Question from http://www.sgforums.com/forums/2297/topics/282744
A steel ball of mass 0.020kg is released from a height of 1.2m above a steel plate and rebounds vertically to a height of 1.1m. The ball is in contact with the surface for 0.9ms
1) What is the velocity of the ball as it arrives at the plate?
2) Calculate the momentum of the ball as it arrives at the plate.
3) Calculate the change of momentum of the ball in its collision with the plate.
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Answer:
1) Assume ball is released from rest
Loss in GPE = Gain in KE
mgΔh = ½ m (v² - u²)
(0.020)(9.81)(1.2) = ½ (0.020)(v² - 0)
v² = 23.544
v = 4.85 m/s = 4.9 m/s (2 s.f.)
2) Momentum of ball as it arrives at the plate = m*v
= 0.020 * 4.85
= 0.097 Ns
3) Taking momentum upwards as positive (there's a need to declare the positive direction for vectors)
Let Vi be the initial velocity just before hitting the plate
Let Vf be the velocity immediately after hitting the plate
Vi = -4.85 (because it is downwards, i.e. negative direction)
Using Conservation of Energy,
mgΔh = ½ m (v² - u²), where v = 0 and u = Vf
(0.020)(9.81)(1.1) = ½ (0.020)(0 - Vf²)
Vf² = 2(0.020)(9.81)
Vf = -4.65 (positive because direction is positive, i.e. upwards)
Hence, change in momentum = (0.020) (4.65 - (-4.85)) = 0.19 Ns upwards
