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Question

A 0.20kg puck travelling right at 6.0m/s collides head-on with stationary 0.40kg puck. The 0.20kg puck bounces back at 1.5 m/s left. What is the velocity of the 0.40kg puck?

Answer 1

Total momentum before the collision equals total momentum after the collision.

Let's assume positive momentum is to the right, Let the 0.2 kg puck be puck #1 (having m₁ and v₁) and let the 0.4 kg puck be puck #2 having m₂ and v₂. Since v₁ and v₂ are different before and after the collision then we'll let v_1b and v_2b be before the collision and v_1a and v_2a be after the collision.

The momentum before the collision is m₁*v_1b + m₂v_2b = (0.20)(6.0) + (0.40)(0) = 1.2

The momentum after the collision is m₁*v_1a + m₂v_2a = (0.20)(-1.5) + (0.40)v_2a = -0.3 + 0.40v_2a

Setting the equal:

1.2 = -0.3 + 0.40v_2a

1.5 = 0.40v_2a

v_2a = 3.75 m/s to the right = 3.8 m/s to the right (considering 2 significant figures)