A falling ball weighing 0.8 kg bounces off the floor. The pre-bounce speed is 10.2 m / s and the post-bounce

| June 10, 2021 | education

A falling ball weighing 0.8 kg bounces off the floor. The pre-bounce speed is 10.2 m / s and the post-bounce speed is 6.4 m / s. Find the force that acts on the ball during the moment of rebound 0.12s?

m = 0.8 kg.

V1 = 10.2 m / s.

V2 = 6.4 m / s.

t = 0.12 s.

F -?

According to 2 Newton’s law, the force F, which acts on a body, is equal to the product of the body’s mass m by its acceleration a: F = m * a.

The acceleration of the body a is determined by the formula: a = (V2 – V1) / t and shows how quickly the speed of the body changes over time t.

F = m * (V2 – V1) / t.

Let’s write 2 Newton’s law for projections on the vertical axis of the OU, directed vertically upwards.

F = m * (V2 – (- V1)) / t = m * (V2 + V1) / t.

F = 0.8 kg * (6.4 m / s + 10.2 m / s) / 0.12 s = 110.7 H = 0.11 * 10 ^ 3 H = 0.11 kN.

Answer: at the moment of impact, a force F = 0.11 kN acts on the ball, directed vertically upward.



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