Question

A 1.5kg object moving with a speed of 2.5m/s strikes a wall and the ball rebounds with a speed of 1.5m/s. The ball is in contact with the wall for 0.045s. What is the magnitude of the average force exerted on the ball by the wall?

Answer 1

F = 133.33[N]

Step-by-step explanation:

This problem can be solved by the principle of momentum conservation, which tells us that momentum is preserved before and after the bounce of the ball on the wall.

In such a way that the movement towards the wall we will take it with a positive sign, and the force of the rebound to the left as negative. The movement to the left will be taken as a negative sign.

`m_(1)*v_(1)-F*t=-m_(1)*v_(2)`

where:

m₁ = mass of the object = 1.5 [kg]

v₁ = velocity of the ball before hitting the wall = 2.5 [m/s]

F = average force [N]

t = time contact = 0.045 [s]

v₂ = velocity of the ball after hitting the wall = 1.5 [m/s]

Now replacing:

`(1.5*2.5)-F*0.045=-(1.5*1.5)\\3.75+2.25=F*0.045\\F=6/0.045\\F=133.33[N]`