A ball of mass1kg is dropped from 20m ht on ground and it rebounds to ht 5 m.find magnitude of change in momentum during its collision with the ground.
Answers
Answered by
2
Velocity of the ball just before hitting the ground : = v
v² = u² + 2 g s
v² = 0 + 2 *10 * 20 = 400
v = 20 meter/sec This is directed downwards. So v = -20 m/sec
Let the velocity of the ball be V after bouncing back.
0 = V² - 2 g S
0 = V² - 2 * 10 * 5
V = 10 m/sec. It is directed upwards. So V = +10 m/s
The change in momentum during the collision = mV - mv = m(V-v)
= 1 kg * (10 - (-20) ) m/sec = 30 kg-m/sec
v² = u² + 2 g s
v² = 0 + 2 *10 * 20 = 400
v = 20 meter/sec This is directed downwards. So v = -20 m/sec
Let the velocity of the ball be V after bouncing back.
0 = V² - 2 g S
0 = V² - 2 * 10 * 5
V = 10 m/sec. It is directed upwards. So V = +10 m/s
The change in momentum during the collision = mV - mv = m(V-v)
= 1 kg * (10 - (-20) ) m/sec = 30 kg-m/sec
Similar questions