Momentum Question 4

A tennis ball of mass 0.058 kg travels horizontally towards a racket at 30 ms^-1. It is hit straight back along the same line at 40 ms^-1. The ball is in contact with the racket for 5.0 ms.(i) Calculate the change in momentum of the ball.(ii) Calculate the average force exerted on the ball by the racket.(i) Take the direction in which the ball leaves the racket as positive. The initial velocity is then -30 ms^-1 and the final velocity is +40 ms^-1.Δp = m(v - u) = 0.058 × (40 - (-30)) = 0.058 × 70 = 4.06 kg ms^-1A common mistake is to use 40 - 30 = 10 ms^-1. Because the ball reverses direction, the two speeds add.(ii) By Newton's second law, the resultant force is the rate of change of momentum. Convert 5.0 ms to 0.0050 s first:F = Δp ÷ Δt = 4.06 ÷ 0.0050 = 812 NBy Newton's third law, the ball exerts an equal and opposite force of 812 N on the racket.The product FΔt is the impulse, and it equals the change in momentum. For the same change in momentum, a longer contact time means a smaller average force. This is why airbags and crumple zones, which make a collision last longer, reduce the force on a passenger.