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Consult Concept Simulation 6.1 in preparation for this problem: A golf club strikes a 0.047-kg golf ball in order to launch it from the tee. For simplicity, assume ...

Question

Consult Concept Simulation 6.1 in preparation for this problem: A golf club strikes a 0.047-kg golf ball in order to launch it from the tee. For simplicity, assume that the average net force applied to the ball acts parallel to the ball's motion, has a magnitude of 7780 N, and is in contact with the ball for a distance of 0.012 m. With what speed does the ball leave the club?NumberUnits

Consult Concept Simulation 6.1 in preparation for this problem: A golf club strikes a 0.047-kg golf ball in order to launch it from the tee. For simplicity, assume that the average net force applied to the ball acts parallel to the ball's motion, has a magnitude of 7780 N, and is in contact with the ball for a distance of 0.012 m. With what speed does the ball leave the club? Number Units



Answers

A golf club strikes a 0.045-kg golf ball in order to launch it from the tee. For simplicity, assume that the average net force applied to the ball acts parallel to the ball’s motion, has a magnitude of 6800 N, and is in contact with the ball for a distance of 0.010 m. With what speed does the ball leave the club?

In this problem we have given a golf club strikes a 0.45 kg golf ball 0.45 kg golf ball moss and it is ready to launch from that T with a magnitude for 4, 600 Newton 6800 Newton. Sorry. And in contact with the ball for a distance. 0.10 m. Hate So with water speed, does the ball leave the club? So here we know that work done by the fourth F work them by the force F is equal to F in two D. This will be equal to 6800 into 0.10 This will be equal to 68 Jule late. Now we know that this wagon this work, this amount of work is done on the wall wagon is equal to change in kinetic energy. So initial energy of all zero final kinetic energy is supposed it has a speed V so and finally, kinetic energy will be equal to half the square. So where then? 60 year Jules is equal to Delta K. That means K Final, which is half m v squared. This is 68 so we square will be equal to 16, divided by the mass of the Gulf War, which is 0.5 kg. No. From here we can see the speed of the ball before leave. The club is to into 68 0.45 Now we will solve this by calculator. This is equal to yeah and the road to and to 68 divided by 0.45 This is 54.97 or we can say students 55 m per second. This is the speed the wall does leave the club.

And this question, the the question covers the work energy theorem. So from the work energy theorem, the network gun is recovering to the changing kind of technology and the network done is by the club on the golf ball. So the welcome by the club is a capability, the final kinetic Energy Minister Michelle kind of technology. And since the goal ball was initially address, we can eliminate the initial cannot technology because it is zero. So the work done by the club, is it colour into the current energy? Okay, so for part a the work done by the club is equivalent to half of the mass of the ball into the final speed of the ball square are welcomed by the club is equivalent to half of the mass of the bodies. 0.045 Kg. Do they ascribe the final speed? That is 41 m/s, script after welcome by the clever ones, 37.82 Through five jewels are in two significant figures to world them by the club. On the ball is 38 jewels. No party, uh, worked on by the club on the ball is equivalent to a rich foods applied by the club on the ball. Into the displacement X. Okay. And from this, the rich forces secure into the work done by the displacement tax. Now, if we substitute the value which falls supplied by the club Is equivalent to 38 jewels upon the displacement and that is 0.01 m or the value of the average forces if you're going to 38, 20 students. So this is the violence for five people.

We're gonna be looking at problems six from Chapter six of the Physics Fifth, the question says, the golfer hits a golf ball in parting with the initial velocity of 52.2 m per second, directly 30° above the horizontal, Assuming that the mass of the ball is 46 g and the club on the border and conduct one point too many seconds. And for part of the impulse imparted on the ball, the part B, the impulse imparted to the club and the party of the average force exerted on the board by the club. Well, this is very simple. The impulse J is equal to the change in momentum. So our momentum of the ball in this case is going to be 46 or not point not for six, um kilograms times by 52.2. And that comes out to 22.4 kg meters per second. So I changed the momentum of a bullet initially isn't moving, So I'll change momentum is therefore 2.4 kg m/s. That's its new momentum. Um, for part B, the impulse that the bull in parts on the club is the same as the impulse that the club in parts of the ball. So, the impulse for the club is also 2.4 kilogram meters per second, Part C. Um uh, the impulse is equal to the force times by the time the force is equal to the change of momentum over time. Therefore putting in two point 4/1 2.0.20 times 10 to the minus three because 1.2 milliseconds, we get 2000 newtons, and this should be sorry to be.

We have great w equals to 1.62 ounce. And uh delta we is equals two B minus we note, which is equal to 1 50 ft per second. And the time the french delta T. Is equal to 0.1 2nd. Okay, so from the equation which is integration of FTT equals two M delta. We and we can right here that F will be equals to uh the force art which is given in the equations are manipulated by delta T. This will be equals two M delta week. So from here we get our equals two M delta. We divided by delta T. So substituting values we get em is 1.62 ounce, which is uh divided by 16 from multiplied by delta V. Is 1 50 delta T is 0.1 seconds. So from here we get our equals to 4 £72. Okay, so this is the answer for the at risk force. So now from the Newton's second law, the force are can be written as M. A. So from here we get eight was to our buy em. And we have articles 2472 divided by M. Will be equal to 1.62 divided by 16. Multiplied by divided by G. So it will reach in the above. So it will be equals two G. Okay, so from here after solving we get acceleration equals two Acceleration equals 215 0000 ft per second is clear. Okay? And this is also equals two 4660 G. Okay so this is the acceleration not solving for the distance D. So we can use dynamic equation so we can right away to square records to be even square be to spare minus we're going to spear equals two A. D. Okay, so from here we get D. Equals two weeks. Two x squared minus leven is square. They were by 28 So we two is equal to uh we square which is 1 50 square and we even is equal to zero. So this is zero squared divided by two manipulated by acceleration. Is this value 15 double zero, double zero? So from here He comes out to be 0.075 ft, which is also equals to 0.9 double zero inches. So these are the different answers to questions. Okay, so this is the answer for the force are and this is the answer holding acceleration A and this is the answer for the distance.


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