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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 Fmax’ the maximum contact force during the collision?

Navjyot Kalra
6 years ago
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)
= 2mv
To 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/s
To 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 ms
To 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 N
From the above observation we conclude that, the maximum contact force Fmax during the collision would be 930 N.