collision

Consider a horizontal surface moving vertically upward with velocity 2 m s-1. A small ball of mass 2 kg is moving with velocity 2i^–2j^ m s-1 If the coefficient of restitution and coefficient of friction are 12 and 13 respectively, find the horizontal velocity (in m s-1) of the ball after the collision.
Figure

Correct answer: 0

Solution

The relative velocity of the particle with respect to the surface, along the normal, is the approach speed of the particle.

vapp=4 m s-1

The separation velocity will be

vsep=evapp=12×4=2 m s-1

∴ v̄m,gr=v̄m,s+v̄s,gr=4 m/s

∫N dt=42+22=12

Now, ∫μN⋅dt=2vxf-2

=-13×12=2vxf-2

vxfinal=0

Hence, the correct answer is 0.

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