2. A 0.30 kg ball is dropped onto a concrete driveway. The ball's velocity before impact is 4.5 m/s and after impact is 4.2 m/s. What is the change in the ball's momentum?

Question

Physics

Shyla Lewis

6 November, 06:09

2. A 0.30 kg ball is dropped onto a concrete driveway. The ball's velocity before impact is 4.5 m/s and after impact is 4.2 m/s. What is the change in the ball's momentum?

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Answers (1)

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Stuart · Answer 1

-0.09 kgm/s

Explanation:

Change in momentum: This can be defined as the product of mass a body and it's change in momentum. The S. I unit of change in momentum is kgm/s.

The formula of change in momentum is

ΔM = m (v-u) ... Equation 1

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

Given: m = 0.30 kg, u = 4.5 m/s, v = 4.2 m/s

Substitute into equation 1

ΔM = 0.3 (4.2-4.5)

ΔM = 0.3 (-0.3)

ΔM = - 0.09 kgm/s

Hence the change in momentum of the body = - 0.09 kgm/s