A soccer ball weighing 400 g was reported to have a speed of 25 m / s on penalty kicks.

A soccer ball weighing 400 g was reported to have a speed of 25 m / s on penalty kicks. If the ball hits the goalkeeper’s chest and bounces back at the same speed, the kick lasts 0.025 s. If the goalkeeper takes a hit on his hands, then after 0.04 s, he dampens the speed of the ball to zero. Find the average impact force in each case.

m = 400 g = 0.4 kg.

V = 25 m / s.

V1 = 25 m / s.

V2 = 0 m / s.

t1 = 0.025 s.

t2 = 0.04 s.

F1 -?

F2 -?

When the ball hits the goalkeeper’s chest, we write 2 Newton’s law: m * a = F.

m * (V – V1) / t1 = F1.

Since after hitting the goalkeeper’s chest the ball bounces in the opposite direction with the same speed V = V1, then F1 = m * (V – (- V1)) / t1 = m * (V – (- V1)) / t1 = 2 * m * V / t1.

F1 = 2 * 0.4 kg * 25 m / s / 0.025 s = 800 N.

When the ball hits the goalkeeper’s hands, we write 2 Newton’s law: m * a = F.

m * (V – V2) / t2 = F2.

F2 = m * V / t2.

F2 = 0.4 kg * 25 m / s / 0.04 s = 250 N.

Answer: when bouncing from the chest, the average force is F1 = 800 N, when hitting the hands F2 = 250 N.