Question

A 0.058-kg golf ball is struck by a club that produces a force of 272 N and gives the ball a velocity of

62 m/s. For how long was the club in contact with the ball?
Variables:
Equations:
Answers

Answer 1

The club was in contact with the ball for 0.013 seconds.

Why?

We can calculate for how long was the club in contact with the ball using the following equation and isolating the time from it:

`Force=mass*aacceleration=mass*(v-v_(o))/(time) \\\\time=(mass*(v-v_(o)))/(Force)`

From the statement we know:

- Initial speed equal to zero.

- Force equal to 272N.

- Final speed (for the moment) equal to 62 m/s.

- Mass equal to 0.058 Kg

So, substituting and calculating, we have:

`time=(mass*(v-v_(o)))/(Force)`

`time=(mass*(v-v_(o)))/(Force)\\\\time=(0.058kg*(62(m)/(s) -0))/(272(Kg.m)/(s^(2)))=0.013s`

Hence, we have:

Variables: Time

Equations:

`Force=mass*aacceleration=mass*(v-v_(o))/(time) \\\\time=(mass*(v-v_(o)))/(Force)`

Answers: The club was in contact with the ball for 0.013 seconds.

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