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Force train problem

  1. May 18, 2013 #1
    1. The problem statement, all variables and given/known data
    A railway locomotive pushed a stationary wagon to accelerate it to a speed of 4.00 m s-1. The engine then applies an equivalent force over the same time on a second wagon to accelerate it from rest to a speed of 6.00 ms-1. The second wagon rolls down the track, collides with the first, and the two lock together. The loco must now push the two stationary wagons to the other end of the shunting yard. What speed could the two railway carriages achieve if the loco applies the same impulse?


    Answer is 4.8ms-1, however I'm not sure how they got it. (answerbook)


    2. Relevant equations
    F=ma


    3. The attempt at a solution
    If F is constant between both equations and you are given A, that means M must change

    so
    6= F/M1
    4=F/M2
     
  2. jcsd
  3. May 18, 2013 #2

    haruspex

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    There's no mention oif acceleration, so F=ma is not going to be useful. Do you know any equations concerning momentum?
    If the engine provided momentum I to the first wagon, how much momentum did it provide to the second? What
    does this tell you about the two masses?
     
  4. May 18, 2013 #3
    m1 u1 + m2 u2 = m1 v1 + m2 v2

    the two masses are the same?
     
  5. May 18, 2013 #4
    What's the relation between force, time and momentum change?
     
  6. May 18, 2013 #5

    Impulse?
    I'm not sure
     
  7. May 18, 2013 #6
    Then, what's the relation between impulse and momentum change?
     
  8. May 18, 2013 #7

    CompuChip

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    I guess ap123 would like to see an equation. You already quoted F = ma but just that is not going to get you there.
     
  9. May 18, 2013 #8
    Impulse = FT= m(v-u) = ΔP

    F is the force
    T is the time
    M is the mass
    v is the final velocity
    U in the initial velocity
     
  10. May 18, 2013 #9
    Good :-)

    We don't know F and t, and the initial momentum is zero, so just write J = P

    Apply this to the first and second wagons individually and to the 2 joined wagons and see what you get.
     
  11. May 18, 2013 #10
    What's J?
     
  12. May 18, 2013 #11
    J is the impulse
     
  13. May 18, 2013 #12
    Sorry but I'm not really following.

    could you give me the formula I should use?
     
  14. May 18, 2013 #13

    haruspex

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    1. You are told that the engine applied the same force for the same time to the second wagon. If it gave an impulse (change in momentum) to the first wagon, what impulse did it give to the second. (Use J = ∫F.dt, or, since we take the force as constant here, J = F*t.)
    2. Another formula for change in momentum is the m (v-u) you quoted. What equations using J, M1, M2 and the given speeds does that give you?
    3. What equation using J, M1, M2 can you write down for the case where the engine pushes both?
     
  15. May 19, 2013 #14
    1. J=F*t
    This could be rearranged as, J = Ft = mv - mu
    The first carriage, m1 : Ft=(m1)(4m/s)
    The second carriage, m1: Ft=(m2)(6m/s)
    So 4m1 = 6m2 or m1= 1.5m2

    now m1u1+m2u2=(m1+m2)v
    so (4m/s)+(6m/s) =(m1+m2)v and incorporating the relationship that m1=1.5m2

    then (1.5)(4m/s)(m2)+(6m2)=(1.5)(m2+m2)v

    (m2 will now cancel), leaving v=12/2.5 = 4.8 m/s

    Thanks a bunch for your help. If you could please show me any other methods for doing this that would be great.
     
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