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Viktor [21]
3 years ago
8

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?
Physics
1 answer:
sergey [27]3 years ago
3 0

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

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Answer:

<em>The initial speed of the car = 11.22 m/s</em>

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Law of conservation of energy: It states that when two bodied collide in a closed system, the sum of momentum before collision is equal to the sum of momentum after collision.

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m₁u₁ + m₂u₂ = (m₁ + m₂)V..................... Equation 1

Where m₁ = mass of the van, m₂ = mass of the car, u₁ = initial velocity of the van, u₂ = initial velocity of the car, V = Common velocity.

<em>Given: m₁ = 2578 kg, m₂ = 825 kg, u₂ = 0 ( the car was at rest), V= 8.5 m/s</em>

<em>Substituting these values into equation 1, and solving for u₁</em>

<em>2578(u₁) + (825 × 0) = (2578 + 825)8.5</em>

<em>2578u₁ = 28925.5</em>

<em>Dividing both side of the equation by the coefficient of u₁</em>

<em>2578u₁/2578 = 28925.5/2578</em>

<em>u₁ = 11.22 m/s</em>

<em>The initial speed of the car = 11.22 m/s</em>

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