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25-10-2022
Physics

Answered

A ball of mass 1 kg is dropped from a height of 3 m. Assume no air resistance. After it bounces off the ground it only reaches a height of 2 m. What is the change in momentum of the before and after the bounce?.

Answer :

kumarravichandra kumarravichandra

04-11-2022

energy at first = mg (3) J = 29.4J

Energy after bouncing: mg(2) = 19.6 J

Energy changed by the ball: 29.4J - 19.6J = 9.8J

Heat energy or sound energy are two ways that this energy is lost.

Work performed during the bounce equals 9.8J.

d. Momentum change = mvf-mvi

29.4 = 1/2mvi2

vf =7.668m/s

19.6J =1/2mvf2

vf=6.261m/s

Change of momentum is equal to 1kg x (-1.407m/s)/1kg = -1.407 kg/s (m(vf-vi)).

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