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

78.6 N

Step-by-step 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`