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Question
A ball weighing 0.01kg hits a hard surface vertically with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on the ball is
A ball weighing 0.01kg hits a hard surface vertically with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on the ball is
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Solution
The correct option is B 10N
Since impulse I = F x Δt
And also Impulse I = Δp (i.e. change in linear momentum)
Change in momentum
= 0.01 x 5 - (-5)
= 0.10 N s
Impulse = 0.10 N s
Therefore 0.10 = F x 0.01
Or F = 10 N
10N
Since impulse I = F x Δt
And also Impulse I = Δp (i.e. change in linear momentum)
Change in momentum
= 0.01 x 5 - (-5)
= 0.10 N s
Impulse = 0.10 N s
Therefore 0.10 = F x 0.01
Or F = 10 N
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