# 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.

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by sukhpreet35

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by sukhpreet35

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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