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Two ice skaters stand facing each other at rest on frozen pond They push off against one another and the 48 kg skater acquires speed of 0.68 m/s. If the other skate...

Question

Two ice skaters stand facing each other at rest on frozen pond They push off against one another and the 48 kg skater acquires speed of 0.68 m/s. If the other skater acquires speed of 0.85 m/s what is her mass in kilograms?Additional MaterialsReading

Two ice skaters stand facing each other at rest on frozen pond They push off against one another and the 48 kg skater acquires speed of 0.68 m/s. If the other skater acquires speed of 0.85 m/s what is her mass in kilograms? Additional Materials Reading



Answers

Two ice skaters stand at rest in the center of an ice rink. When they push off against one another the 45 -kg skater acquires a speed of 0.62 $\mathrm{m} / \mathrm{s}$ . If the speed of the other skater is $0.89 \mathrm{m} / \mathrm{s},$ what is this skater's mass?

Okay, so we've got these these ice skaters, right? Um, and they do a lift. Okay, So first we've got this skater. This, I guess, is the guy, right? And they are 50% more massive than the leave skater who is a gal, right? Don't you said there they are. So and this person is going 6.2 meters per second, and they have 50% war mass, So they are 1.5 times the mass of this person. This person is 1.0 times their own mass. They're going a little bit slower, right? Right. And then in the second picture, we've got this guy holding up the second person. There we go. All right. So this, by the way, it would be 1.5 times the mass of this person plus 1.0. So that's gonna this is gonna be two point five times the mass, and we want to know what is that velocity? Right. So they noticed that they haven't told us with the massive this person, So you could either make up a mass. I hope you're cancels, or in this case, it actually will just cancel out right okay, cause it'll be EMS on both sides. Right. Okay, so we were just gonna treat this as conservation of momentum. Remember, momentum is M V. Right. So if we add up all the momentum here, that illegal the momentum there. Right. Okay. So if I go, um, 1.5 AM times 6.2 plus 1.0 m times 5.5, that equals 2.5. And that 2.5, of course, is the 1.5 plus the 1.0 right combined times V. Right. Okay. Noticed that the, uh, we get the red pen of cancellation. You the mass cancels from both terms. Right, Cause have you divided both sides by I am your divide every term on the side in every term, on the side by and remember terms of things with what pluses reminds his behind between between. Right, So it's a 1.5 times. Ah, 6.2 plus one plus zero times 5.5. Devise equal to. And then that's 14.8. Divided by 2.5 5.92 Um, and then you really have to sick things. Seems like maybe we do Yeah, because they've been Yeah. So it's really to six. So we have to say 5.9 meters per second. Yeah. There we go. All right, that's it.

In from rest, Two skaters push off against each other on smooth level ice, where friction is negligible, One is a woman and one is a man. The woman moves away with the velocity of positive 2.5 m/s relative to the ice. The math of the woman is 54 kg and we know the velocity of the woman is 2.5 m/s. The mass of the man is 88 kg, assuming that the speed of light is three m/s, which is not true. So that's the relativistic momentum must be used to find the recoil velocity of the man relative to the ice. Okay, so they start from rest so their initial momentum is zero, so their final momentum, the momentum of the man plus the momentum of the woman has to be equal to zero. So that means the momentum of the man is equal and opposite to the momentum of the woman. So the moment remember the woman is mass times velocity over the squared of one minus V squared over C squared, Solving for this. And we get the momentum as 244 .22 kgm/s. So that means the momentum of the man is negative 244 .22 kgm/s. So we need to find the recoil velocity. So if we have the momentum and we have this equation, then we can solve for the velocity. And we get that the velocity is equal to the momentum over the square root of the mass squared. What's the momentum squared overseas squared Lacking in values. And we get a recoil velocity of negative 2.0 meters per second.

This weapon is going to look at conservation of momentum in a closed system and a closed system, as you may recall Means the total forces acting are zero. So from our course, in close relation we have uh delta T. Equals change momentum at zero. So our total change in momentum mobile zero. Yeah. So what we have for our clues, the system is to ice skaters there originally at rest. So their initial blast equals zero. They're in contact and they push off of each other. Um we know one skaters mass and final velocity. We know the other skaters mass. And we want to find her final velocity. So we have skater one, sorry, mass equals 79 kg. And her velocity is an unknown for skater too. We have her mass equal to 57 kg and her velocity after the push is mhm 2.4 m/s in the positive X. direction. All right, so now we have our initial momentum equals Matthew the first skater time for velocity. And we said they're both at rest, you know, was massive. The second skater times for velocity which is also zero. So change in velocity is going to be zero. I'm Sorry. Change in Momentum zero. Now for the final momentum we have mhm equals mass of the first skater time to initial velocity. The final velocity mhm. Which is question Mark zero but in two Time to her final velocity which is 2.4 at minus initial velocity which is europe, we get. All right, okay. So yeah. Mhm. Yeah, mm. Never. Yeah. Yeah. Mhm. Yeah. Yeah. Yeah. Yeah. And that makes sense when you push off, one's going positive, ones going negative. The heavier skater will have a smaller velocity Julian. Yeah.

In this problem. The mass of the woman is given as 54 kilogram, and the mass of the man is given as 88 kilogram. So initially there at rest and they push each other off. And then they started gaining some velocity. So the velocity of the woman gains is 2.5 meter per second and we want to find velocity. The man gains. So now the first thing is that since they're pushing off each other and we're neglecting the friction from the conservation off linear Mormon, too, we're gonna have the initial momentum, which is zero the clothes, some of mo mentum. Final moment. Um, which is P one plus p two, right? No, that means the magnitude of P one has to be called to manager of P two. All right, now, let's first find the magnitude of P one related mystical, more momentum. The woman gains, which is pure Nichols MV over one minus from one p. 1/1 minus. If you wanna square over c squared and take the square root of this. No, I'm gonna just do this. So m one is 51 54 kilogram and be one was 2.5 meters per second over one minus. V. One was 2.5 meter per second squared over. So C is given as three meter per second. All right, so that's gonna be a three meter per second squared. So a square root of everything here. All right, so this one is going to give us P one. He calls to 44.22 kilogram meter per second to 44.22 kilogram meter prayer second, and this is gonna be cold, two p to write in their magnitude. So pew on it. Close P two now also p two because M two V two over square, root off one minor fee to square over C squared and squaring and rearranging. This one is going to give us p two square minus p two scurvy two squared over C square. He calls I'm to a square the two x squared and let me reread this everything from here we can take the two coming from this to here. We're in a really is and take you to come in here. He two square I'm to square plus p two square over C squared he calls p two squared, right? And from here, if we take if we take this to the right inside is going to go to the denominator, and we're gonna take the square root of this. So this is gonna get V two equals key to taking the square. Would p two over square root of M two squared plus P two square. Oversee two squared. All right, Now, all we need to do it. Just plug in the value off whatever we got. And this is going to give us 2.0 for meter per second. And this is the magnitude of the velocity. Now, since the man is going in opposite direction to the woman so that the total momentum is zero, that means V to is actually called to negative 2.4 Peter per second.


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