How much momentum will a dumb-bell of mass $10 k g$ transfer to the floor if it falls from a height of $80 c m$ ? Take its downward acceleration to be $10 m s^{- 2}$ .

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Solution

Given : $m = 10 k g$ $u = 80 c m = 0.8 m$
Velocity before it strikes the ground $v = \sqrt{2 g h}$
$⟹ v = \sqrt{2 \times 10 \times 0.8} = 4 m / s$
Momentum transfer $P = m v = 10 \times 4 = 40 k g m / s$

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How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 ms−2.

Given : m=10 kg u=80 cm=0.8 mVelocity before it strikes the ground v=√2gh⟹ v=√2×10×0.8=4 m/sMomentum transfer P=mv=10×4=40 kg m/s

How much momentum will a dumb-bell of mass $10 k g$ transfer to the floor if it falls from a height of $80 c m$ ? Take its downward acceleration to be $10 m s^{- 2}$ .

Given : $m = 10 k g$ $u = 80 c m = 0.8 m$
Velocity before it strikes the ground $v = \sqrt{2 g h}$
$⟹ v = \sqrt{2 \times 10 \times 0.8} = 4 m / s$
Momentum transfer $P = m v = 10 \times 4 = 40 k g m / s$