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 b
all 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.
