The goalkeeper kicks a soccer ball weighing 400 grams, falling at a speed of 15 m / s

The goalkeeper kicks a soccer ball weighing 400 grams, falling at a speed of 15 m / s, after which the ball begins to move at a speed of 20 m / s. The goalkeeper’s contact with the ball lasts 0.02s. Determine the force acting on the ball.

m = 400 g = 0.4 kg.

V1 = 15 m / s.

V2 = 20 m / s.

t = 0.02 s.

F -?

According to 2 Newton’s law, the impact force F is equal to the product of the ball’s mass m and its acceleration a: F = m * a.

Let us write the definition for the acceleration of the ball a in vector form: a = (V2 – V1) / t.

Since the direction of the ball before and after the impact has the opposite direction, the acceleration will be determined by the formula: a = (V2 + V1) / t.

The force of impact on the ball will be determined by the formula: F = m * (V2 + V1) / t.

F = 0.4 kg * (20 m / s + 15 m / s) / 0.02 s = 700 N.

Answer: the force of impact on the ball is F = 700 N.