Understanding Momentum: Solving Simple Problems with Rocks on a Railroad Car

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The discussion centers on the principles of momentum as applied to a railroad car loaded with rocks. When rocks are thrown horizontally from the car, it speeds up due to the conservation of momentum. However, when rocks fall vertically through a hole in the car, the car's speed remains constant because the system is not closed; gravity acts on the rocks, but no external force affects the car's horizontal motion. The key takeaway is that momentum is conserved in a closed system, and since the car does not experience a net force in the horizontal direction, its velocity remains unchanged.

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1. A railroad car loaded with rocks coasts on a level track without friction. A worker on board starts throwing the rocks horizontally backward from the car. Then what happens?
(a) car slows down
(b) car speeds up
(c) first the car speeds up and then slows down
(d) car's speed remains constant

2. Which answer would you choose if the rocks were to fall out through a hole in the floor of the car one at a time?

So the answer to number 1 is (b) speeds up and the answer to number two is (d) car's speed remains constant. I understand number one but I can't understand why number two would be (d) using the following equation for momentum:




Pi=Pf
Here's what I was thinking please tell me where I'm going wrong:
(c=car and r=rock)

McVc + MrVr = McV'c + MrV'r

Initially Vc=Vr=Vi
and since rocks are being dropped directly downward, final momentum of the rocks (being dropped one at a time) should be zero because their falling down, out of the railroad car.
So:

Vi( Mc+Mr) = McV'c

V'c = [Vi(Mr+Mc)] / Mc


But wouldn't that increase the velocity? I know logically thinking it wouldn't but I can't seem to get this equation to show that.

Someone please help explain this to me!
 
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You have to remember, momentum is a vector quantity, so having rocks fall down in the vertical direction doesn't change the car's velocity in the horizontal direction.
 
sinclair18 said:
and since rocks are being dropped directly downward, final momentum of the rocks (being dropped one at a time) should be zero because their falling down, out of the railroad car.
They're dropping straight down relative to the car. To an observer on the ground, they have the same forward speed as the car (until they hit the ground, of course, but that doesn't affect the car's motion).
 
Your second equation must include the final momentum of the rocks and of the Earth/objects the rocks hit. Remember, momentum is always conserved in a CLOSED system, meaning there are no external forces. So, if your system only includes the train and the rocks, it is not a closed system (gravity is acting on the rocks as they fall, speeding them up and momentum "disappears" once they hit the ground). But if the Earth is included, the system is closed and the momentum lost by the train as the rocks fall is preserved by a miniscule gain in velocity by the Earth. So the train loses mass (rocks) and thus momentum, the Earth and rocks attract each other equally but in opposite directions, and the fallen rocks transfer momentum to the Earth itself or objects on the ground once they hit.
 
Last edited:
The car itself doesn't change velocity in the second situation because no force is applied to it.
 
McVc+MrVr=McV'c+MrV'r+P'

Where P' equals any momentum gained by the objects hit by the rocks. So, there is no need for the car to speed up: the momentum lost by the car/rocks is gained elsewhere.
 
BrainSalad said:
The car itself doesn't change velocity in the second situation because no force is applied to it.

This is enough to answer question 2. Why is there a need to use momentum formula at all? No force no acceleration, no change in velocity.
 

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