A 0.2 kg baseball is pitched with a velocity of 40 m/s and is then batted to the pitcher with a velocity of 60 m/s. What is the magnitude of change in the ball's momentum?

The magnitude of change in the ball's momentum is 4 kgm/s

Explanation:

Given;

mass of the ball, m = 0.2 kg

initial velocity of the ball, u = 40 m/s

final velocity experienced by the ball, v = 60 m/s

Therefore, the change in momentum of the ball is given as final momentum minus initial mometum;

ΔP = mv - mu

ΔP = m(v-u)

ΔP = 0.2 (60 - 40)

ΔP = 4 kgm/s

Therefore, the magnitude of change in the ball's momentum is 4 kgm/s

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multiply 2kg by latent heat of vap