Ask Question
31 July, 01:03

A golf ball of mass m = 0.21 kg is dropped from a height h. It interacts with the floor for t = 0.115 s, and applies a force of F = 16.5 N to the floor when it elastically collides with it.

Write an expression for the ball's velocity, v, just after it rebounds from the floor. (Hint: The fact that the collision is elastic is important when solving this problem.) Use only t, F, m

+2
Answers (1)
  1. 31 July, 01:04
    0
    Elastic colision = > conservation of kinetic energy = > constant speed: |V1| = |V2| = V

    Momentum before collision: - mV

    Momentum after collision: mV

    The signs of the two momenta are different because the direction of the ball is reversed during collision.

    Change in momentum: mV - (-mV) = mV + mV = 2mV

    Impulse = F*t = Change in momentum

    F*t = 2mV

    V = F*t / (2m)

    That is the expression requested.

    Now, to find V, you just have to plug in F, t and m values.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A golf ball of mass m = 0.21 kg is dropped from a height h. It interacts with the floor for t = 0.115 s, and applies a force of F = 16.5 N ...” 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