Question

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 collision with the floor, and the ball rebounds to a height of 1.45 m. What impulse was applied to the ball during the collision?

Answer 1

J=0.211kg.m/s

Step-by-step 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=√(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=√(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`