rce versus time during the collision of a 44- tennis ball with a wall. The initial velocity of the ball is 36 perpendicular to t
he wall; it rebounds with the same speed, also perpendicular to the wall. What is the maximum value of the contact force during the collision
1 answer:
Answer:
Maximum contact force = 487.4 N
Explanation:
The graph showing force vs time is attached to the solution of this question.
According to the an interpretation of Newton's second law of motion, the magmitude of the change in momentum is equal to the magnitude of the impulse experienced duting this impact.
F.t = Δ(momemtum)
Δ(momemtum) = 0.044 [36 - (-36)] = 0.044 × 72 = 3.168 kgm/s
Meaning, Impulse = 3.168 kgm/s too.
The area under the curve above is the impulse we just obtained.
If the maximum contact force is called F
The impulse = Area under this graph
= (1/2)(F)(4-0) + (F)(7-4) + (1/2)(F)(10-7)
= 2F + 3F + 1.5F = (6.5F) mN.s = (0.0065F) N.s
0.0065F = 3.168
F = (3.168/0.0065) = 487.4 N
Hope this Helps!!!
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The given data in the problem is;
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h is the distance of fall = 200 m
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t is the time period = 10 second
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