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A golf ball strikes a hard, smooth floor at an angle of 30.0 and, as the drawing shows, rebounds at thesame angle. The mass of the ball is 0.047 kg, and its speed is 45 m/s just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball’s momentum changes during impact with the floor, and ignore the weight of the ball.)

3.7 $\mathrm{kg}. \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 7

Impulse and Momentum

Moment, Impulse, and Collisions

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Rutgers, The State University of New Jersey

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everybody Well, here in problem 12. We have our 1st 2 dimensional momentum problem, but they have made it a little easier. Easier for us and just worrying only about the why component of this momentum. So we're gonna break this down and we'll get to it. Main thing we're looking for here is what is the magnitude of impulse applied to the golf ball by the floor? All right, so that's what we're looking for. Let's start with what we know. Massive. This golf ball is 0.47 kilograms. We know that the total velocity of the golf ball coming in and going out is 45 meters per second. So let's talk about the eye coming in. We'll make that one. Or negative, since it is going down is a negative 45 meters per second. And we'll call the final our outgoing velocity, our positive 45 meters per second. Okay. And when it strikes the ground No, we don't need to worry about time because we're just finding impulse. Never mind. Okay, so I said here, we need to worry about the why component of this. So we need to go in and find the y component. Now, if you look at diagram drawn. Oh, that wasn't it. I'm trying my best here toe. Recreate what they have in the book. We have this kind of V shaped pretend this is symmetrical. I know it doesn't look like it, but the book has a much better picture or both of these air 30 degree angles. Okay? And we have our X and Y access here, X axis. Why access? So what we need to do is we're going to use our trig functions, and we're going to find the why component of our, um, ball here. Try switching from colors. We're gonna find the why component. So we're looking right along this axis here. Okay, So do that. We will use our trig functions knowing that these are 30 degree angles. Okay, To find the why components of the why is going to equal the hypothesis, which is our 45 meters per second and again remember, incoming is negative. Outgoing is positive here, and we're going to take that times the co sign of 30 degrees. Since the component that we're looking for is adjacent to the angle there, So co sign of 30 degrees and that gets us a why component of our velocity of 39 meters per second. So this is actually the velocity that we're going to use Okay, Getting into solving the problem. Now we are looking for impulse So Capital J, which is our mo mentum final minus our mo mentum Initial remembering momentum is just mass times velocity so J equals sometimes be final minus m times v initial. In this case, the mass stays the same so we can factor out the mass. So am I times the final minus the initial and we're ready to substitute our numbers in. So our impulse here equals the mass which waas zero point zero for seven kilograms And then now remember, with these velocities these are the why components okay were only worried about the white complaint here we found that earlier are white components were 39 meters per second. We said that final was positive and our initial was negative. So 39 meters per second for the final in the wind direction minus and negative 39 meters per second which was our initial and the impulse that must be exerted on the golf ball by the floor comes out to be 3.7 kilogram years per second. Okay, there we go already.

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