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Standard IX

Physics

Conservation of Momentum

A ball weighi...

Question

A ball weighing $10 g m$ hits a hard vertical surface with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for (0.01) sec. The average force exerted by the surface on the ball is:

100 N

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10 N

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1 N

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0.1 N

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Solution

The correct option is A $10 N$
Velocity of ball initially $u = 5^i m / s$
Velocity of ball finally $v = - 5^i m / s$
Mass of ball $m = 10 g = 0.01 k g$
Change in momentum of the ball $Δ p = m (v - u) = (0.01) (- 5 - 5)^i = - 0.1^i k g m / s$
Magnitude of change in momentum $| Δ p | = 0.1 k g m / s$
Force $F = \frac{Δ p}{Δ t} = \frac{0.1}{0.01} = 10 N$

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A ball weighing 10gm hits a hard vertical surface with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for (0.01) sec. The average force exerted by the surface on the ball is:

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A ball weighing $10 g m$ hits a hard vertical surface with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for (0.01) sec. The average force exerted by the surface on the ball is: