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Calculate the velocity of the system After the collision

  1. Feb 16, 2015 #1
    1. The problem statement, all variables and given/own data
    http://prntscr.com/661d5d

    2. Relevant equations
    m1v1 + m2v2 = m1v1p +m2v2p

    3. The attempt at a solution
    I asumed that the momentum in the y direction is constant, and that in the x direction it is conversved.
    Also, i thought of it as 2 parts to this problem. The momentum of the first cart plus that of the rock has to equal the combined momentum.
    This combination plus the negative momentum of the 2nd cart, must equal the desired velocity of the entire system.
    I dont think this logic is correct since i didnt get the right answer, so clarification on that should help me solve the problem.
    anyway here's the math
    (x direction)
    Vrx = 1.20cos30
    Mc1Vc1 + MrVr = McVc1f + MrVrf * here i assume that the final velocity is going to be the same for both of them.
    plugging in the values i get this
    Vf=300/(Mc1 + Mr)
    Vf = .75m/s

    Now, the the momentum of the entire system should be equal to the rock and the first cart minus the second.

    (Mc2+Mr)Vf - Mc2Vc2 = M(system)V(system)

    I get V = .359
    Ans : .205 m/s

    Any help is appreciated, thanks!
     
  2. jcsd
  3. Feb 16, 2015 #2

    BvU

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    Is it clear to you what all these symbols stand for ? A list would be helpful.

    First collision: what is the 300 ? Fill in the correct momentum to get a better value for vf.

    Second collision: if you don't show what you are doing, only give the result you get, it is very difficult to guess what goes wrong. Most physicists have very limited telepathic capabilities. The expression is correct and the answer I get is indeed 0.2049 m/s.

    I tried some possible errors but couldn't reverse-engineer the .359 whatever I tried !
     
  4. Feb 16, 2015 #3
    Are you sure that is correct? That expression would apply in the case of a perfectly elastic collision. I would assume that the rock collides with the cart and stays in the cart.
     
  5. Feb 16, 2015 #4
    Oh. Sorry. If you assumed that the velocities of the cart and the rock are the same after the collision then that should be correct.
     
  6. Feb 16, 2015 #5
    BvU, my apologies for not expressing things cleary enough

    so i did (300)(.6) +(100)(1.2) for the initial momentum of the first cart and the rock, which gives
    300 = Mc1Vc + MrVr
    so 300=(Mc1+Mr)Vf
    300/(Mc1+Mr) = Vfx
    Vfx=.75m/s

    As for the equation, aleph, are you saying that the coefficient of restitution comes into play here? if so, im a little confused on how to implement that, however i do know how to find it.

    The second calculation i did this

    400(.75) - 400(.3) = MfVf

    (300-120)/800 = .225 m/s

    Looks like i did the calculation wrong at the end there BvU and for any grief that might have caused.
    Regardless the answer is still wrong!
     
  7. Feb 16, 2015 #6
    I got the same answer as BvU. It would appear that you rounded a little too heavily. The value for the velocity of cart 1 and the rock that I got is .709808. Other than that everything is correct.
     
  8. Feb 17, 2015 #7

    BvU

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    Dear Jose,

    No need for apologies. I now understand the .75 m/s, and it's not good. Reason: momentum is a vector. To get (horizontal) speed of car + rock, you can add horizontal momentum of car and only the horizontal component of momentum of rock. I though that's what you intended to do, because you did write Vrx = 1.20 cos30 .
    The vertical component of the momentum of the rock is somehow absorbed by the car, either in shock absorbers or by denting the floor.

    So that settles the first part. Forgot the numbers, but you get some horizontal momentum (in fact you don't need the speed of the first car after rock has settled an before the connecting collision with car 2). That one you did just fine. IF you re-do it, you'll get the book (?) answer :smile:
     
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