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Momentum problem due this Tuesday

  1. Dec 2, 2006 #1
    Momentum homework problem due this Tuesday

    1. The problem statement, all variables and given/known data
    Here's the question:
    The surfaces are frictionless. The tracks are 60 degrees from horizontal. A 365 kg mass is released from rest on a track at a height 4.2 m above a horizontal surface at the foot of the slope. It collides elastically with a 480 kg mass initially at rest on the horizontal surface. The 480 kg mass slides up a similar track. The acceleration of gravity is 9.8 m/s^2. (A.) What is the speed of the 480 kg block immediately after the collision? Answer in units of m/s. (B.) To what maximum height above the horizontal surface will the 480 kg mass slide? Answer in units of m.


    2. Relevant equations
    (I'm assuming by relevant equations, you mean the ones that I think are relevant. If you mean equations given in the problem, then disregard these two equations)
    GPE= mgh
    KE=(1/2)mv^2


    3. The attempt at a solution
    (A.) I found the gravitational potential energy of the first block mgh=365 * 9.8 * 4.2 and then set that equal to (1/2)(365)(v^2) and solved for v. Is that the correct way to solve for the velocity of the 480 kg block since they collided elastically? (v=9.073 m/s)

    (B.) I would assume that since the collision is elastic, the height the 480 kg mass would reach would be 4.2 (the height that the 365 kg mass started at). I feel like that is way to easy and that I must not be seeing something. (h=4.2m)
     
    Last edited by a moderator: Dec 2, 2006
  2. jcsd
  3. Dec 2, 2006 #2

    radou

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    For (A): by using energy conservation and solving for v, you get the velocity of the 365 kg mass when it reaches the horizontal plane. Now you know the velocity of that mass before the collision. Use conservation of ________ and _______ ______ to solve for the velocities after the collision.

    For (B): use energy conservation again.
     
  4. Dec 2, 2006 #3

    Hootenanny

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    Hi FizzixIzFun, welcome to the forums,

    Just for future reference there are Homework forums dedicated to homework questions.
    Your on the right track, but not quite there yet. You calculated the potential energy correctly, and you are correct to assume that this energy is transferred to the kinetic energy of 480kg object. However, you have note solved correctly for the velocity. What you have found is the velocity of the 365kg object not the 480kg object as you are required to do.
    Indeed you are missing something. Compare the gravitational potential energy of the 360kg object with that of the 480kg.
     
  5. Dec 2, 2006 #4
    Ok, I think I got it. (A.) I use (365)gh = (1/2)(480)v^2 and solve for v. (B.) I use (1/2)(480)v^2= (480)gh and solve for h. Is that correct?
     
  6. Dec 2, 2006 #5
    Ok, I think I got it. (A.) I use (365)gh = (1/2)(480)v^2 and solve for v. (B.) I use (1/2)(480)v^2= (480)gh and solve for h. Is that correct?
     
  7. Dec 2, 2006 #6

    radou

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    The equation you wrote implies that the velocity of the 365 kg object equals zero after the elastic collision, which is not true.
     
  8. Dec 2, 2006 #7
    OK, I'm lost. I don't know what to do.
     
  9. Dec 2, 2006 #8

    radou

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    Ok, let's slow down. Let the 365 kg mass be m1 and the other one be m2. What we can do first is use conservation of energy to find the velocity of the mass m1 when it reaches the horizontal plane. The other mass, m2, is at rest, right? The they collide. Since the collision is elastic, you know that momentum and kinetic energy must be conserved. You must use these facts to calculate the velocity of the mass m2 after collision. (v1 has some other velocity, which isn't of interest to us.) After you do so, you can use energy conservation again to calculate the height the mass m2 climbs up to.
     
  10. Dec 2, 2006 #9
    Here's what I've come up with. Using m1gh=(1/2)m1v1^2, I solve for the velocity of m1 right before the collision. Then, I do m1v1=m2v2 and solve for v2 which would be the velocity of the block immediately after the collision. I then do (1/2)m2v2^2=m2gh and solve for h which gives me the height to which m2 rises. Please tell me I've got it this time.
     
  11. Dec 2, 2006 #10

    radou

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    Everything is OK except the equation marked red. Conservation of momentum tells you that m1v1 = m1v1' + m2v2'. Further on, you have to use conservation of kinetic energy, i.e. 1/2 m1v1^2 = 1/2m1v1'^2 + 1/2m2v2'^2. Solve the system of equations for v2' (v1' is of no interest to us.), and proceed as you already did. To see why that is so, here's a useful link: http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html#c1 (look at the 'Head-on Elastic Collisions' part.).
     
  12. Dec 2, 2006 #11
    Thank you so much for the help. I've got it now.
     
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