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ASArOuOah Ouceno mE (Show all work)10 kg mass slides dow frictionless incline from Tec CTOssc m wide sutface with coellicient of friction of 0. and goes back up ano...

Question

ASArOuOah Ouceno mE (Show all work)10 kg mass slides dow frictionless incline from Tec CTOssc m wide sutface with coellicient of friction of 0. and goes back up another frinctionless inclineWhat thc velocity at the bottom of the first incline? How high up the other side will the mass go?50 kg astronaut, freely floating 10 m/s is hit by arge 200 kg lemon cream pie moving oppositely 10 ms;Mi= 200 kR 10 m/sM2 50 k V2 = 10 m/sBeforeAfterAssuming the pie sticks to the astronaut what is his final velo

ASAr OuOah Ouceno mE (Show all work) 10 kg mass slides dow frictionless incline from Tec CTOssc m wide sutface with coellicient of friction of 0. and goes back up another frinctionless incline What thc velocity at the bottom of the first incline? How high up the other side will the mass go? 50 kg astronaut, freely floating 10 m/s is hit by arge 200 kg lemon cream pie moving oppositely 10 ms; Mi= 200 kR 10 m/s M2 50 k V2 = 10 m/s Before After Assuming the pie sticks to the astronaut what is his final velocity? Which force is bigger? The force 0n the astronaut from the pie 0r the force the Astronuu hits the pie with?



Answers

The Astronaut and the Cream Pie* A $77 \mathrm{~kg}$ astronaut, freely floating at $6 \mathrm{~m} / \mathrm{s}$, is hit by a large $36 \mathrm{~kg}$ lemon cream pie moving oppositely at $9 \mathrm{~m} / \mathrm{s}$. See Fig. 10-70. How much thermal energy is generated by the collision?

Yeah. In this problem we have a solid cylinder um I'm afraid there's 10 cm. Let's see here, uh mass of 12 kg. And it starts to rest without and without slipping roles at distance six m down a roof that is inclined at an angle of 30 degrees. So we have um you have something like this, we have this disk here rolling down here and then it um leaves. And then, you know, we want to figure out how far how far it travels in the X. Direction before it hits the ground. So that was that was here. So but we have then is uh the conservation of energy. And so the first thing we need to do is um figure out the velocity when it leaves the roof. So when when you basically the transitions from rolling without slipping to free fall here, um Obviously there'll be still rotating um whatever velocity, assuming that's no, you know, we're assuming that every energy is conserved here. So, you have the initial kinetic energy are the initial potential energy, sorry. And I'm measuring that potential energy from here from the height where their roof ends. Um So that's this distance times. Um The sign of this angle here, it starts from rest. So we have no kinetic energy initially. Um It ends up here again I guess. I'm assuming that the the references at the center when it leaves the Um so the center of mass is that um zero height at that point? And so we have no potential gravitational potential at that point. And we but we have some translation um some kinetic energy that's both translational and rotational. Yes, again, for a solid disc that um mask moment about the center of mass is one half M. R. Squared. Um If it's rolling without slipping, the angular velocity is the velocity of the center of mass divided by the radius. No let's see here we have conservation of energy, basically winds up with the final kinetic energy. You must equal the initial potential energy. You can plug that in do a little algebra and we get the velocity the velocity of this thing at the when it comes leaves, the roof is the square root of four third G. L. Sign of fada. And if we look at one to calculate the angular velocity um that is winds up being and plugging in all the numbers we have that winds up being 63 radiance per second. Um Yeah I guess yeah they asked us for that kind of an intermediate thing. So it would be it would be still spinning at that calm rate as it hits the ground. Now once it once it leaves the roof were in free fall and we know that the velocity um is in is in you know on the is parallel to the to the roof. So the velocity in the X. Direction is minus V. Cosign data and in the Y direction is minus V. Science data. So it's that way that means we plugging in the numbers, we get these two these components of velocity. Now we know that we know the distance here, H the height of the roof. Um And that is um five m. So it's gonna fall five m from where it left the roof. So we know that that's five m and then just gravitate under gravitational acceleration. And with an initial velocity we have that the change in Y is one half Gt squared plus B. Y. T. Um that gives us a quadratic equation for tea. And the route that we're interested in is the one that is 0.74 seconds. So it takes 0.74 seconds to follow this far. And so now we want to figure out how far it went, um in the X direction. And so Delta X is just the velocity of the extraction times, time And plugging everything. And we get about -4.0 m. So it went as a magnitude, it went about four m. I'm away from the edge here.

Okay. So in this question, I told that at Iraq and Jack's gas at 50 km/s and The the mass rate is 1500 kg per second. The velocity is 15 treatment times per second. My sister is going to be the mass of all the time per second and this goes to be the final velocity and in charlotte is going to be zero because it starts from rest inside the rocket and then injects at the speed. So I want to find the amount of force that allows you to eject. And our end goal is to find the the maximum mass or the rockets that can be supported by this amount of force created by this. So we're going to use a physical too. And over T bracket V man issue. So F is the force M is the mass Tuesday time final velocity and used in shot blasting. So F. B. Called to we can convert this. This two. Okay, no notice. We can convert this to me. That's the second by multiplying by 1,050,000 meters per second and We get 50,000 times 1500 crypt. And mine is terrible danger at that. So this would be called to seven 35 times 10 to the power six mutants. No, since we know this force is this this force can support the weight of the rockets of exactly this because the forces have to balance for the rockets to be able to stay to me to be supported by the way, by this force. And so Mhm. We need to actually find the mass from this way. So the weights or the maximum weight of the rocket is going to be So on five times standard about six newtons. And for the mast obviously go to energy. JM recalled to W. G. MB coats in 75 times 10 to the power of six. Whatever I. G, is nine point it's one seven squared, and this will be called to 7.7 Time stands at Power six kg. So this this is the mass that the rocket can have is the maximum master, or it can have for this rocket to actually be able to right.

This question were given the force of gravity formula with a few constant and values were asked to find the work required to launch a 500 kilogram rocket up by 2500 kilometers. So first, we do want the distant to be in meters. So I would just rewrite this as 2.5 times 10 to the six meters because I'm adding three zeros in or convert two meters and next we want to write their work formula, which is work equals into grow from A to B Athol x dx and we are we have us if we have this F function with a lot of letters, but only variable is X. So this is from 0 to 2.5 times 10 2 to 6 g m um, over X plus our square dx. And since G m and M are all Constance, we can be put those outside of the integral so over g Castro and small M in to grow from 0 to 2.5 times 10 to 6 and it's no running one over x plus r squared. I can write this ass x plus R to the power of negative too. So now I can just used a power role for the integral. So g m no b X plus are to the negative one to find a of one. And I'm evaluating from 0 to 2.5 times 10 to the six. The negative one. I can this denominator Negative one. I could just move it to the front. And this exponents negative one. I could just change this back to one over. Explicit are So this is negative key Bigham Small, um, one over x plus r. You know, I'm evaluating from 32.5 times 10 2 to 6, which is just this. And then times one over to a 10.5 times 10 to the six. I was here minus went over. And then when I put in zero into X, my second term would just be one over r. And if I were to put in all the Value Star given the G capital M 500 for small M and our I would get a 0.87 times 10 to 9 jewels. Second question asked us to find a work to launch 500 to 500 kg rocket up by ex kilometer without specifying what x is. So we have basically done not already in this formula. It's are in this calculation. The only difference is that we're not using X were using Sorry, we're not using 2.5 10. 76 were using tax. So we just rewrite that here. Just negative, G um, one over X plus r And we are evaluating this from zero to accidents that so this would be negative, g um, one over so X replacing X, which is still just being X and zero. Putting in there would be again just one over r. And that's basically all you need to do. Um, if you want, you can have the second part of us combined so that it's a common denominator. So first term I would multiply but are over are so are over, uh, expose our second term out multiply by X plus are over x plus are so expose our our experts are which you can see if I can have r minus are and be left with just accepted top. So I'll be negative. G and um will be X guarded by here X plus bar and Of course, you can substitute 500 into em. And the other values into G, M and capital are if you want next question access to find the work required to launch the rocket to space. In other words, excess approaching infinity. So we again can just look back at this one and to work. We found B, but now X is approaching infinity. So that means this work would be the limit of X approaching infinity of G Um, um one over X plus R minus one over R So the sorry, I forgot to negative. So, um, the second term one over are multiplied by negative G. M and M is not really affected by the live it. So the only term affected by the limit is the one plus our which I will just work out on the side here. So this part, there's only what part that's affected by the limit. So the limits of X approaching infinity of one over X plus R. So if excess approaching infinity, we have one over infinity, which we know is just zero. So final answer without the limit is negative. G m. And this whole term, when this one over exploits are entirely is just zero, which it means we just have negative one over our remaining. And of course you don't want to negative. So it's just g m. And over our last question, we are asked to find escape velocity. And we're told that that's where work is equal to the kinetic energy form, which is 1/2 MV square. So we have G um um, over our, which we found in part C is equal to half MV squared minutes, just algebra, so you can see it at the EMS. Cancel up small arms. Cancel. So that's a massive rocket cattle, and we can multiply by two on both sides. So to GM over R equals the square and square with both sides, which means V is equal square root of two G and over our that's it.

I agree. One, This is the problem based on conservation of energy and Bogdan here it is given them Ah, block off! Last 10 kg. Is that cut off off in plant plane having the height to meet him Had inclination 30 degree It can slight on and plantain ended with horror. Genter frictionless Track on at the end. Off it the ridges Brig Off spring Constant 500 Newton perimeter It is hurt. We have took fire. Yeah, first part. Where? The city at the bottom. If his spring gets compressed Well 175 meter. And be part We have to fire Bogdan by friction and she part We have to fight velocity at the base as it requests by Spain. Yeah, and deeper distance moved over than client plane. Let us start solving the first part. Applying the contribution off energy energy at C. Bell B. It's got to energy at Be and see. There is only kind of technology. Sorry XY There is only potential energy off this brick on at it, having one leader kind of technology. So from here you can get the velocity at any point that is he x upon them que is given 500 X is 575 My sister on solving it the veracity You will get five point 3033 m per second. Yeah, no second part. Yeah, E yeah. Here to calculate. Yeah, Yeah. Speak. I'm velocity at peace. I played conservation off energy. So he is called two years. That is half 500 in tow. 175 That is 1 40 point 6 to 5 years old. At eight. It is M G h A, but is 10 into 9.8 and two Group, that is 1 96 Jude. So government from A to B is going to change in kinda technology. That is 1 40.6 to minus 1 96. It is simply five negative. 55 3752 No, I'm e v does is scare toe TV. So it will require with the same velocity. That is 5.3033 meter per second. Okay, Yeah, yeah. Distance dia is too upon sign or 30 that is 4 m. So welcome from a to B. Nearly a full half the x and the servant. It is forever. Well done. You know minus 55.367505 It's called toe forever. So frictional force, you will get minus 13.8438 Newton. Yeah, well, confront Vidas today it's got toe total energy entity minus total energy at leaders, that is and deep daily my sign off 30 TV. This it will be 10 in 29.8. About two. Yeah, Just government. Yeah. No, this is equal toe averaging. Yeah, mhm. So you can like 10 and to nine. Point is upon to minus 1. 40.6 to 50 It's good to minus 13.84 for sort of input. Thanks. Distance 2.23 77 m. Yeah. Mm. And the height attended. You confined? Do decide off 30. So it is to be 1.11 89 m. That so? Thanks for watching it


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