🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning Numerade Educator ### Problem 35 Easy Difficulty # Two cars, one a compact with mass 1200$\mathrm{kg}$and the other a large gas guzler with mass 3000$\mathrm{kg}$, collide head on at typical freeway speeds. (a) Which car has a greater magnitude of momentum change? Which car has a greater velocity change? (b) If the larger car changes its velocity by$\Delta v,$calculate the change in the velocity of the small car in terms of$\Delta v .(\text { c) which car's occupants }$would you expect to sustain greater injuries? Explain. ### Answer ##$2.50 \Delta v\$

#### Topics

Moment, Impulse, and Collisions

### Discussion

You must be signed in to discuss.
##### Andy C.

University of Michigan - Ann Arbor

##### Marshall S.

University of Washington

Lectures

Join Bootcamp

### Video Transcript

in this question, a compact car with a mass off 1000 and 200 kg collides had to head with a big car that has a mass off 3000 kg. Both cards were traveling a typical freeway speeds, say V. Then in the first item off this question, we have to deter mined. Which car has the greater change in the momentum and which car has the greater change in the velocity? For that, we have to use the principal off momentum conservation, which tells us the following the net momentum before the collision is he goes to the net momentum after the collision. So in equations, it tells us that p net before the collision is equals to P net after the collision. This is a very simple principle. Okay, now, to use that principle, we have to think about what is the contributions to the net momentum before the collision. So before the collision, both cars were moving, one car was moving to the right and another car was moving to the left. Let me choose a reference frame such that everything that points to the right is pointing to the positive direction in that reference frame. We can say the following the momentum off car number one will be positive because its velocity is pointing to the positive direction off my reference frame and the momentum off car number two will be negative because its velocity is pointing to the negative direction off my reference frame. Then the calculations goes as follows. So the net mo mentally for the collision is given by the momentum off car number one, which is just it's mass. 1000 and 200 kg times its philosophy, which is V plus the momentum off car number two before the collision, which is given by its mass 3000 kg times its velocity. The velocity of car number two is minus V because it's going to the left on its the negative direction off my reference frame. Then, after the collision, what happens is that both cars will be traveling together. Then, after the collision, we can say the following the net momentum is given by the momentum off this composition. So it's given by the mass off car number 1 1000 and 200 kg, plus a mass off car number two, 3000 kg times the velocity off both cars after the collision. I don't know. What is that velocity? Okay, now we start to solve this equation to get a relation between Big V and small V. So we want to get a relation between the velocity after the collision and the velocity before the collision. Now, let's do it. So we're using these equations. We can say the following. So adding these two terms leads to minus 1000 and 800 times small V is equals to 4000 and 200 times Big D the reform. We can say that big V is equal to minus 1000 and 800 divided by 4000 and 200 times small V. Now we can cut out the zeros from this fraction like this. So we conclude that big V is equal to minus 18 divided by 42 times small V. Okay, what now? So now we can calculate what is the change in the velocity off both cars? So let's calculate what is the change in the velocity off car number one? So Delta V number one don't have you. Number one is equals to the velocity off car number one After the collision and that velocity is equals to big V, which is minus 18 divided by 42 times We So the velocity after the collision, minus 18 by 42 times we minus the velocity before the collision which was a cost to V then this gives us a following. This is minus 18 divided by 40 to minus one times we on this is equals to minus 18 divided by 42 Plus. Now we multiply and divide one by 42 so we can get 42 divided by 42 times small V. Now all we have to do is at 18 with 42 on these results in 60. So the variation in the velocity off car number one is because to minus 60 divided by 42 times is more V. Okay, Now, what is the change in velocity off card number to the change in velocity off car number two follows the same logic it is given by the velocity off our number two. After the collision which was minus 18 divided by 42 times, we minus velocity off car number two before the collision before the collision, the velocity off car number two was minus V. So here we have minus minus V on diseases minus 18. Divided by 42. Plus one on this wolfing is multiplying V. Okay, Now again we multiply and divide one by 42 so you can get the following minus 18. Divided by 42 plus 42 divided by 42. This is multiplying V. Then we get the following minus 18. Plus 42 is equals to 24. So here we have 24 divided by 42 times V. This is the change in the velocity off car number two than I ask you. Doubt of you. One is bigger 10 or smaller than out of it. You. Well, you might tell me that it is smaller because we have a minus sign here. Okay, I agree with that. But then comparing the magnitudes off the change in velocity so the magnitudes ignore that minus sign. So the magnitude in the change of velocity off car number one is 60 divided by 42 times we and the magnitude in the change of velocity off car number two is equals to 24 divided by 42 times. We now by looking at the magnitudes, it's clear that car number one suffered a change in the velocity that is much bigger than the change in the velocity that was suffered by car number two. So the conclusion is car number one suffers the biggest change in the velocity now. The next point is the following. Which car has a greater magnitude of momentum change? The momentum change is given by the following Delta P is equals to the mass times Delta V. This is because the mass will not change the reform. That change in the momentum is just equals the mass times the change in the velocity. So the change in the momentum off car number one is equal to the mass off car number 1 1000 and 200 kg times they changed in the velocity off car number one, which is minus 60 divided by 42 times V. The reform. The change in the momentum off car number one is equals to minus 72,000, divided by 42 times we. Now the changing the momentum of car number two is given by 3000 times the change in the velocity off car number two, which is 24 divided by 42 times V For the change, the momentum off car number two is equals to 72,000, divided by 42 times. We then, as you can see, the change in the magnitude off the momentum off both cars is the same notice that I'm talking about. Magnitudes and the magnitudes ignore that minus sign. This was unexpected conclusion because momentum is conserved. So if the momentum off car number one decreases by an amount, say 15 it means that the momentum off car number two must increase by 15 to compensate for that change in the momentum off car number one. And this is the first item off this question. So car number one so first, the biggest magnitude in the change of velocity. But the changing the magnitude of momentum off cars number one entry is the same. Now we go to the next item this question. So in the second item, we suppose that the change in the velocity off car number two is equals to Delta V. Now in terms off Delta V. What is the change in the velocity off car number one? That is what is Delta V one. For that, we can use that fact that I just mentioned about the change in the momentum. So the momentum is conserved. Therefore, any change in the momentum off car number two must be compensated by a change in the momentum off car number one such that the net change in the momentum is equal to zero because the momentum is conserved. Now let's do it. Given that the change in the velocity of car number two is Delta V, what is the change in the mo mentum off car number two. So the change in the momentum off car number two delta P two is equal to the mass off car number two times it's change in the velocity. Okay. Now, as you can see, the momentum off car number two will increase if out of this positive on decrease if out of its negative. So the opposite must happen with the momentum off car number one. That is the change in the momentum off our number one must be equals to the opposite. Change the momentum off card number two. This must be the case if you want that, the next change in both momentum's is equals to zero. And this must be the case because momentum is conserved. Okay, then, using that fact, you can say that the change in the momentum of car number one is a close to minus 3000 times Delta V. But you can also say that the change in the mo mentum off car number one is equals to the mass off car number one, which is 1000 and 200 kg. Times they change in the velocity off car number one, the reform, where you can say that 1000 and 200 Delta V one is equal to minus 3000 Delta V. The Reform Delta V one is equal to minus 3000, divided by 1000 and 200 times Delta V By simplifying the zeros you get. The following Delta V number one is equals to minus 30 divided by 12 times Delta V. You can divide both 30 and 12 by six, and by doing that, you get five year and to hear. So Delta V one is minus 5/2 times Delta V on. This is minus 2.5 times Delta V and this is the answer to the second item off this question. So Delta, if you want, is a question minus 2.5 times Delta V. This is the relation that we want. Now we go to the next item off this question. It asks us about the damage that the car's occupants would suffer during this collision. So both occupants off cars number one, and you should suffer some damage. But who suffers the most? Well, you have to think about Newton's second law. Newton's second law tells us that the net force that is acting on something is equal to the mass off that thing times the acceleration off that thing. The acceleration is directly related to the change in the velocity. Acceleration, in fact, is given by the change in the velocity divided by the time it took for that change to happen. Okay, so as you can see, the net force is proportional to the change in the velocity. Therefore, we can say that the occupants off car number one will suffer the most because the change in the velocity off car number one is 2.5 times bigger than the change in the velocity off car number two the reform. The force that the occupants off car number one should feel will be 2.5 times bigger than the force that the occupants off car number two would feel the reform. The occupants off car number one will suffer the most injuries on this is the answer to this question.

Brazilian Center for Research in Physics

#### Topics

Moment, Impulse, and Collisions

##### Andy C.

University of Michigan - Ann Arbor

##### Marshall S.

University of Washington

Lectures

Join Bootcamp