ball of mass 200 gram falls from a height of 20m on a floor and rebounds to a height of 5 metre the magnitude of change in momentum of a ball is ( take g as 10 m/s^2
Answers
Given : ball of mass 200 gram falls from a height of 20m on a floor and rebounds to a height of 5 metre
g = 10 m/s^2
To Find : the magnitude of change in momentum of a ball
Solution:
Initial Potential energy = mgh
m = 200g = 0.2 kg
g = 10 m/s²
h = 20 m
0.2 * 10 * 20 = 40 J
Hence KE before hitting = 40J
= (1/2) * 0.2v² = 40
=> v² = 400
=> v = 20 m/s
Momentum = 0.2 * 20 = 4 kgm/s
Final PE = mgh
= 0.2 * 5 * 10
= 10 J
KE after rebound = 10 J
= (1/2) * 0.2 * v² = 10
=> v = 10 m/s
Momentum = 0.2 * 10 = 2 kgm/s
Change in Magnitude of Momentum = 2 kgm/s
as velocities are in opposite direction Hence
Change in momentum = (4 - (-2)) = 6 kgm/s
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