physics

A 60 g golf ball leaves the face of a golf club with a velocity of 75 m/s. If the club exerted an average force of 3.0 x 10^4 N, what was the time of impact between the club and the ball?

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  1. f=m*a

    30000N = 60g * a

    a = (change in velocity) / (change in time)

    30000N = 60g * (Vf - Vi) / (change in time)

    500 = (75 - 0) / (change in time)

    change in time = .15 sec

    the change in time is the time that the ball was in contact with the club

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  2. the mass you use is not 60g, it is .06kg. Sorry. Answer is .00015 sec.

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