Question

Puck 1 of mass m 1 = 0.20 kg is sent sliding across a frictionless lab bench, to undergo a one-dimensional
elastic collision with stationary puck 2. Puck 2 then slides off the bench and lands a distance d from the base
of the bench. Puck 1 rebounds from the collision and slides off the opposite edge of the bench, landing a
distance 2d from the base of the bench. What is the mass of puck 2?

Answer 1

Since both pucks are at same height ,they both falls for same time t

H =1/2gt²

Horizontal distance covered by Puck 2 in time t = d

so v₂ =d/t -----------(1)

Horizontal distance covered by Puck 1 in time t = 2d

so v₁ =-2d/t -----------------(2)

From (1) and (2)

v₁ =-2v₂ ---------------(3)

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In elastic colliision

Velocity of Approach Equals Velocity of Recession

From (3)

--------(4)

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Take Conservation of energy so we dont need to worry about signs

In elastic Collision Kinetic energy is conserved

Kinetic energy before collision =Kinetic energy after collision

Given u₂ =0

Cancel v₂²

ANSWER_:

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