1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mining carts momentum problem

  1. Nov 15, 2011 #1
    1. The problem statement, all variables and given/known data
    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 lets 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.


    2. Relevant equations
    p = mv where p = momentum
    pbefore = pafter



    3. 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 im not sure how the final momentum of the carts changes.
     
  2. jcsd
  3. Nov 15, 2011 #2

    Delphi51

    User Avatar
    Homework Helper

    Welcome to PF!
    Cart B has more PE to convert into KE on the way down.
     
  4. Nov 15, 2011 #3
    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: Nov 15, 2011
  5. Nov 16, 2011 #4
    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 dont think i can just drop the mass of the snow from that side of the equation completely.
     
  6. Nov 16, 2011 #5

    Delphi51

    User Avatar
    Homework Helper

    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?
     
  7. Nov 16, 2011 #6
    I believe it doesnt 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.
     
  8. Nov 16, 2011 #7

    Delphi51

    User Avatar
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Mining carts momentum problem
  1. Cart Problem (Replies: 5)

Loading...