# A metal ball weighing 20 g, falling at a speed of 5 m / s, strikes elastically on a steel plate and rebounds

**A metal ball weighing 20 g, falling at a speed of 5 m / s, strikes elastically on a steel plate and rebounds from it in the opposite direction with the same speed in absolute value. Find the change in the momentum of the ball and the average force that caused this change if the impact lasted 0.1 s**

m = 0.8 kg.

V0 = 5 m / s. m = 20 g = 0.02 kg.

V1 = 5 m / s.

V2 = 5 m / s.

t = 0.1 s.

Δp -?

Fср -?

The momentum of the ball p is the product of the mass of the ball m by its velocity V: p = m * V.

Let us express the change in the momentum of the metal ball Δp by the formula: Δp = p2 – p1 = m * V2 – m * V1.

Let’s draw a vertical coordinate axis directed vertically upward. For projections onto this axis, the change in momentum will take the form: Δp = m * V2 – (- m * V1) = m * (V2 + V1).

Δp = 0.02 kg * (5 m / s + 5 m / s) = 0.2 kg * m / s.

We express the average impact force Fav according to 2 Newton’s law: Fav = m * a = m * Δp / t.

Fav = 0.02 kg * 0.2 kg * m / s / 0.1 s = 0.04 N.

Answer: the change in the momentum of the ball is Δp = 0.2 kg * m / s, the average impact force is Fav = 0.04 N.