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Homework Help: Question involving conservation law and springs!

  1. Aug 20, 2010 #1
    1. The problem statement, all variables and given/known data
    A freight car of mass 1800kg is timed at 4.2m/s just after it runs into a spring-loaded bumper at the end of the track. The bumper consists of an 800kg mass that the car runs into, and a pair of large springs. The car travels 2.4m before coming to rest.

    a)using an appropriate conservation law, find the speed of the freight car before it struck the bumper.

    2. Relevant equations

    i'm not sure, conservation of momentum? or kinetic energy?

    3. The attempt at a solution
    Before it struck the bumper, it's moving at a certain speed(unknown) at a given mass. hmmm...looking at what's given to find the initial speed we have to use conservation of momentum: m1v1 +m2v2 = 1800v1 + 800(4.2)

    Does that look right? Can someone guide me please?
     
  2. jcsd
  3. Aug 20, 2010 #2

    kuruman

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    It is indeed conservation of linear momentum, but you are not applying it correctly.

    You need to say Pbefore = Pafter.

    Can you find expressions for Pbefore and Pafter?
     
  4. Aug 20, 2010 #3
    ok. so p before = p after

    m1v1 + m2v2 = m1v1 + m2v2

    It's an elastic collision, right?

    i calculated it and i got speed before = 10.9m/s

    There's a second part to this asking for the time taken for the car to be brought to rest. do i have to consider the springs?? or can i just use kinematics equations
     
  5. Aug 20, 2010 #4

    kuruman

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    This equation is an identity. Since the terms on each side are the same it basically says 0 = 0. Can you show me how you got 10.9 m/s?

    We will worry about the second part after it is certain that you got the first part correctly.
     
  6. Aug 20, 2010 #5
    pbefore = pafter

    m1v1 +m2v2 = m1v1 + m2v2
    1800v1 = 1800 *4.2 +800v2
    1800v1 = 7560 +800v2

    using conservation of energy:
    v1i - v2i = -v1f + v2f
    v1i - 0 = -4.2 + v2f

    solve for v2f. v2f = v1i + 4.2

    1800v1 = 7560 + 800(v1i + 4.2)
    1800v1 = 7560 + 800v1 + 3360
    1000v1 = 10920
    v1 = 10.9m/s
     
  7. Aug 20, 2010 #6

    kuruman

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    Picture the situation right after the collision: The car pushes on the bumper which pushes on the springs that get compressed until the car stops. If the car is moving initially at 4.2 m/s just after the collision, what do you think is the speed of the bumper?

    energy is not conserved because the masses stick together and stay stuck. Besides, this is not an expression of energy conservation. What is the equation for kinetic energy? Check your textbook.
     
  8. Aug 21, 2010 #7
    the speed of the bumper would b 4.2m/s?

    This is an inelastic collision then?

    m1v1 + m2v2 = (m1 + m2)v2
    1800v1 = (1800 + 800) 4.2
    v1 = 2600/1800
    v1 = 1.44m/s
     
  9. Aug 21, 2010 #8

    kuruman

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    Yes.
    Correct, but to avoid confusion, use one symbol to denote one and only one variable. Here v2 stands for the velocity of the bumper before the collision (which is zero) and for the common velocity of the masses after the collision (which is 4.2 m/s). This habit might get you in trouble in the future.
    Correct.
    What happened to the "4.2" in the previous equation?
     
  10. Aug 21, 2010 #9
    ooops, i forgot to multiply 4.2. The answer i get is 6.01m/s.
     
  11. Aug 21, 2010 #10

    kuruman

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    Good. For the second part, you cannot use the kinematic equations because they are good for constant acceleration. Masses at the ends of springs do not experience constant acceleration. The quickest way to get the second part is energy conservation.
     
  12. Aug 21, 2010 #11
    So using PEg + PEs + KE = PEg + PEs + KE

    but we need to find time.

    How do we do that?
     
  13. Aug 22, 2010 #12

    kuruman

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    What fraction of the period T is the time that you are looking for?
     
  14. Aug 22, 2010 #13
    I don't quite understand what you mean about what fraction? All i know is that the car travels 2.4m before coming to rest. Do i have to find the final velocity and find time from that?
     
  15. Aug 22, 2010 #14

    kuruman

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    When it comes to rest, does it remain at rest or is it instantaneously at rest and then goes back where it came from?
     
  16. Aug 22, 2010 #15
    it goes back and forth since the bumper is attached to the spring
     
  17. Aug 23, 2010 #16

    kuruman

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    OK, in that case it has a period T which is the time required for a complete round trip. Here the masses go from the point of collision to maximum compression of the springs, which is not a round trip, just a piece (fraction) of it. What piece (fraction) of the period T then is the corresponding time it takes the masses to go from the point of collision to the maximum compression of the springs?
     
  18. Aug 23, 2010 #17
    we want half of the period...is that right?
     
  19. Aug 23, 2010 #18

    kuruman

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    A period is the time for a round trip. Therefore

    From maximum stretching of the springs to maximum compression it is ______ of a period. So ...
     
  20. Aug 23, 2010 #19
    isnt' it half? because one period would be from stretch to compression to stretch...

    sorry if i'm not catching on right away =(
     
  21. Aug 23, 2010 #20

    kuruman

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    Half a period is from maximum stretch to maximum compression. When the car hits the block, the springs are relaxed, not at maximum stretch. So ...
     
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