From a height of h1 = 2 m, a ball with a mass of m = 200 g freely falls onto the steel plate and jumps to a height

From a height of h1 = 2 m, a ball with a mass of m = 200 g freely falls onto the steel plate and jumps to a height of h2 = 0.5 m. Determine the momentum received by the plate upon impact.

h1 = 2 m.

m = 200 g = 0.2 kg.

h2 = 0.5 m.

g = 10 m / s2.

ΔР -?

The change in the momentum of the plate ΔР is equal to the change in the momentum of the ball when it hits it.

Let us find the change in the momentum of the ball upon impact: ΔP = P1 – P2, where P1 is the momentum of the ball before the impact, P2 is the momentum of the ball after the impact.

Р1 = m * V1, where V1 is the speed of the ball before hitting.

P2 = m * V2, where V2 is the velocity of the ball after impact.

V1 = √ (2 * g * h1).

V2 = √ (2 * g * h2).

Since the speed of the ball is directed downward before the impact, and after the impact is upward, then ΔР = Р1 – Р2 = m * V1 – (- m * V2) = m * V1 + m * V2 = m * (V1 + V2) = m * (√ (2 * g * h1) + √ (2 * g * h2)).

ΔР = 0.2 kg * (√ (2 * 10 m / s2 * 2 m) + √ (2 * 10 m / s2 * 0.5 m) = 0.62 kg * m / s.

Answer: upon impact of the ball, the plate received an impulse ΔР = 0.62 kg * m / s.