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Figure shows an approximate representation of force versus time during the collision of a 58-g tennis ball with a wall. The initial velocity of the ball is 32 m/s perpendicular to the wall; it rebounds with the same speed, also perpendicular to the wall. What is the value of F max ’ the maximum contact force during the collision?

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5 years ago Navjyot Kalra
654 Points
```							As the tennis ball strikes with the wall having velocity v and then rebounds with same velocity, thus the change in momentum Δp of the tennis ball will be equal to,Δp = (mv) – (-mv) = 2mvTo obtain the change in momentum Δp of the ball, substitute 58-g for m and 32 m/s for v in the equation Δp = 2mv, we get,Δp = 2mv = 2 (58-g) (32 m/s) = 2 (58-g×10-3 kg/1 g) (32 m/s) = 3.7 kg.m/sTo obtain the impulse we have to find out the area under force-time graph for the trapezoid.So, J = Fmax (2 ms+6 ms)/2 = Fmax (4 ms)So, Fmax = J/4 ms = Δp/4 msTo obtain the maximum contact force Fmax during the collision, substitute 3.7 kg.m/s for Δp in the equation Fmax= Δp/4 ms,Fmax= Δp/4 ms = (3.7 kg.m/s) / (4 ms) = (3.7 kg.m/s) / (4 ms×10-3 s/1 ms) = 930 kg. m/s2 = (930 kg. m/s2) (1 N/1 kg. m/s2) = 930 NFrom the above observation we conclude that, the maximum contact force Fmax during the collision would be 930 N.
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5 years ago
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