A 0.2 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30m/s and rebounds up at 20m/s. The impulse on the ball during the collision is:
Correct answer: A. 10 Ns upward
- A. 10 Ns upward
- B. 2.0 Ns upward
- C. 10 Ns downward
- D. 2.0 Ns downward
Explanation
Pi = mvi = (0.2)( -30) = - 6 Ns Pr= mvr= (0.2)(20) = + 4 Ns △P = mvr- mvi = + 4 - (- 6) = + 10 Ns Taking downward velocity is -ve and upward is +ve.
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