A 2.5 kg ball and a 4.5 kg ball, each moving at 0.90 m/s, undergo a head-on coll

Question

A 2.5 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. Part B: How much mechanical energy was lost in this collision? Express your answer in J. 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 to two significant figures and include the appropriate units.

Explanation / Answer

A) conserving momentum 2.5*(-0.9) + 4.5*(0.9) = 2.5*0.9 + 4.5*v so, v = - 0.1 m/s so the heavier ball will move with a velocity of 0.1 m/s in a direction opposite to its initial direction. B) mechanical energy lost = inial energy - final energy = 0.5 (2.5+4.5)*(0.9^2) - 0.5*(2.5)*(0.9^2) - 0.5*(4.5)*(0.1^2) = 1.8 J C) initail energy = 0.5 (2.5+4.5)*(0.9^2) = 2.835 J so fraction of energy lost = 1.8/2.835 = 0.634 or 63.4%

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