Like

Report

At one instant, the center of mass of a system of two particles is located on the $x$ -axis at $x=2.0 \mathrm{m}$ and has a velocity of $(5.0 \mathrm{m} / \mathrm{s})$ ) One of the particles is at the origin. The other particle has a mass of 0.10 $\mathrm{kg}$ and is at rest on the $x$ -axis at $x=8.0 \mathrm{m}$ . (a) What is the mass of the particle at the origin? (b) Calculate the total momentum of this system. (c) What is the velocity of the particle at the origin?

(a) $m_{1}=0.30 \mathrm{kg}$

(b) $\overrightarrow{\mathrm{p}}_{\text {tot}}=2.0 \hat{\mathrm{i}} \quad \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$

(c) $\overrightarrow{\mathrm{v}}_{1}=6.67 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}$

You must be signed in to discuss.

{'transcript': "So this time we're given a new problem. Uh, it's, uh you have to party colds. Think about the ex oxes. So this happens to be the origin. And there's a particle at the origin and and also the and sent off muss off the two particles. So, you know, this is this is one party call, and then another one is here. I'm just going to call this one a the one of the origin where X equals to zero. That's the position. And then this second one, I'm going to call it B. Um, this is going to be at X equals two, eight point zero meters. This's just zero meters. And then this is the scent of muss that happens to be a exit was to two point zero meters. And the, um this is the second party call. You know, the B party call, huh? Happens to have the moss off only zero point one zero kilograms and then the a particle. We don't know what the masses for that one. So that's the information we've given. We're also told that the velocity off the system at the centre of muss happens to be a five point zero meters per second. I heard you know, that's the velocity of the system. Well, sign put A We want to find the the moss off the particle at the origin we want to find him, eh? And then and B, we want to find the total momentum. Pete, Pete, total off the system. You know, that's the momentum off the system and then in. But see, I wanna find the velocity off the particle at the origin. So the calling it the This is the velocity off the particle, but origin And so you know, we're going to do a step. Voice crosses. Remember, I will use the scent off months to resolve this problem. The center ofthe muss. It happens when I'm going to call that this should be called XY. So excessive sent off muscle XY because too m a close no x e plus plus m b times x b. So just to make sure the levels of right this is X B and this is x eight and this is sexy. And so this is divided by em and me, plus m b Good. That's what we're given. And so our goal is to solve for the mass off, eh? We'Ll multiply both sides by i by this value. So we get a taxi Emmy plus m b He wants to Emmy plus XY plus and the Bixby and so may we use distributive property right there. M a x e plus m b r exceeded was to Emmy plus X plus m b x b Then our goal again is to find M s o. We will move This made this what? This side. And then we'LL move this to the right So we have Emmy X E plus or rather miners because it's moving to the left. So, you know, this is, uh this is going to be minus. This will be minus. Emmy goes to x e nous em be xb minus and be sexy. So we pull out the A Then you left with XY minus one. Recourse to x e plus them the xB finest ex. See, those are the positions we have. Um, so we're getting closer to solving for wonderful for Emmi. So we'll just shift to the next page and make sure that we have our numbers, right? So m e Boone XY minus one equals two x a plus em be xB minus sexy. Recall that X is the dimension of the origin. Right here. So that means it. Zero so x a zero. So we're left with No, You know, we could divide both sides by XY minus one over XY minus one. And so we end up with M A being called to M V X B minus X e off XY minus one. And so, if you go back to see that expertise eight point zero meters. Uh, sorry. M b M M B zero point one zero kilograms. So we need to change that rail. Quick. We need to change this one. The missus MBf wants to be owned zero zero point zero point one zero kilograms and then x b is eight point zero meters and then XY is two point zero meters and then exceeds two point zero meters minus minus one two panzer leaders minus one. And so, if we simplify this problem again, uh, point one times six. So we get points. Six. The mask being a point six kilograms, so m e equals to zero point six kilograms. So we just want to go back and make sure that we had Our answer is right. So we had Wei had we had m m one. Oh, sorry. M a no x e So that would've made it x zero. So this one? Yes. So we need to change some things. Here s O. You know, we had a plus. It shouldn't be a plus. It should be a a multiplication his state of ah instead of a plus So this one right here it's not a plus. This is Ah sorry. This is a multiplication. So this one right here it's a multiplication tze times and then this one is times and then this one is gonna be times and then we'll move it to the other side. We don't have this one. I don't have this. So this becomes, um this becomes x A yes has become extra right here. Okay, Just remember that. That's Exley M a X. Someone was simplified. This is gonna be exe, eh? Yes. And then we don't have this right here. Okay, so this is just a minor corrections just to make sure that the mark is right. And then when we go to the next page, this's X a This is extra right here. And this is exterior here. This is X ray right here. Okay. So you know when we when we divide that we end up with extra right here, but then x zero point zero meters. And so this is going to change to zero point three kilograms. Because of this zero point three kilograms on That happens to be the mass off A. So that's the first part of the problem. Were able to get the mass of the second part of the problem. Will return. Next raid in this time was supposed to get the velocity. Oh, the origin. Well, Steve. But the origin we do know in terms ofthe momentum is that of the momentum. Um, off the system. Em one plus him too. This is calling momentum because too. Ah, this is actually a m a and M B. That's the velocity of the senator. Must this one right here, This one right here. So if you come back, we see that we have Emmy of a thie A Remember the air wass the velocity that we're looking for. This one right here and then, um V v is the velocity of the second one, plus, uh, M b b. And so, you know, V velocity at center of Mass is equivalent to the individual mo mentum over the sum of the masses. And their goal is to get the velocity of hey, so going to solve that and so easy em A plus A and B because too m a d a plus tembe of Phoebe remember something else they say about the problem and this was given. I think we should have mentioned that, uh, they are The particle is at rest. So that means I wanna make sure that V b equals to zero meters for seconds that you want to remember that we just don't know what it is, and that's what we're trying to find out. So coming back to this, we get to see using distributive property. Emmy of the CM have lost its center must plus in the club, V. C M equals to Emmy. Hey, plus M B V b of going to the next page one of simplified the problem because our target is to get the so put the M e m is get the the inputted by itself, so m e the a has t equal. If you look back, it has the equal. Maybe see him like that plus m B of the cm and then minus is going to be a minus because this one moves all the way to the left. So it's going to be a minus. M b B B. I'm being Phoebe. Okay, So vi a has to equal to m e the cm plus m b c m minus tembe be off of a member of the C m Is the velocity at the centre ofthe muss I don't know And then you want to recall that so peace v c m is the lost it at the centre of muss and then ah thie v V is the velocity of B and es is the last yet a This should have been Emmy. Sorry. So when you plug in the numbers the mass off a verified We already got that in the second part. Zero point three kilograms, zero point three kilograms our times The velocity of the senator must which is five point zero meters The second I had plus muscle be muscle be happens to be if you go back to your point One killed rooms, you put one zero kilograms times the velocity at center must switch is meters per second. I hunt minus mass off being zero point one zero kilograms, Oh times the velocity of B which is zero meters a second. And so this part of the equation will just disappear because gonna be zero and then all that is divided by myself. Hey, which is your point three kilograms. When you simplify that, you get that the lost in off a happens to be a six point seven made us second. I had just one clarification thiss problem. This part should actually be a party. This is but see the velocity off off off the particle at the origin. So we skip the step on Barbie. But we're going to go back, remember? But he was asking about the total momentum off the system. And so we'll do that on a new page. And and so the the total momentum off the system, which I'm calling p total hus to be em one rather m A plus m b and then of C M. And so this zero point muscles, eh? We got that was there a point of coming back to the zero point three kilograms plus the mass of B which was your print one zero kilograms times the velocity of the center must which happens to be five point zero meters per second. I heart Oh, you know, that's that's total. So if you plug in those numbers, you get that the total momentum it happens to be, you know, three point three point one. That's point foreign five. That gives us two kilogram meters per second. That's the total momentum off the system. Hope you enjoyed it. Problem are looking forward to the next question. And if you have any questions, you can feel free to send them my way. Aunt, have a wonderful day. Okay, Thanks. Bye. On DH. See the final. The velocity of able six point seven meters per second."}

California State Polytechnic University, Pomona