# 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 need the answer with steps .....plsss.....fast...

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by g2000

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by g2000

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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

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

hence momenum transferred= massx velocity

p=10kgx40cm/sec=400kg-cm/s