Ask Question
7 September, 22:09

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?

+2
Answers (1)
  1. 7 September, 22:35
    0
    -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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers