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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About Force and Motion
Displacement, velocity and acceleration describe motion, including the horizontal and vertical components of projectile motion. The chapter connects these quantities with Newton's laws, momentum and impulse, then applies conservation of momentum to collisions, distinguishing elastic collisions from inelastic ones.
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