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Relative motion

  1. Nov 16, 2008 #1
    Hello everyone,
    My teacher posted this question and I've been having problems solving it. Please help.

    1. The problem statement, all variables and given/known data
    3 kids are standing on top of a cart. The cart is on top of a frictionless surface. The kids run WEST at a maximum of 10m/s. As such, the cart will be pushed in the EAST direction. So all three kids run at this maximum speed, at which time one of the kids jump off, then the 2nd, then the 3rd. I need to know the velocity of the cart after the 1st, 2nd, and 3rd kid jumps off. The kids each weigh 60kg and the mass of the cart is 120 kg. Conservation of momentum is applied here.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 17, 2008 #2

    gabbagabbahey

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    You need to show an attempt to get assistance here... You say in your post that conservation of momentum should be applied, so what do you get when you try to apply it?
     
  4. Nov 17, 2008 #3
    I'm going to use the following denotations (it's easier to type)
    x= gVc (initial Velocity of cart relative to ground)
    y = gVc' (final velocity of cart relative to ground)
    Mc = mass of cart
    M3k = mass of 3 kids, 2 kids, etc
    gVk = cVk + gVc (velocity of kids relative to ground = kids velocity relative to cart + cart velocity relative to ground)

    Mcx + M3k gVk = Mcy+ m2k gVk'

    I know that the cart will be moving faster as each of hte kids jump off, because less mass, the kids all run @ 10 m/s, therefore the cart must be moving faster.
    this is the part where i'm stuck. Do I need to caluclate the initial velocity from rest or from the max speed the system can move with all3 kids on it b4 the kid sjumps off?
     
  5. Nov 19, 2008 #4
    Could somebody please point me in the right direction?
     
  6. Nov 19, 2008 #5

    gabbagabbahey

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    Using your notation, I think it is best to divide the problem into two parts:

    (1) What can you say about Mcx + M3k gVk (the total initial momentum of kids and cart system relative to the ground)?
     
  7. Nov 19, 2008 #6
    Based on the numbers, I find that 120kg * 6m/s + (180 kg * (-10m/s-6m/s) = zero

    which doesn't make any sense and it's where i'm stumped. Based on this, the system is initially not moving?
     
  8. Nov 20, 2008 #7

    gabbagabbahey

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    Individual components of the system (each kid, the cart) are moving relative to the ground, but the system as a whole isn't....in fact this is what you should have used to determine that x=6m/s; which makes me wonder where you actually did get that number from (it's right, but how did you get it)?
     
  9. Nov 20, 2008 #8
    I just assumed that the system starts @ zero, so...

    Mcx + M3k gVk = Mcy+ m3k gVk'
    0 = Mcy + m3k gvk'
    0 = 120kg y + m3k (cVk + y)
    find that the system is moving @ 6m/s.

    Where i'm stuck is....how do I find the speed of the cart after one of the kids jump off? I figured that since mass is decreased after, velocity should increase, but I don't know the math to prove it
     
  10. Nov 21, 2008 #9

    gabbagabbahey

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    Well, if the total momentum of the system relative to ground is zero before each kid jumps off, then it is zero after each kid jumps off too....use that.
     
  11. Nov 21, 2008 #10
    Mcx + M3k gVk = Mcy+ m2k gVk'

    0 = 120 kg y + 120kg (cvk' + y)
    y = 5 m/s

    which is wrong, cause the cart's supposed to be moving FASTER. which is where i'm stuck
     
  12. Nov 22, 2008 #11

    gabbagabbahey

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    Isn't Mcy+ m2k gVk' just equal to 120kg y+ 2*60kg*(-10m/s)?
     
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