Consider a horizontal surface moving vertically upward with velocity 2 m s - 1 . A small ball of mass 2 kg is moving with velocity 2 i ^ – 2 j ^ m s - 1 If the coefficient of restitution and coefficient of friction are 1 2 and 1 3 respectively, find the horizontal velocity (in m s - 1 ) of the ball after the collision.

Question

Consider a horizontal surface moving vertically upward with velocity 2 m s - 1 . A small ball of mass 2 kg is moving with velocity 2 i ^ – 2 j ^ m s - 1 If the coefficient of restitution and coefficient of friction are 1 2 and 1 3 respectively, find the horizontal velocity (in m s - 1 ) of the ball after the collision.,

MARKS App · Accepted Answer

The relative velocity of the particle with respect to the surface, along the normal, is the approach speed of the particle. v app = 4 m s - 1 The separation velocity will be v sep = e v app = 1 2 × 4 = 2 m s - 1 ∴ v ̄ m , g r = v ̄ m , s + v ̄ s , g r = 4 m/s ∫ N d t = 4 2 + 2 2 = 12 Now, ∫ μ N ⋅ d t = 2 v x f - 2 = - 1 3 × 12 = 2 v x f - 2 v x final = 0

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