Ask Question
10 September, 12:58

A steel ball with mass 0.347 kg falls onto a hard floor and bounces. Its speed just before hitting the floor is 23.6 m/s and its speed just after bouncing is 12.7 m/s.

(a) What is the magnitude of the impulse of the ball?

(b) if the ball is in contact with the floor for 0.0139s, what is the magnitude of the average force exerted onthe ball by the floor?

+1
Answers (1)
  1. 10 September, 13:01
    0
    (a) The magnitude of impulse of the ball is 12.5951kgm/s

    (b) F=906.19N

    Explanation:

    m=0.347kg, u=23.6m/s, v=12.7m/s

    we will take the direction of final velocity as negative.

    Impulse (ft) = m (v-u)

    =0.347 (-12.7-23.6)

    Impulse=-12.5951 kgm/s

    = 12.5951kgm/s

    (b) t=0.0139s

    F=m (v-u) / t

    F=12.595/0.0139

    F=906.19N
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A steel ball with mass 0.347 kg falls onto a hard floor and bounces. Its speed just before hitting the floor is 23.6 m/s and its speed just ...” 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