i-a-7150-kg-railroad-car-travels-alone-on-a-level-frictionless-track-with-a-constant-speed-of-150-ms

Name: i a 7150 kg railroad car travels alone on a level frictionless track with a constant speed of 150 ms
Uploaded: 2019-05-26
Duration: 3 min 4 s
Channel: Numerade
Description: (I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?

Question

(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?

Kai Chen · Accepted Answer

seems this question. We initially had a car traveling at a certain speed is an office. Such a load is job off your car and we want to know what will be. The cars near speed so usually wins the massive czar card increases. It cannot keep the same speed if certain on your energy attitude, because then it&#x27;s against. You can see that it&#x27;s kinetic energy will increase its mass increases, so that violates the conservation of energy because it has, a car must slow down. So the rule you will follow his conservation of momentum and conservation momentum is just this one from women&#x27;s. Some momentum before anything will be equal to his momentum. It it&#x27;s the end, and this is actually this is assuming that there is no external force to it, because in Section One, we also know that there is formula is that we also know that in situ and there&#x27;s formulas that&#x27;s a force applied to. It is the change of momentum over situation of time. So this&#x27;s assuming that there is no force applying to it is India. Tha t will be zero so a momentum would accept, and the reason I&#x27;m saying there is no external force applied in this question is because this load is dropped out to the car. So I am assuming that when&#x27;s the car is driving inthe e horizontal direction, a load is jumped onto it in the vertical direction. So it does not have any horizontal component off forces. So it&#x27;s that is it was a context. That is why we can use the conservation of momentum here. So Pete in the show equals P fi? No. So we can write it down is and this causes car A and his causes load b initially a V a service. So no mention his car and he has no momentum because it wass we&#x27;re assuming this initially it regressed. Um, yes. Oh, what does it say? Can you see around it? Yeah, it does not say the mass of the salon, so I just say Ciro afterwards, some car answer lord travels together at a different speed. Um, I just quite fee. So we need to find V. We need to find the new speed, and that&#x27;s gonna be in a feet, eh? Over in a class and be. And with those numbers, someone 50 kilogram terms if you consider rock meters per second um 73 kilogram plus study 3 50 clover that gives us turn 0.2 meters presented. So that&#x27;s how much this car was throw down due to the added mass.

Farhanul Hasan · Answer

momentum conservation of this problem. So initial momentum will be equal to final moment. Initial momentum is on, not times be not mass and velocity of the car. Um uh uh And so and then So that&#x27;s equal to final momentum Which will be and not plus and one car massive car plus massive. Uh ah, the load times final velocity. And so Visa f and so are yeah, it&#x27;s called me survive. So visa half is equal to and not do not over the sum of the two masses Ah, and so that&#x27;s, uh, 9150 kilograms times 15 meters per second over 91 50 plus the load 43 50 kilograms velocity. The final velocity of the two is 10.2 meters per second.

Alexandria Calder · Answer

Well, hello. So today we have this question where we have a literate buggy, a car of a mass of 12,600 kilograms. It&#x27;s moving on its own on a fraction less rules. So we can make let any friction anything which can assume it. All images put into this car. It&#x27;s going to the velocity of 18 meters per cent. So we have this card is moving and we have this mast initially at rest. So we save velocity of zero meters per second on this mass is placed on top of this car and then they go together, they&#x27;re stuck together and they start moving at his new velocity. Now we want to find out what is that new velocity. So it&#x27;s just going to cover a concept which is known as momentum. So mo men come. The momentum is signifying what P is equal to simply muss times velocity. So going back to this question, we have the mass moving on its own with a velocity. And then when this mask the second muss, his blue box is loaded onto his car. The two masses are stuck together and they move together at this new velocity. So we&#x27;re using this this concept of momentum mass and velocity to solve with a second velocity. So we have that first muss move in with its own velocity. Plus that second Merce moving up the other velocity, which we are denied zero so they can cancel this and it&#x27;s all conserved because again frictionless room all area that goes in it&#x27;s going to come out. Ana, we use said earlier that those two masses are stuck together so we can add them together under both move in at the same velocity the two, which is what we want to find. So let&#x27;s put it in a different color. No, I just have to really plugging numbers because we know what these two masses are. You know, they&#x27;re both losses before, and we just want to find a velocity after. And we already have this equation, this ratio of hell to conceptualize the school question. So no, we&#x27;re gonna put inner my salon, which is 12,600 it&#x27;s moving out of the last year 18 meters for seconds. So we have our 12,600 kilograms. Don&#x27;t forget your units on. It&#x27;s moving at a velocity of 80 year meters. Person, we got to remember that that first you know that that masters loaded on its initially arrested has no bills. So we can just put on a zero plus zero. It&#x27;s you and then it&#x27;s gonna be equal to island. Those two masses together because you remember, they&#x27;re stuck together. They&#x27;re both moving together. You know, that must additional months is 5350. Just wanna remember, remember that 12,600 We&#x27;re adding that two 5350 mass. It&#x27;s loaded on. And this, oh, times that second velocity. So, you know, plug into center a calculator. We&#x27;re going to get that the tubes. The two is equal twelves 0.6 meters. That&#x27;s our answer.

Kristela Garcia · Answer

so a car initially traveling well. The car, the last one initially traveling at 30 meters per second, has a massive 900 kilograms. So the England of it is 30 meters per second and then the deer which will call Mess too, is 150 kilograms with an initial velocity of 12 meters per second. But it&#x27;s also going in the same direction, so we&#x27;re gonna quickly sketch the situation are dear car strikes dear and they travel off together. So this is a conservation of momentum problem. So the moment of the car, plus the moment of the deer and together they&#x27;re going to have a perfectly in elastic collision, so they&#x27;re gonna have the same final velocity. So we get 900. I was 30 WAAS 150 times 12 divided by 900 plus 150 years is a total final velocity off 27.4 meters per second.

Mayukh Banik · Answer

We will use the conservation off the near moment. Um m one v one plus m does and Jovito equals total mass and want us m two times the final velocity V. From here, we can find the final velocity ease m one view on. So that&#x27;s 900 kilogram times 30 meter per second plus MTV do who use 150 kilogram times 12 meters for a second. The program will cancel out down below with his 900 plus 1 50 kilogram. So altogether we find the final velocity to be 27.4 major for a second.

Ryan Williams · Answer

for this problem. We have our card with certain mass moving in a certain velocity and our snow falling at a given rate per minute. And after a certain amount of time, the snow will accumulate inside of the car and it will slow the car down to some the F and we want. If we want to figure out what that is, we have to figure out how much mass was added to the car. So the snow as falls at a rate of 3.5 kilograms per minute and after 90 minutes, we therefore have to multiply by the time interval. We will get the puddings in your calculator that the cart will gain 315 kilograms of snow after this time interval. So if this is our and one in the sour and to and initially moving at a V, I using conservation of momentum, the initial momentum, which would be m one times be I as equal the final momentum which will be M two times our GF when you want to solve for VF. So our m one times v i over em to will be equal Teoh VF our m one was just the massive the cart, which was 3800 kilograms and also to note initially, we don&#x27;t have to worry about the momentum of the snow since is falling directly, vertically, and here we&#x27;re calculating the horizontal momentum. So therefore the snow has no horizontal initial momentum, so that&#x27;s zero. So going back to your calculation, that&#x27;s our mass times. Our initial velocity, which was 8.6 meters per second, divided by M two, which will be the mass of the cart, which is 3800 kilograms plus the amount of snow that accumulated, which was 315 kilograms. Plugging this all into a calculator will end up giving US 7.94 meters per second for a final velocity.

Problem

(I) A 110-kg tackler moving at 2.5 ms meets head…

Problem 3 Easy Difficulty

(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?

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