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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?

Respuesta :

onyebuchinnaji

Answer:

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

asuwafohjames

The magnitude of the change in the momentum of the baseball is 4 kgm/s.

To solve the problem above we need to use the formula of change in momentum

Change in momentum: This can be defined as the product of the mass and the change in velocity of a body.

The formula of change in momentum is

M = m(v-u).................. Equation 1

Where M = change in momentum, m = mass of the baseball, v = final velocity of the baseball, u = initial velocity of the baseball.

From the question

Given:

  • m = 0.2 kg
  • v = 60 m/s
  • u = 40 m/s

Substitute these values into equation 1

  •          M = 0.2(60-40)
  •          M = 0.2(20)
  •          M = 4 kgm/s

Hence, The magnitude of the change in the momentum of the baseball is 4 kgm/s

Learn more about change in momentum here: https://brainly.com/question/11789105

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