A ball weighing 10gm hits a hard surface vertically with a speed of 5ms−1 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 ball is :
100N
10N
1N
0.1N
A
100N
B
1N
C
0.1N
D
10N
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Solution
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Given,
v=5m/s
m=10g=0.01kg
t=0.01s
Newton 2nd law,
F=ΔPt
F=mv−(−mv)t=2mvt
F=2×0.01×50.01=10N
F=10N
The correct option is B.
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