# Mining carts momentum problem

• chrizzd21
KE" mean that cart B would have to put in more work than cart A?No, cart B would still have the same amount of work done because the PE is still being converted into KE. In summary, Cart B has more PE to convert into KE on the way down.f

## Homework Statement

Hello all,

I was given a problem which states that there are two identical mining carts, A and B, traveling down a distance L with equal velocities on two identical tracks. There is an identical man in each cart. As the carts travel, it begins to snow, which in turn increases the mass of both carts and leaves snow on the tracks. The man in cart A begins to shovel the snow off the cart, while the man in cart B let's the snow accumulate. Ignoring friction, cart B arrives at its destination before cart A. Using Kinematics, Energy-work theorems, and momentum equations, explain how this could be possible.

## Homework Equations

p = mv where p = momentum
pbefore = pafter

## The Attempt at a Solution

I had tried accounting for an increase in mass for the second cart, cart B, as the snow simply accumulates and isn't brushed off to the side like in cart A. The initial momentum of each cart is equal, but I am not sure how the final momentum of the carts changes.

Welcome to PF!
Cart B has more PE to convert into KE on the way down.

Welcome to PF!
Cart B has more PE to convert into KE on the way down.

Thank you! I had thought about using Energy methods but was unsure how to express it; how would i go about explaining that in an expression? My thoughts were:

PEf + KEf = PEi + KEi
0 + 1/2(Mcart+Msnow)*Vf^2 = (MCart + Msnow)*(Gravity)*(L) + 0

Edit: After looking at it, doesn't the change in mass cancel out the change in potential energy? I've just run some test numbers and no matter how i change it i get the same velocities for cart A and B.

Last edited:
Okay, after talking with my professor quickly he hinted that to actually solve the problem, i need to use conservation of momentum equations and that since the motion is along the x-axis (which i forgot to include in my explanation >_<), the potential energy equation isn't necessarily true :(.

I had started by using the momentum before=momentum after equation, but the velocities for each cart end up remaining the same. The only thing i had thought might work was if i didn't include the added mass of the snow in the final momentum of the cart, which would give Cart B a larger velocity, but i don't think i can just drop the mass of the snow from that side of the equation completely.

Sorry, I gave you a bad idea.
I don't see why energy conservation doesn't apply.
Momentum conservation does not apply because the cars go down; gravity is an external force that increases momentum on the way down.

On the face of it, this is an old Galileo problem. The bigger mass falls at the same rate as the smaller one; if you imagine the masses broken up into equal sized little chunks, they all fall at the same rate and hooking them together doesn't change anything.

Could it be something to do with the fact that snow mass is gradually added?
No, because it is gradually added to both cars equally. Only the shoveling out is different. Shoveling out should not change the speed unless it is thrown forward or backward.

Maybe just that the heavier car with more energy and momentum suffers a smaller % loss in pushing aside the snow on the track?

Sorry, I gave you a bad idea.
I don't see why energy conservation doesn't apply.
Momentum conservation does not apply because the cars go down; gravity is an external force that increases momentum on the way down.

On the face of it, this is an old Galileo problem. The bigger mass falls at the same rate as the smaller one; if you imagine the masses broken up into equal sized little chunks, they all fall at the same rate and hooking them together doesn't change anything.

Could it be something to do with the fact that snow mass is gradually added?
No, because it is gradually added to both cars equally. Only the shoveling out is different. Shoveling out should not change the speed unless it is thrown forward or backward.

Maybe just that the heavier car with more energy and momentum suffers a smaller % loss in pushing aside the snow on the track?

I believe it doesn't apply because there is no change in the y direction, as the carts are going along parallel, level tracks. Basically in the x-y coordinate plane, the two tracks would just be moving to the right, or along the positive x-axis. I told my professor that the heavier cart would have to convert more PE into KE over the same distance, which he agreed with, but he said that wouldn't explain the problem in full detail.

traveling down a distance L
implies going down.
So does "convert PE into KE".
The only answer I see is the bit about pushing snow aside, which would absorb the same amount of momentum from both carts, but would slow the lighter one more.