A 0.15 kg ball moving at 20 m s−1 is hit straight back at 30 m s−1. The bat touches the ball for 0.010 s. Find the average force on the ball.
- Take the final direction as positive: u = −20 m s−1, v = +30 m s−1.
- Change in momentum = m(v − u) = 0.15 × (30 − (−20)) = 0.15 × 50 = 7.5 kg m s−1.
- Force = change in momentum ÷ time = 7.5 ÷ 0.010 = 750 N.
Answer: 750 N, in the direction the ball leaves the bat.