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Hockey puck $B$ rests on a smooth ice surface and is struck by a second puck $A$ , which has the same mass. Puck $A$ is initially traveling at 15.0 $\mathrm{m} / \mathrm{s}$ and is defiected $25.0^{\circ}$ from its initial direction. Assume that the collision is perfectly elastic. Find the final speed

of each puck and the direction of $B^{\prime}$ s velocity after the collision. [Hint: Use the relationship derived in part (d) of Problem 8.96 .

$6.34 \mathrm{m} / \mathrm{s}$

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Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

{'transcript': "So in this exercise you have to hockey bucks A and B off Rico masses. And in the beginning, the hockey puck A has initial velocity V one off 15 m per second and hockey puck too is at rest right after the collision. Hockey Puck Chew gains a velocity, uh, pointing upwards and making a 25 degree with its initial direction. Access, which is this X axis here, and the hockey book be also gained a below Stevie B. So, in the security size, what we want to find is what is the magnitude off the final velocity off book? A. We also want to find the magnitude off, uh, the velocity of book be and we want to know what angle does the book be makes with book A. So I want to know this that I'm drawing here in green, which is what is five plus 25 degrees. Okay, so the exercise states that the collision between the bucks is perfectly elastic, so we have conservation off linear momentum and kind of energy. So it's starting with the conservation off linear momentum. We have that the components ex and why off the linear movement are conserved. So it's starting with the X component we have that the mass of Pak Wan times initial velocity V one has to be equal to the mass off back to Sorry, the mass of pack Juan times the X component off the final velocity which I'll call V H two x. Sorry if this is confusing plus the mass off rugby times the final velocity off book be in the X direction. Okay, so from here we have that the projection off V off D eight books velocity into the X Direction V A X two Uh, it's just the magnitude of the H two times the consign off 25 degrees and we can also cancel out this term. So we have that, um, the A one So is equal to V H. Two times Cool sign off 25 degrees plus B the big X, which is the be times consign off I okay now going thio the conservation on the Y direction we have that there is no initial velocity in the Y direction on in the X. So we have that zero has to be equal to am V A to why so the y direction off the final of velocity off book. Um, A plus M v b y. So this cancel and we have that v b y has to be equal to, um V A to sign off 25 degrees. Okay, so let's keep this expression for a second. Let I'll call it one. Now we have we can apply the condition off conservation off kinetic energy. So we have that the initial kinetic energy is, um the A one squared over two is equal to also put em in evidence. So over to p A two squared plus V b squared. And from this, we can cancel out am over to, and we have another relation between the three velocities, which is being a one squared is equal to V two squared, plus the big squares. Okay, so way can expand this. So have that expression to Zvi a one squared because you ve a two squared plus so I can express V b squared in terms off the X component off P and the Y component. So this is equal to V b X squared plus V b y squared. And we know that Phoebe, why is just this expression one so we can express it has in terms off v A and the angle that we know. So this is a good choo create You squared sine squared off 25 degrees. Okay, We can also rewrite be vb X from the conservation off the X component to remember that this is just V b X, and we can express it s o isolating this V a one minus v a to good sign. 25 degrees. Okay, so let's put it this here. So I'll write it in red. So we have Sorry. The A one squared plus V H two squared. Who sign squared off 25 degrees plus two times V A. One V two. Because sign off 25 degrees. Okay. This is just Do you want Bliss G two Good. Sign off. 25 degree squared. Okay. Just this thing squared. Sorry is the minus. So we have a minus here minus And also you might assign here. Okay, so So from expanding vb x and V I X. We have that this term. Very one squared from B X squared will cancel out with this. Okay? And we also have that this term here will sum up with this so we can rewrite everything as V A two squared plus v A to so squared sign squared off 25 degrees plus consigned squared off 25 degrees. Um, sorry miners to V A. One V A to consign off 25 degrees. So notice that from the trigonometry populations we have that this is equal to one. So this term sums up with this term, and we have that, uh, and all of this is equal to zero. Sorry, I forgot to write, so we can put, um so we have to v a one V H u go sign of 25 degrees is equal to two V A to squared so we can divide both sides by V two and we also divide by two. So if I found a expression off V A to in terms off things that we know, which is the initial velocity, which we know is 15 m per second and could sign off 25 degrees. And from this, we find that the magnitude of the final velocity of book A has to be equal to 13.6 meters per second. Okay, so we can use this result to find the magnitude off books Book B's Final Velocity. So going back to conservation off kinetic energy we had we had the expression that is V A one squared is equal to V A two squared plus V b squared. Okay, now we know what v a two squared iss so we can isolate VB. So you have that VB is equal to the square root off 15 squared which is V one squared minus vhe squared. So 13.6 squared and we find that the magnitude off the final velocity off books be after the elastic collision has to be equal to 6.34 m per second. Now the only thing left is to find the angle between the velocity of book A and the velocity of book. Be on. Well, since we already know V h u M v b, we can just go back to this expression here, for instance, so expression one. So we still to the values we already know. So let's go back from this is from conservation off the Y component off linear momentum so we can find the angle that puck bees makes with the horizontal from the conservation off the components off the momentum or the X component. But I'm going to take the way components So p y we had the following equality B b. Why equal to v H u sign off 25 degrees. We know vey chew, and we can express V B y as the magnitude of the B times the sign it makes with the horizontal. And for the more we can isolate uh, the angle by taking the ark sign off this term. So I have that fire is going to be deep arc sine of V A to over Phoebe time sign of 25 degrees on which VH you is just 13.6 and BB we found to be six point 34. So we find Dingle to be equal to 65 degrees. Now I want to find what is the angle phi plus 25 degrees. So 65 plus 25 is you go to 90 degrees. So we found that particles be that the book be makes a 90 degree angle with books A after the collision. So this concludes the exercise"}