A ball weighing 10g hits a hard surface vertically with a speed of 5 ms -1 and rebounds with the same speed. The ball remains in contact with the surface for 0.01 s. The average force exerted by the surface on ball is.
Text Solution
Verified by ExpertsThe correct answer is:
B
Impulse = Ft = change in momentum = mv – (-mv)
= 2mv = 2 × 0.01 × 5 = 0.1
∴ ∴ F = 
= 10 N
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