two-frictionless-carts-mass-500-g-each-are-sitting-at-rest-on-a-perfectly-level-table-the-teacher-ta

Question

Two frictionless carts (mass $=500$ g each) are sitting at rest on a perfectly level table. The teacher taps the release so that one cart pushes off the other. If one of the carts has a speed of 2 $\mathrm{m} / \mathrm{s}$ , then what is the final momentum of the system (in $\mathrm{kg} \cdot \mathrm{m} / \mathrm{s} ) ?$
(A) $2,000$
(B) $1,000$
(C) 2
(D) 0

Khoobchandra Agrawal · Accepted Answer

two cars each of months. 500 grand initially at rest. Yeah, on the smooth horizontal surface. If one has to be pushed big speed off who meet up for a second, we have to find the final momentum of the system since external force on the system is zero. Yeah, so initial momentum must be equal to final momentum. Initial momentum is zero, so final momentum will be also zero hence of Cindy&#x27;s correct.

Ze-Han Lee · Answer

so in this problem. Initially, we have two cards. Ah, So this two cars there? Ah, they&#x27;re taped it together. So the initial moment of zero Ah, soapy initial zero. And then we gave ah. Then the two cars. They put the push each other and the one of them obtain velocity of two meters per second, two meters per second and the other one off course. It&#x27;s negative. Two meter the second. So the total momentum in this case final, it&#x27;s still zero. Okay, because the moment, um, off this two cars, they have the same magnitude, but opposite. So the final the final moment. Moment of me. Still zero. So this is the moment the moment of preservation.

Nick Auwerda · Answer

question 84 is kind of a fun problem that involves conservation of momentum and conservation of energy. So we have to, uh, air carts mass of pointing floor and mass of 0.42 And the smaller cart is initially so. Cart one, the larger Cardizem rest Sylvia zero and then the initial. The other card is initially pushed towards it with a speed of 00.68 meters per second. Headed this direction. There is a spring on this that&#x27;s going to allow these two have an elastic collision and that spring has a constant of 690 Newtons per meter. Question A says What is the speed of the two carts at the moment when their speeds are equal? So a If they have equal velocities, what is that velocity? And so this is a conservation of momentum problem. And so we have 20.68 times point for two plus zero because that&#x27;s no momentum they have at the beginning, is equal to the momentum they have at the end. But this is not the end after the collision. It&#x27;s the end when they have the same velocity so I can go ahead and write 00.4 to the massive cart one plus 10.84 The massive car, two times their common velocity V and so 0.68 times point for two is 20.28 56 and this has to equal 1.26 times V. So if we divide by 1.26 we get a final velocity of 0.227 meters per second on dhe. That&#x27;s the velocity the two cards have when they have the same velocity. Part B of this question says how much energy is stored in the spring when the cards have the same speed. So be we want the elastic potential energy. In the spring, this one were actually gonna rely on the law of conservation of energy for the carts started with some kinetic energy. The gravitational potential never changes on this level track, so we can ignore it. So the kinetic energy we start with is constant. Well, the mechanical energy we start with is going to be converted into the kinetic energy of both of the carts, plus whatever elastic potential energy is stored in the spring. And so I&#x27;m going to have 1/2 the mass of the velocity or 1/2 the massive 0.42 times the velocity of 0.68 squared, that is, the kinetic energy we started with has to equal half the mass. Well, at the end of this problem, we have both of these carts. So 1.26 kilograms, with their final velocity of 0.227 squared plus whatever elastic potential energy is stored in that spring. So the energy we started with has got to be distributed everywhere and so 0.5 times 0.42 times 0.68 squared is 0.97 0.97104 This is equal to 0.5 times 1.26 times 0.227 squared or 0.3 246 plus the elastic potential energy in the spring. So all we have to do is subtract that 2460.32 from that 0.97 and we see that are elastic. Potential. Energy is 0.646 point 0646 Jules and then part see on this says what is the final speed of the carts after the collision This one, we can go ahead and use conservation of momentum together with conservation of energy on, we know that 0.68 times 0.42 has to be equal to 0.42 times the final velocity of Cartman, plus 0.84 times the final velocity of cart too. But remember that the difference in their velocities that they start with the one minus V two that they start with is equal to the opposite of their difference in their velocities. Be one f minus V two f that they end with. And so I do know that initially cart one had of the lost of 10.68 this minus zero because the other was a rest is equal to the opposite of I&#x27;m using one and two for my finals. So v one minus V two. If I simplify this a little bit, I get 20.68 minus equals negative view Iran plus V to war 0.68 plus V one equals V two. I&#x27;m gonna go ahead and substitute this into that equation. And so I yet 20.68 times point for two, which is 20.2856 point 2856 is equal to 28560.42 times V one, plus 0.84 times V too, which is really 0.68 plus V one. So we have to distribute that. So 10.2856 equals 0.4 to V one. Plus pointing four times 40.68 is 0.5712 0.5712 plus 0.8 for eveyone. And so I have a 0.2856 minus 0.5712 which is negative 0.2856 equals 0.42 plus 0.84 or 1.26 of the one. So then we can divide by 1.26 and get V one. Final velocity of cart one is negative 10.227 meters per second. And, uh, final velocity of cart too is given to us by 0.68 plus V one so minus uh, that 0.227 his 0.906 0.9 Whoops. Sorry. Uh, 0.68 I added that minus 0.227 is 0.453 0.453 meters per second

Susan Lu · Answer

in this example, we have two cards that have attractive magnetic ends that are facing each other. So with the absence of and in external force, these two cards can still be pulling together, so they&#x27;re still going to attract each other. We have one cart that is stationary at one point, and cart one is going to be coming towards it after CART one comes towards it at some unknown velocity. What&#x27;s going to happen is because they are being attracted to one another. We have car to that&#x27;s going to travel at a velocity of negative 1.9 years percent and cart one that&#x27;s going to end up traveling at a velocity of 4.5 years for sex. So just gonna draw a frame of reference to show what these positive and negative signs tell us. So we have carte walk that&#x27;s moving to the right, and we have cartoon that&#x27;s moving to the left. So what we&#x27;re going to need to do to determine this speed is going to find the total momentum off the system now, since there&#x27;s a absence of a net external force and then your mo mentum is concerned, so what we could do is use this instance to find the total momentum and then work our way backwards to determine this velocity here. That&#x27;s a fine Pete Total. We&#x27;re going to me to find p one and add it to P two p one close to um one e one, which is our 2.3 kilograms. Multiply by the velocity of 4.5 kilograms. I&#x27;m sorry for from five meters per second and that gives us a total of 10.35 kilograms. Peter&#x27;s percent. Then we can do the same thing with P two, but with its corresponding mass AM velocity, though for two we have mass of 1.5 kilograms velocity of negative 1.9 meters per second back. Use us negative too 0.85 So P total tells us we&#x27;re going to need to add these up or determine that. So we&#x27;re going to get a total momentum of 7.5 kilograms meters per second. So this is our P total. Now we&#x27;re going to use that to determine what the mo mentum of cart one it. So you noticed here that in the very beginning, cart to a stationary So because V is equal to zero p two is that also equal to zero, which then brings us back to here on this side. So that means that cart one makes up the total momentum in the beginning before the whole attraction occurred. So this is going to also give us key one. So now that we know P one, this is P one before attraction, so be a little more specific. So now that we know p one, we can plug it into the equation for mo mentum. The one is equal to, um one you want. So we get 7.5 kilograms. Meters per second is equal to 2.3 killer grounds. The one Div Aibel sides by 2.3 and you will find that the one going to give you 3.3 meters per sec. In the case, you guys want this for reference. P total is going to be 7.5 kilograms used for sex

Robert Daine · Answer

and we have three cuts moving, colliding and then moving together. No, we have two cars moving to the right on one car, moving to the left. Our situation is on an elastic collision. So the equation for an elastic collision will bay and one V one plus to be to plus and three V three. No, because they stick. We just have to add the three masses together, multiplied by a final common speed. From here, it&#x27;s just a case of plugging in the numbers and figuring out what our final speed has got. A big so four times five plus 10 times three plus three times negative for because it&#x27;s moving to the left, we have to make sure we include that negative. Yeah, and then the three matters together for plus 10 plus three. Multiply it by the final speech. So four times five is 20 10 times three is 30 three times four is 12 on its months because of that negative, and that is equal to 17 via so dividing both sides. Now, by 17 we can see that the is equal to 38 over 17 which is 2.2 full. Meat is per second. Now, our second part of the question asked those what happens if they collide in a different order? Well, it actually doesn&#x27;t matter what order they collide in because momentum is indiscriminate, won&#x27;t save all collided. They will be moving at the same speed and in the same direction regardless. So I will just right here. Collision order does not matter.

Kevin Shryock · Answer

Welcome back to New Ride. My name is Kevin Chirac. Let&#x27;s take a look here at a 15 kilogram carts. So right then and AM is equal to 15 kilograms, moving with a velocity of 7.5 meters per seconds. Will write that in 7.5 meters per second down a level hallway. So we know there&#x27;s not companion change, potential energy. The constant force of 10 Newtons acts in the carts. Would you have a force? We&#x27;re told that&#x27;s 10 Nunes acts in the cart and its velocity becomes 3.2 meters per second. So it&#x27;s gonna be a final velocity now, three points you meters per second. So you know this is gonna be an opposing force because I&#x27;ve decreased my speed. What is the change in kinetic energy for the cart? Well, we know that we need to consider that some of energy equation of energy initial close. Any work done is equal to the sum of energy final. We know that we&#x27;re only going to have kinetic energy in both the initial and final states so we can put that in instead. We can replace this with the equations. We know 1/2 M V initial squared Let&#x27;s work equals 1/2 M V Final squared and then from here, weaken dio a little bit of rearranging. So let&#x27;s make this that since we&#x27;re looking for the change in kinetic energy, let&#x27;s get both the kinetic energy on the same side. So this will be work is equal. Teoh 1/2 m v final squared minus 1/2 m The initial squared which we know is going to be equal to the change in kinetic energy. We can plug in a couple different things here, and in fact, we can even make the math simple for ourselves We can do. This is 1/2 times M, which is 15 times final velocity squared, which is going to be this 3.2 squared minus of the 7.5 squared. Okay, working, just go ahead and plug that in. So let&#x27;s take just a moment, a lot or a calculator. And then we get 1/2 times 15 times and then in parentheses, 3.2 squared minus 7.5 squared. See, that&#x27;s a change in kinetic energy of negative 345 jewels, so it&#x27;s going to be equal to our changing kinetic energy, which, as it turns out, is also our work. So when it ask us, what&#x27;s the work that&#x27;s done in the cart is gonna be the exact same thing. Work is responsible for what&#x27;s changing my kinetic energy for part C. How far to the cart move while the force acted upon it? Well, for that, all we need to do is remember that work is equal. Teoh force times a distance. And since we know that the work here is gonna be negative 345 and the forces and negative 10 Newtons were just solving out for that distance. Okay, so it&#x27;s going to be three or 34.5 meters once you divide away that 10 hope enjoyed this video. If you did, please take a moment and click that heart at the bottom of the screen to let me know as well as the help me reach out to more students like you.

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