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An iron sphere of mass 10 kg is dropped from a height of 80 cm , if 'g' = 10cm/s² . Calculate the momentum transferred to the ground by the body . I

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M = 10 Kg
Velocity just before striking Earth = V
V² = u² + 2 g h
V² = 0² + 2 * 9.8m/sec² * 0.80 m
V = 3.96 m/sec

Momentum before striking earth = 10 kg * 3.96 m/sec = 39.6 kg m /sec
After collision, both the iron sphere and the earth have the same momentum. The total momentum is = 39.6 m/sec.

 momentum conservation:
    39.6  =  Momentum of ground + momentum of iron sphere
             = (mass of ground + mass of iron sphere ) * velocity of both
             = mass of ground * velocity of ground    as sphere is negligible as
                                                                 compared to mass of ground.
 Momentum transfered to ground = 39.6 kg m /sec

0 0 0
Velocity= 2gh^1/2= (2x10x80)^1/2=40cm/sec
hence  momenum transferred= massx velocity
0 0 0
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