A steel ball with a mass of m = 10 g fell from a height of h1 = 1 m onto a steel plate and jumped after the impact

A steel ball with a mass of m = 10 g fell from a height of h1 = 1 m onto a steel plate and jumped after the impact to a height of h2 = 0.8 m. Determine the △ p-Change in the momentum of the ball.

m = 10 g = 0.01 kg.

g = 10 N / kg.

h1 = 1 m.

h2 = 0.8 m.

△ p -?

Let us express the change in momentum vector: △ p = p1 – p2.

p1 = m * V1, p2 = m * V2, where V1 is the speed of the ball before hitting the plate, V2 is the speed of the ball immediately after hitting the plate.

Since V1 is directed vertically downward and V2 is directed vertically upward, then △ p = p1 – p2 = m * V1 – (- m * V2) = m * V1 + m * V2 = m * (V1 + V2).

According to the law of conservation of total mechanical energy: m * g * h1 = m * V1 ^ 2/2, m * g * h2 = m * V2 ^ 2/2.

V1 = √ (2 * g * h1).

V2 = √ (2 * g * h2).

△ p = m * (√ (2 * g * h1) + √ (2 * g * h1)).

△ p = 0.01 kg * (√ (2 * 10 N / kg * 1 m) + √ (2 * 10 N / kg * 0.8 m)) = 0.0847 kg * m / s.

Answer: the change in the momentum of the ball upon impact is △ p = 0.0847 kg * m / s.