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Elastic collision find max height

  1. Sep 30, 2006 #1
    I am having a lot of difficulty with this problem:
    Consider a frictionless track as shown in Figure P6.48. A block of mass m1 = 4.20 kg is released from A. It makes a head on elastic collision at B with a block of mass m2 = 10.5 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision.
    [​IMG]
    My plan was to use (KE+PE)i=(KE+PE)f, somehow.
    I know that at the time of release block A has 205.8 J of potential energy, from PE=mgy, but I need the final velocity in order to use the conservation equation.
    I found the velocity before the collision using (delta y)=.5(Vyo+Vy)t, and got 9.9m/s for Block A, B is at rest.
    Then I tried to figure out the velocity after the collision, using (m1v1+m2v2)i = (m1v1+m2v2)f
    I had (4.2 x 9.9)= (m1v1+m2v2)f, so m2v2f=41.58-m1v1, and that is where I seem to be stuck, I can't figure out how to carry on or if I am even right up to this point. Any tips are EXTREMELY appreciated!
     
  2. jcsd
  3. Sep 30, 2006 #2

    radou

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    Homework Helper

    In the case of an elastic collision, except linear momentum, kinetic energy is also conserved. Use that fact, too. After finding the speed of the first mass after the colision, use the fact that the work m1g*h done by gravity equals the change of kinetic energy from the moment after the colision to the moment when v1 equals 0 (i.e. the moment the mass m1 obtains its maximum height h).
     
    Last edited: Sep 30, 2006
  4. Oct 1, 2006 #3
    I don't think I understand what the next step would be, I haven't been able to get as far as finding the velocity right after the collision. The initial PE (mgy) was 205.8 J, so I know that the KE at the bottom (right after collision) must equal that, but how do I tell the difference in that from the KE it had right before the collision, when it was also at 0 elevation? I tried setting it up with the values right before the collision as (.5mv^2 + mgy)i, which is (.5 x 4.2 x 9.9^2 + 0) equal to at where the max height would be, (.5mv^2 + mgy)f, which would be (.5 x 4.2 x 0 + 4.2 x 9.8), and when I solved for y I got 5.00, which isn't right, that was my initial height, I seem to be doing something circular. Can you explain what my next step should be?
    Sorry!
     
  5. Oct 1, 2006 #4
    Nevermind, I just got it, sorry. I used v1i + v1f = v2i + v2f and set v2f equal to (v1i + v1f), since v2i is zero, and solved to get -4.24 m/s for the velocity after collision, then used the KE=PE rule to solve for the height, 0.92 m.
    Thanks for the tips!
    One question: the equation above, v1i + v1f = v2i + v2f, was given in lecture and appears in the book, but I really don't understand it, how do the masses cancel out of the blocks are inequal?
     
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