A ball with a mass of 0.1 kg moves at a speed of 5 m / s. After hitting the wall, it began to move in the opposite

A ball with a mass of 0.1 kg moves at a speed of 5 m / s. After hitting the wall, it began to move in the opposite direction at a speed of 4 m / s, which is equal to the measurement of the ball’s impulse as a result of hitting the wall.

m = 0.1 kg.

V1 = 5 m / s.

V2 = 4 m / s.

ΔP -?

The momentum of the body P is called a vector physical quantity equal to the product of the mass of the body m by the speed of its movement V: P = m * V. The momentum of the body P is directed just like the speed of movement of the body V.

The change in impulse ΔP is expressed by the formula: ΔP = P2 – P1 = m * V2 – m * V1 = m * (V2 – V1) – vector.

Since the speed of the ball before impact is directed in one direction, and after impact in the opposite direction, the vector difference (V2 – V1) = (V2 – (- V1)) = V2 + V1.

ΔP = m * (V2 + V1).

ΔP = 0.1 kg * (4 m / s + 5 m / s) = 0.9 kg * m / s.

Answer: the change in momentum upon impact of the ball is ΔP = 0.9 kg * m / s.



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