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Force of a Golf Swing. A $0.0450-\mathrm{kg}$ golf ball initially at rest is given a speed of 25.0 $\mathrm{m} / \mathrm{s}$ when a club strikes. If the club and ball are in contact for 2.00 $\mathrm{ms}$ , 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?

$0.441 \mathrm{N}$

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Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Numerade Educator

Hope College

{'transcript': "well, we have the most of the golf ball. Well, most of the golf ball is equal to zero point 045 kilo crown. Its initial velocities. Zero initially, the lost is equal to zero. It's finally lost your starting five meters per second. Finally been most is equal to 25 meters per second and the time interval. Delivering the bull and the club are in contact His Delta T that is equal to find zero milli second. Well, we know that momentum is given by most times. Lost e Let's quality see question number one. And according to Newton's second law, the total net force acting giving a time interval Delta T is given by. So the total force is equal to be to minus B one divided by a Delta T. Let's call this equation number two. Now we do park a Let us to party. Well, first we plug our values for M and are we into equation number one? We I implicate the number one, so we get the initial momentum off the ball. So initial momentum of the world, uh, is equal to zero point and 015 times zero. That is equal to zero ancient Momenta Musical 20 and final momentum Be Do is equal to 0.45 ah, 0.45 times 25 meters per second 25 and that is equal to 1.1251 point 1 to 5. Uh, could look round meter for a second, right? Okay. And further, uh, lettuce, a soft city of these values in I encourage number two. Then our total force is equal to 1.1 to 5. Minuses. It'll divided by 2.0 times 10 to the dollar minus three. So total force is equal to 562 point five Newton right now. Let us to Part B. Well, first, we calculated a gated ball. Will weight of the ball is m times G. Well, jeez, acceleration due to gravity. Well M E 0.45 times nine point it. Therefore, bait is equal to 0.441 on Newton. Well, since the weight off, the bomb is very small compared to the average net. Force acts on ball, differing its contact with the club and can be neglected. So the weight off the ball does not have a significant effect during the con"}