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Changing Mass. An open-topped freight car with mass $24,000 \mathrm{kg}$ is coasting without friction along a level track. It is raining very hard, and the rain is falling vertically downward. Originally, the car is empty and moving with a speed of 4.00 $\mathrm{m} / \mathrm{s}$ . What is the speed of the car after it has collected 3000 $\mathrm{kg}$ of rainwater?

(a) $v_{2}=3.56 \mathrm{m} / \mathrm{s}$

(b) The rain is falling downward and collecting the wateron the car increasing its total mass.

since the momentum is conserved in the horizontal direction and we know that the momentum depends on the mass and the velocity, so when the mass increases, the velocity must decrease so the momentum stays constant.

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in this question and open topped fright car with a mass off 24,000 kg is moving to the right. But it's moving to the right in the rain and it's moving initially with a velocity of 4 m per second. But after some time it will accumulate an amount of water that is equivalent to 3000 kg, and now it will be moving to the right with some other velocity v off course. It accumulates water because it's raining and it's open topped. Now all we have to do is evaluate. What is that new velocity? For that, you have to remember that the net momentum is conserved, so the net momentum before is equal to the net momentum after the net momentum before conceits off the momentum off that empty fried car on the momentum off the empty fried car is given by the mass times the velocity and now, So let me say that everything that is moving or that it's pointing to the right it's positive. So this is my reference frame. Then, in my reference frame, the momentum off the fright car before is given by the mass off the empty fried car 24,000 kg times the velocity which is 4 m per second now after it accumulated 3000 kg off water. We have another momentum that is given by the mass off the car, plus the mass off the water. So 24,000 plus 3000 and then they multiplied by the new velocity. So, in order to evaluate the new velocity, all we have to do with solve this expression for V and this is very easy to do. And as we can see, V is given by 24,000 times four divided by 27,000. Then, by solving that calculation, you get a velocity off approximately 3.56 m per second on this is the answer to this question.

Brazilian Center for Research in Physics