A 2.5 kg ball and a 3.5 kg ball, each moving at 0.90 m/s, undergo a head-on coll

Question

A 2.5 kg ball and a 3.5 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.

Express your answer using two significant figures.

1.) Find the post-collision velocity of the heavier ball. Assume the initial direction of the lighter ball as positive.

2.) How much mechanical energy was lost in this collision? Express your answer in J.

3.) How much mechanical energy was lost in this collision? Express your answer as a fraction of the system's initial mechanical energy.

Explanation / Answer

1)

m1 = mass of lighter ball = 2.5 kg

m2 = mass of heavy ball = 3.5 kg

V1i = initial velocity of lighter ball = 0.90 m/s

V2i = initial velocity of heavy ball = - 0.90 m/s

V1f = final velocity of lighter ball = - 0.90 m/s

V2f = final velocity of heavy ball

using conservation of momentum

m1 V1i + m2 V2i = m1 V1f + m2 V2f

2.5 (0.90) + (3.5) (- 0.90) = 2.5 (- 0.90) + (3.5) V2f

V2f = 0.39 m/s

2)

Lost energy = (0.5) m1 V21i + (0.5) m2 V22i - (0.5) m1 V21f - (0.5) m2 V22f

Lost energy = (0.5) [2.5 (0.90)2 + (3.5) (- 0.90)2 - 2.5 (- 0.90)2 - (3.5) (0.39)2 ]

Lost energy = 1.15 J

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