Question

a 50.0 grams ball is released from rest from initial height of 100 meters, the ball falls vertically downward to the ground and rebound to a lower height of 81% which mean the ball rebound to 81 percent of previous height for the first bound. 81 meters. a)what is the speed of the ball immediately after the first bounce? b) what height does the ball reach after the fifth bounce? c) what is the impulse from the ground on the object after the fifth bounce? (include magitude and direction)

Answer 1

The potential energy of the ball at height of 100 m is mgh = 0.05*9.81*100 = 49.05 J

81% of 100 = 81m

The potential energy at this height is 0.05*9.81*81 = 39.73 J

This is the kinetic energy immediately after the first bounce.

(1/2)mv^2 = 39.73

=> v = 39.86 m/s

b) Height after second rebound will be 81% of 81 = 65.61 m

third: 81% of 65.61 = 53.14 m

fourth: 43.04

5th rebound: 34.86 m (Answer)

c) KE just after the fifth bounce is 0.05*9.81*34.86

(1/2)*0.05v^2 = 0.05*9.81*34.86

v = 26.15 m/s

KE just before the fifth bounce = 0.05*9.81*43.04

(1/2)mu^2 = 0.05*9.81*43.04

=> u = 29.05

Hence impulse = change in momentum = m(v-u) = 0.05(26.15 - 29.05) = -0.145 Kg m/s

Hence the magnitude of impulse is 0.145 Kg m/s

Negative sign indicates the downward direction.