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California State Polytechnic University, Pomona

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Problem 18 Easy Difficulty

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a $1750-\mathrm{kg}$ car traveling to the right at 1.50 $\mathrm{m} / \mathrm{s}$ collides with a $1450-\mathrm{kg}$ car going to the left at 1.10 $\mathrm{m} / \mathrm{s}$ . Measurements show that the heavier car's speed just after the collision was 0.250 $\mathrm{m} / \mathrm{s}$ in its original direction. You can ignore any road friction during the collision. (a) What was the speed of the lighter car just after the collision? (b) Calculate the
change in the combined kinetic energy of the two-car system during this collision.

Answer

(a) $v_{B 2}=0.409 \mathrm{m} / \mathrm{s}$ to the right.
(b) $\Delta K=-2670.26 \mathrm{J}$

Discussion

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Video Transcript

{'transcript': "once again welcome to a new problem at this time. We have two cars will given two cars, driving in opposite directions. You have a bigger car with the mass going to call it em. A happens to be seventeen fifty kilograms, and then it has of last e v a, which you call the initial velocity moving in the, uh in the right direction. It's a positive velocity, so that's gonna be one point five meters per second. Positive. That's what we're given. And then we're also given a second car. I'm going to call it B aunt. It's driving towards the left, has a mass off one thousand four hundred and fifty kilograms, and then it has an initial velocity in the left off one point one meters per second. But I'm gonna have it is negative because it's moving in the opposite direction. So this is positive and that's negative. But the other thing that were given is that thes two cars, they're gonna bump onto each other so there's gonna be some bumping going on, though they'LL collide. Okay, so this is kind of like the before on then, after you know, after what's they collide So this is the second car. When they collide, the final velocity off the fast car is still in the positive direction, but this time it's reduced to zero point two five meters per second for the other car after collision. So we're saying this is what happens after collision. It's going to be going. So both of them are now traveling in the same direction. So the final velocity of the second car happens to be. It happens to be if you check that, you know, we don't know what the fine of lost is going to be. So that's what we need to find out. Look at want. I want to find out what happened afterwards. Ah, and then also, we want to find the change in kinetic energy. Okay, remember, there's no friction in this scenario, so that means we don't have any losses, losses of energy in the system. So the the fact that there's no friction, the momentum is conserved. Okay, we have, um we have conservation of momentum, you know, that's that's what's going on in this case. Change in Connecticut and you simply is the final kinetic energy of the system, minus the initial kinetic energy of the system on DH. That means we have to account for both the kinetic energy off the two, the two cars. So gonna kick off one part, eh? And the sea that we're applying the so momentum mentum after a collision happens to be exactly the same as the momentum before condition. And so the momentum after collision is the Marcel's eight times the final velocity off, eh? Plus the muscles. The times, the final velocity of B, which is a second car and then before condition. We have mass off eight times the initial velocity of a plus the Marcel's B times. The initial blast. You'LL be or goes to find the final velocity of B, which is a second car, and we know it's it's moving, going to see how it's moving. It could be moving towards left. It was right. Just depends on the sign off off the solution. So this is m A ve a plus M B of the FBI minus. So moving this one to the right m a the a final. We have all that information and we want to divide everything by the Marcel's beat that gives us our final velocity of B for purposes of writing the numbers in the problem. We're going to jump into the next page to get the V fi final velocity of B of the second car. And if you can see from that, we have the mass off a seventeen fifty kilograms. Okay, we have seventeen fifty kilograms times the velocity off a which is one point five meters per second. The initial velocity of a one point five zero meters per second on DH then also want to add the muscle B fourteen, fifty kilograms times the velocity initial velocity of B one point one zero meters per second on DH. Then there's also another portion ofthe problem which leads to be subtracted and that happens to be the mass off A. On the final velocity of hay, the mass of a and the final velocity of sa was subtracting that seventeen fifty kilograms times. The final velocity of a happens to be if you go back on this side zero point two five meters per second, zero point two five meters per second on all of that is going to be divided as you can see by the mass will be always a good habit to leave the variables up until the end. So this is fourteen fifty kilograms. When you plug in the numbers, you find that the velocity of light a car ends up being zero point for zero eight, eight six, eight, seven. So let's let's dislike sure that this is this is the right the right process. So we have seventeen, fifty time sworn point five zero plus fourteen fifty no times negative one point one zoo, then minus seventeen fifty times point two five. So we and then we have to divide by fourteen fifteen. That gives us a zero point force, your eight six. That's the final lost yield off the smaller car. And it's positive, meaning that it's moving towards the right meters per second Member was still under the law ofthe, ah, conservation of momentum. Okay, and so that's That's what we're using in this case in the second part of the problem. That's but be I'm going to see how the change in Connecticut and happens. That's just the Connecticut and the final off the system minus Connecticut initial off the system and so the final kinetic energy it is um, the Connecticut. The final kinetic energy. Both the bigger car, which we're calling a final plus the final kinetic energy off the smaller car minus the final. Our initial kinetic energy off the big car, plus the initial Connecticut Angie off the a smaller car. So that's that's the process right here. So this is one half Emmy V a final squared plus one half M b v be final squared minus Always used rockets when you have to one half em the initial squared plus one half em be Phoebe initial squared. So that's our change in Connecticut. Do we can simplify the whole thing by moving things around. So we have one half emi The final squared this one right here. I want to put them together minus one half Emmy be a initial squared and then bring this on Plus one half em be the the final squared minus one half em be be nation squared. So what we're doing right now we're doing algebra, were doing a distribution of the miners to both of them and then putting all the airs and all the bees together. So these are there's a big one half outside And then we have Emmy of the final squared minus the initial squared. Plus Oh, you have to change this part right here because it's a single we pulled. We pulled out the one half outside because it shows up in all of them. So we want to have that. So this is plus em be vehbi. Um, I should have called this V b final. So, you know, it looks like something a little bit different. It's a subscript Oh, so we have to bring it back down their final squared like that. Get rid of this. This one right here. This square will need a single one. So in this case, Vehbi final squared minus V b initial square. Don't forget that square right there. And that's our whole problem. Next part is we just need to plug in numbers. We could use the difference between two squares if you wanted to. But again, that's optional up up to you. So the change in kinetic energy becomes equivalent to one half. I will plug in the numbers muscle, a hiss seventeen fifty kilograms and then the plug in the final velocity of A, which is zero point two five meters a second. You know you want to square. I want to square that results. So we have to be very focused on how we simplify this. So you know this this whole thing is spread and then minus the initial velocity was one point five. That's for a could see. The initial velocity was one point five meters per second on That's also going to be squared and then go back. We see we made a plus must be which is forty two fifty kilograms on then the final velocity off B. We need to get the final blast of future zero point four zero eight six, zero point four zero eight six meters per second, squared minus. If you go back initial velocity off of B, that's at the beginning. Where we see it's negative one point one zero meters per second, only one for it, one zero me this second and we also need to square that one. We plug in the numbers, plug in the numbers, you get to see that the final final. Um so we have seventeen. Fifty. My start was simplify. This is one point five and so the final final becomes negative. Twenty six. Seventy round, About point three jewels. Okay, so this is the change in kinetic energy. That's the change in kinetic energy That happens when you have to, Because moving in the opposite directions and then they bump into each other. Okay. Uh, so negative. Two six, seven point three. Jewels are kilograms. Hope you enjoyed the video. Feel food to share any questions. Comments? Looking for tea, uh, talking to you in the next video and have a wonderful day, okay."}