Question

A 1.5 kg ball strikes a wall with a velocity of 8.9 m/s to the left. The ball bounces off with a velocity of 7.7 m/s to the right. If the ball is in contact with the wall for 0.19 s, what is the constant force exerted on the ball by the wall

Answer 1

Answer:

78.6 N

Explanation:

Considering the change in kinetic energy is equal to force*time then knowing that kinetic energy change is given by

$\triangle KE= 0.5m(v_f^{2}-v_i^{2})=ft$ where m is the mass, v is velocity, subscripts f and i for final and initial respectively, f for force and t is time

$\triangle KE=0.5*1.5(7.7^{2}-8.9^{2})=0.19f\\F\approx 78.6N$