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Elastic Collisions

  1. Apr 7, 2007 #1

    cepheid

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    Hi,

    I'm wondering if I'm forgetting some key point about elastic collisions from my basic mechanics. If you don't want to read my lengthy solution, just read the problem statement and then answer the following: can this problem be solved essentially at a glance, without the lengthy algebra I used? This is a Physics GRE problem, which means that I would have ~ 1.5 min to solve it under actual test conditions.

    1. The problem statement, all variables and given/known data
    A ball of mass m, suspended from a wire, is released from height h and collides elastically, when it is at its lowest point, with a block of mass 2m at rest on a frictionless surface. After the collision, to what final height does the ball rise, in terms of h?

    2. Relevant equations

    See below

    3. The attempt at a solution

    Conservation of momentum holds. Given that this is an elastic collision, kinetic energy is also conserved. Finally, at the instant of collision, all velocities lie along one coordinate direction, reducing this to a one-dimensional problem. Using all of this info, we arrive at two equations:

    [tex] mv_0^2 = mv_1^2 + 2mv_2^2 [/tex]


    [tex] mv_0 = mv_1 + 2mv_2 [/tex]

    where v0 is the initial velocity of the ball at the instant it collides with the block, and v1 and v2 are the velocities just after the collision of the ball and block, respectively. For the nitpickers, every instance of the term 'velocity' above should probably be replaced with 'speed' or 'magnitude of velocity.' After some lengthy algebra, I obtain the formula:

    [tex] v_1 = \frac{m - 2m}{m+2m}v_0 [/tex]


    [tex] = -\frac{1}{3}v_0 [/tex]

    Using T and U for kinetic and potential energies respectively, the answer follows immediately (since only conservative forces act on the ball):

    [tex] \frac{T_{\textrm{after}}}{T_{\textrm{before}}} = \frac{v_1^2}{v_0^2} = \frac{U_{\textrm{after}}}{U_{\textrm{before}}} = \frac{mgh_{\textrm{final}}}{mgh} [/tex]

    [tex] h_{\textrm{final}} = \frac{1}{9}h [/tex]​

    This is the correct answer. Again, is there some way to solve this problem in about a minute without resorting to this derivation or to memorizing the results for the final velocities of the two colliding particles? I agree that the GRE does call for memorization of certain results and formulas beyond the fundamental ones, but "the velocity just after an elastic collision of a particle having collided with another stationary particle" seems like far too particular a result to memorize, especially for an exam that will test one's knowledge of all of undergraduate physics.
     
    Last edited: Apr 7, 2007
  2. jcsd
  3. Apr 7, 2007 #2

    cepheid

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    bump for this thread
     
  4. Apr 7, 2007 #3
    you might want to just remember the results for one D elastic collisions, which really its not that bad cuz of symmetry.

    http://en.wikipedia.org/wiki/Momentum (scroll down to the relevant section).
     
  5. Apr 8, 2007 #4

    daniel_i_l

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    maybe you can remember the equation for two objects of different mass colliding with the same speed, you could use that by looking at the problem from a moving frame of 1/2V0.
     
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