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Problem 81 Easy Difficulty

An $8.00-\mathrm{kg}$ ball, hanging from the ceiling by a light wire 135 $\mathrm{cm}$ long, is struck in an elastic collision by a $2.00-\mathrm{kg}$ ball moving horizontally at 5.00 $\mathrm{m} / \mathrm{s}$ just before the collision. Find the tension in the wire just after the collision.

Answer

$102 \mathrm{N}$

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Video Transcript

{'transcript': "So in this exercise, we have a ball off mass and B that is hanging by a mess. Lis wire off length are and this must be is just about to get hit by a mass A that has initial velocity V a one. Okay, in the velocity is on the horizontal right after the collision between the particles A and B. The collision is perfectly elastic. And the particle be games velocity V B to deposit to the negative X direction and the particle, um, A will have a final velocity that points, um to the opposite direction off its initial velocity. So what we want to know in this exercise it what it is what is going to be the tension on the wire right after the collision. Okay, so So let me race this. So it is easier to see. We have that there is the gravitational force, so I'll call it F g f g. Here. We also have the tension that I'll call t one. Okay. In this case, 21 is equal to F. G. So the information we have is the massive particles, A and B. We have the initial velocity off particle a and we have the length off the wire. Okay, so, um, since it is a lasted collision, we can use, uh, the books equations 8.25. And this equation relates, uh, the velocity off particle be with the initial velocity off park. Oh, wait by, uh, two times a m a plus m b. So by this factor, this is equal to V A one. Okay. And why do we want the B? Since Because we want to find t. So why calculate vb? Well, we have to calculate, uh, vb because, um because in Newton's second Law, so second law, we have that the net force eyes going to be the mass times, the acceleration and in our case, since the forces on Lee point into the radio direction. So we'll look to the net force in this case. So because the forces only point point to the radio direction, we can express this by so and be times the velocity off B squared over two. Sorry. Over our on which this is the radio acceleration off particle be and this is going to be equal to the tension on the wire that we want to find t minus the gravitational force which is equal to M B times G. Okay, so we can isolate t in this case, and we have that tea is going to be equal to be, uh, times v b squared over r plus g. So this is why we want to find VB. Okay, So using this expression and the values we have, we can calculate VB So we have that this is equal to two times to over two plus eight times the velocity, the initial velocity off particle A which you know it is equal to five. So we find that the initial velocity off particle we that is the velocity of particle. We right after the collision is just equal to 2 m per second, so we can use this. And so the students in the expression from detention and we have that attention is going to be equal to the mass of particle piece. So eight times the velocity squared, so two squared over the radios 0.135 plus 9.8, and this is going to be called to 102 Newton's. And with this, you conclude the exercise"}