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Momentum questions part 2

  1. Jun 14, 2013 #1
    1. The problem statement, all variables and given/known data

    fy1fsm.jpg

    2. Relevant equations



    3. The attempt at a solution

    My guess is that, the with the modelling clay the pellet is absorbed initially by the clay, (it sicks in) some of the energy is lost through the clay but some is also transferred to the car giving it the means to move forward. With the metal plate, I feel it could be the transfer in energy to car is quicker and much less is absorbed by the plate (possibly minimal vibrations).

    for part 2, I wrote it bounced directly backwords and at the same speed, it means the impact was made exactly at the horizontal meaning. But how do I link this in with it being able to travel further?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 14, 2013 #2

    Simon Bridge

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    Look at the problem in terms of the conservation of momentum and the conservation of kinetic energy.
    The term you want to know more about it "elastic collision".
     
  4. Jun 14, 2013 #3
    Oh so, with the metal plate, it appears the collision was elastic meaning, kinetic energy is conserved, therefore the car will travel further as no loss of energy, whereas the clay absorbed some of the energy before it reached the car, meaning not all of the total initial energy was transferred to the car, thus why it traveled less.
     
  5. Jun 14, 2013 #4

    Simon Bridge

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    Don't forget about the conservation of momentum part.
     
  6. Jun 14, 2013 #5
    The momentum is conserved for both.
     
  7. Jun 14, 2013 #6

    Simon Bridge

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    But the momentum changes are different - how?
     
  8. Jun 14, 2013 #7
    it changes as in, the pellet is going backwards = -velocity therefore even more velocity in the forward motion by the car to balance out the momentum total?
     
  9. Jun 14, 2013 #8

    collinsmark

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    There ya go. :approve:
     
  10. Jun 14, 2013 #9

    Simon Bridge

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    Well done - you see the important parts are in the details.

    if the pellet is mass m and the car mass M, and the initial velocity of the pellet is +u,
    then the first case has mu=(m+M)v - so v=mu/(M+m) as the final velocity of the car.

    In the second case, if the rebound velocity is -u, then v=2mu/M

    2mu/M > mu/(M+m)

    You could work out what the rebound velocity would have to be to make v come out the same - also compare what happens to the total kinetic energy.
     
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