A 0.0450-kg golf ball initially at rest is given a speed of 25.0 m/s when a club strikes it. If the club and ball are in contact for 2.00 ms, what average force acts on the ball? Is the effect of the ball's weight during the time of contact significant? Why or why not?

Answer

$\left(F_{\mathrm{av}}\right)_{x}\left(t_{2}-t_{1}\right)=m v_{2 x}-m v_{1 x}$ gives $\left(F_{\mathrm{av}}\right)_{x}=\frac{m v_{2 x}-m v_{1 x}}{t_{2}-t_{1}}=\frac{(0.0450 \mathrm{kg})(25.0 \mathrm{m} / \mathrm{s})}{2.00 \times 10^{-3} \mathrm{s}}=562 \mathrm{N}$ $w=(0.0450 \mathrm{kg})\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right)=0.441 \mathrm{N} .$ The force exerted by the club is much greater than the weight of
the ball, so the effect of the weight of the ball during the time of contact is not significant.
Forces exerted during collisions typically are very large but act for a short time.

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Video Transcript

So this is 8.7 have a 45 a gram golf ball. It's starting from rest, and then it's undergoing a two milliseconds impact from a golf club. This then causes it, uh, near the ground at 25 meters per second. And what we need to find out is what is the average force that the golf club is exerting on the golf ball during the impact with it. Now we know that the impulse is equal to both the change in momentum of an object and the average force times the length of time that the force was being exerted. If we're allowed to use the average force, we happen to know what the dependence on time of the forces or whatever. Uh, this is an integral, but here is just Yeah, this is the expression we need to use for this. So it's straightforward enough to rearrange this, and this shouldn't be surprising because this is the expression. Oh, if you take the limit of this, his expression is Delta T goes to zero, then this is just Newton's second law stated in the general momento form or ethical. It's the amateurs. Now the ball starts with zero momentum because it's addressed, so change of momentum is just equal to its final momentum. This is 25 meters per second times the 45 grams mass that it has divided by our two milliseconds to say, and somehow this force works out to these 500 and 62. So that's all well and good now. The other thing the question asks us to consider is whether the weight of the ball is significant. Uh, and this problem we found the force that the golf club is exerted on here. But this isn't going to work if the weight is actually significant, because that would be contributing to the forces acting on the ball. Uh, you'll be in trouble, but it's easy enough to check what the weight of the ball is on the surface of the Earth. It's just equal to its mass times, the standard acceleration of gravity. So this is our 45 grams ties. Let's talk 10 because because no meters per second squared and this is less than half of you, considering that this is very small compared. So the answer we got assuming that the way it didn't matter You know, assuming making that assumption works because, yeah, this is gigantic impediments. So here's how you can figure out what the average horse on the golf ball was. If you happen to know how long the club is in contact with it, and then you don't need to worry about of the golf ball for this.

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