A ball weighing $10 g m$ hits a hard surface vertically with a speed of $5 m / s$ and rebounds with the same speed. The ball remain 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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150 N

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Solution

The correct option is B $10 N$
As after rebounds speed didn't change so change in momentum will be just double of the initial momentum i.e., $Δ p = m Δ v = .01 \times 10 = 0.1 K g m / s$
Average Force will be $Δ p / Δ t = 10 N$
Option B is correct.

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

The correct option is B 10NAs after rebounds speed didn't change so change in momentum will be just double of the initial momentum i.e., Δp=mΔv=.01×10=0.1Kgm/s Average Force will be Δp/Δt=10NOption B is correct.

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

The correct option is B $10 N$
As after rebounds speed didn't change so change in momentum will be just double of the initial momentum i.e., $Δ p = m Δ v = .01 \times 10 = 0.1 K g m / s$
Average Force will be $Δ p / Δ t = 10 N$
Option B is correct.