A 0.0450-kg golf ball initially at rest is given a speed of 25.2 m/s when a club strikes. part a part complete if the club and ball are in contact for 1.95 ms, what average force acts on the ball? f = 582 n submitprevious answers correct significant figures feedback: your answer 581 n was either rounded differently or used a different number of significant figures than required for this part. part b is the effect of the ball's weight during the time of contact significant?

We are given information:

m = 0.0450 kg

Δv = 25.2 m/s

Δt = 1.95 ms = 0.00195s

To find force we use formula:

F = m * a

a is acceleration. To find it we use formula:

a = Δv / Δt

a = 25.2 / 0.00195

a = 12923.1 m/s^2

Now we can find force:

F = 0.0450 * 12923.1

F = 581.5 N

To check the effect of the ball's weight on this movement we need to calculate it and then compare it to this force.

W = m * g

W = 0.0450 * 9.81

W = 0.44145 N

We can see that weight is much smaller than the applied force so it's influence in negligible.