A 2.0 kg ball and a 4.5 kg ball, each moving at 0.90 m/s, undergo a head-on coll

Question

A 2.0 kg ball and a 4.5 kg ball, each moving at 0.90 m/s, undergo a head-on collision. The lighter ball rebounds opposite its initial direction, with speed 0.90 m/s. Part A Find the post-collision velocity of the heavier ball. Assume the initial direction of the lighter ball as positive. Express your answer to two significant figures and include the appropriate units. Part B How much mechanical energy was lost in this collision? Express your answer in { m J}. Express your answer using two significant figures. Part C How much mechanical energy was lost in this collision? Express your answer as a fraction of the system's initial mechanical energy. Express your answer using two significant figures.

Explanation / Answer

(a)

We shall sove this using conservation of momentum

Initial Momentum = 2(0.9) - 4(0.9) = -1.8

(velocity of lighter ball is +ve)

Final Momentum = -2(0.9) + 4v

v is velocity of heavier ball

When we do

Initial Momentum = Final Momentum

we get v = 0

The post-collision velocity of the heavier ball = 0

(b)

Initial Kinetic Energy = 1/2 x 2 x0.92 +  1/2 x 4 x0.92 = 2.43 J

Final Kinetic Energy = 1/2 x 2 x0.92 = 0.81 J

Therefore, loss in  Kinetic Energy = 2.43 - 0.81 = 1.62 J

Therefore, the answer is 1.62 J

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