A 1.6 kg ball strikes a wall with a velocity of 9.2 m/s to the left.

Question

The ball bounces off with a velocity of 8 m/s to the right. If the ball is in contact with the wall for 0.21 s, what is the constant force exerted on the ball by the wall?

Answer 1

Hi Charlotte A.!

Our solution to this question derives from the impulse-momentum theorem. This is a special case of the impulse-momentum theorem where the impulse of the force acting on the object (ball) equals the change in momentum of that object (ball). As long as we select the time interval to be taken as arbitrarily small, then this equality is true even when the force is not constant but varying.

According to the impulse-momentum theorem:

J = F*t = Δp = mΔv

We want to select the terms that are relevant to our problem, rearrange and then solve:

J = F*t = Δp = mΔv => F = mΔv/t = m(v_f - v_o)/t

<=> [(1.6kg)(8 m*s^-1 - (-9.2 m*s^-1))]/0.21s

<=> [(1.6kg)(8 m*s^-1 + 9.2 m*s^-1)]/0.21s

F = 131.0 N

Hope this helps!

Cheers