Jennifer hits a stationary 200 gram ball and it leaves the racket at 40 m/s. In
ID: 1977355 • Letter: J
Question
Jennifer hits a stationary 200 gram ball and it leaves the racket at 40 m/s. In time lapse, photography shows that the ball was in contact with the racket for 40 ms:
(a) What average force was exerted on the racket?
(b) What is the ratio of this force to the weight of the ball?
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I need help with this.
Please, don't answer it unless you're going to explain every single step and equation you use, as clearly as possible.
I would, for sure, give full Karama Points to a clearly-explained, and fully-described answer.
Thanks.
Explanation / Answer
The mass of the ball, m = 200 g = (200 g)(10-3 kg/1 g) = 0.200 kg The speed of the ball, v = 40 m/s The contact time, t = 40 ms = (40 ms)(10-3 s/1 ms) = 0.04 s -------------------------------------------------------------------------------------- a) From the definition of the impulse, we have Impulse (J) = The change in momentum Ft = P Ft = mv Therefore, the average force was exerted on the racket is Favg = mv/t = (0.200 kg)(40 m/s)/(0.04 s) = 200 N ------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------- b) The weight of the ball, W = mg = (0.200 kg)(9.8 m/s2) = 1.96 N The ratio between the force and weight of the ball is Favg/W = 200 N/1.96 N = 102.04 = (0.200 kg)(9.8 m/s2) = 1.96 N The ratio between the force and weight of the ball is Favg/W = 200 N/1.96 N = 102.04Related Questions
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