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Cart Thrust and Momentum?

  1. Dec 3, 2015 #1
    1. The problem statement, all variables and given/known data
    An empty cart travels at 15 m/s as the rain begins to fall. The rain goes into the cart, and adds 10 kg of mass per second.

    1) Will the rain cause the momentum of the cart to change?
    2) Write an equation for the speed of the cart as a function of time.
    3) How fast will this cart be moving after 200 seconds?

    2. Relevant equations
    dp/dt = 0
    delta p = 0 = m(dv/dt) + v(dm/dt)
    a = dv/dt

    3. The attempt at a solution
    1) No, a change in momentum in this system = 0 (as given by my teacher during the quiz). So mass will change due to rain, but velocity will adjust and there will be negative acceleration that will slow down the cart. Momentum, overall, will not change and will stay the same.

    2) v(t) = 15 m/s - 0.15t m/s^2

    0 = m(dv/dt) + v(dm/dt)
    -150 mkg/s^2 = 1,000 kg (dv/dt)
    -0.15 m/s^2 = dv/dt = a
    at = v

    3) v(200) = 15 m/s - 0.15 m/s^2 (200s) = 15 m/s - 30 m/s = -15 m/s
    *If the cart is not on a slope or hill, it is probably already at rest by 200 s. IT has slowed considerably due to the rain.

    This was a quiz my teacher gave, and I only got the first question right. I now have a 33% grade (unfortunately, it's all or nothing => no partial credit...) :/ She also refuses to post a solution sheet or give us the answers, and insists that we find the solutions ourselves to enhance our understanding of the material... Can someone please guide me through what exactly I did wrong on this 3-part quiz? Thanks for any help!
     
  2. jcsd
  3. Dec 3, 2015 #2

    haruspex

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    Where does the acceleration of -.15m/s2 come from?
    To answer this question, you need to know the mass of the cart.
     
  4. Dec 3, 2015 #3
    Oh sorry! The cart is 1000 kg, I forgot to include that. My bad.
     
  5. Dec 3, 2015 #4

    haruspex

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    Ok, so use the result in part 1. What is the initial momentum? What is the mass at time t?
     
  6. Dec 3, 2015 #5
    The initial momentum is m(vi) = 1000 kg (15 m/s) = 15,000 kgm/s, and I'm assuming that stays the same because momentum change = 0.

    As for the mass at time t, there's 10 kg per second, so 200 seconds later, there's 2,000 more g. So the mass at 200s is 3,000 kg, I woud suppose.
     
  7. Dec 3, 2015 #6

    haruspex

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    Yes.
    Yes, but (2) requires this expressed as a function of t, not at a specific time.
    When you have that, use it and the conservation of momentum to find the velocity at time t.
     
  8. Dec 3, 2015 #7
    So wouldn't it be v(t) = 15 - .15t m/s?
     
  9. Dec 3, 2015 #8

    haruspex

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    Why do you think it would be that? I asked you earlier where you got the 0.15 from.

    (Ask yourself what this would make the velocity after 200 seconds.)
     
  10. Dec 3, 2015 #9
    I got -0.15 from looking for dv/dt in the thrust equation. It's also acceleration, so multiplying that by time gives me velocity. So the original speed (15 m/s) minus the potential change in speed (-0.15t) would be what I assume gets the number. Multiplied out, however, it gets to -15 m/s, which is it's either at rest or going backwards.

    um...... v(t) = ... I don't know how to bring mass into that. Should I use a momentum equation?
     
  11. Dec 3, 2015 #10

    haruspex

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    But you used the initial mass and initial velocity. Both change over time, so the acceleration will change.
    If the initial mass is m0 and mass is being gained at rate ##\lambda##, what is the mass at time t?
     
  12. Dec 3, 2015 #11
    ah okay

    m(t) = 1,000 + 10t kg

    ^would that be it?
     
  13. Dec 3, 2015 #12

    haruspex

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    Yes. What equation does that allow you to write using conservation of momentum?
     
  14. Dec 4, 2015 #13
    p(t) = (1,000 + 10t kg) (15 - .15t m/s)?
     
  15. Dec 4, 2015 #14

    haruspex

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    No, the acceleration is not a constant -.15m/s2. Please abandon that.
    You know the initial momentum, you know momentum is conserved, you know the mass at time t. So what is the velocity at time t?
     
  16. Dec 4, 2015 #15
    If we think about conservation of momentum, then the rain water also do have momentum. In actual world the cart will be lowered with increased load, thus changed in direction.

    Just like in inelastic collision.
     
  17. Dec 4, 2015 #16

    haruspex

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    For the purposes of the question, only horizontal momentum is interesting. The rain arrives, we assume, vertically.
     
  18. Dec 4, 2015 #17
    Hmmm..... Initial momentum is 15,000 kgm/s. m(t) = 1,000 + 10t kg.

    v(t) = 15,000 kgm/s / (1000 + 10t kg)

    Is it this one?
     
  19. Dec 4, 2015 #18

    haruspex

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    Bingo.
     
  20. Dec 4, 2015 #19
    BOO-YAH!

    And as follows:

    v(200) = 15,000 kgm/s / (1000 + 10(200) kg) = 5,000 m/s?
     
  21. Dec 4, 2015 #20

    haruspex

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    Right idea, wrong arithmetic.
     
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