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Jobisdone [24]
3 years ago
7

A rubber ball of mass 18.5 g is dropped from a height of 1.90 m onto a floor. The velocity of the ball is reversed by the collis

ion with the floor, and the ball rebounds to a height of 1.45 m. What impulse was applied to the ball during the collision?
Physics
1 answer:
kati45 [8]3 years ago
3 0

Answer:

J=0.211kg.m/s

Explanation:

The impulse-momentum theorem states:

J=\Delta p\\J=m(v_a-v-b)\\where:\\m=mass\\v_a=velocity\_after\\v_b=velocity\_before

The velocity before the impact is given by:

(v_b)^2=2.a.\Delta y\\v_b=\sqrt{2*(9.8m/s^2)1.90m}=6.10m/s(-\hat{j})

For the velocity after the impact:

(v_f)^2=(v_a)^2+2.a.\Delta y\\(0)^2=(v_a)^2+2.(-9.8m/s^s).(1.45m)\\\\v_a=\sqrt{2*9.8m/s^2*1.45m}\\v_a=5.33m/s(\hat{j})

so:

J=18.5*10^{-3}kg(5.33m/s(\hat{j})-6.10m/s(-\hat{j}))\\\\J=18.5*10^{-3}kg(5.33m/s(\hat{j})+6.10m/s(\hat{j}))\\\\J=0.211kg.m/s

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