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Elastic collision problem?

Discussion in 'MCAT Study Question Q&A' started by Fighter127, Sep 4, 2014.

  1. Fighter127

    Joined:
    Jan 21, 2014
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    Two blocks are released from rest on either side of a frictionless
    half-pipe. Block B is more massive than
    block A. The height HB from which block B is released is less
    than HA, the height from which block A is released. The blocks
    collide elastically on the flat section. After the collision, which
    is correct?
    A. Block A rises to a height greater than HA and block B
    rises to a height less than HB.
    B. Block A rises to a height less than HA and block B
    rises to a height greater than HB.
    C. Block A rises to height HA and block B rises to
    height HB.
    D. Block A rises to height HB and block B rises to
    height HA.
    E. The heights to which the blocks rise depends on where
    along the flat section they collide.

    Can someone please explain how to figure this out?
     
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  3. CelinaBelle

    5+ Year Member

    Joined:
    Jan 8, 2012
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    Status:
    Pre-Medical
    Hi!

    The best way to intuit this problem is to draw a picture of the situation. Once that's done, take some time to thank yourself for choosing science over visual arts...let's move on.

    In an elastic collision, both kinetic energy and momentum are conserved.

    We can find the kinetic energy of each block at the flat part of the half-pipe before the collision by using basic conservation of energy principles: the blocks are initially at rest so KE = 0 for both. However, there is potential energy for each block: PEa = (mass of block A)*g*(initial height of A), PEb = (mass of block B)*g*(initial height of B).

    At the flat portion of the half-pipe, before the blocks collide, each block has converted all potential energy into kinetic energy. (PE = 0, KE = (1/2mv^2)
    To find the velocity of each block, we set these equations equal to each other to find that va = sqrt(2g*height of A) and vb = sqrt(2g*height of B).
    Now, for the collision: because momentum is conserved, block A transfers all of its momentum to block B and vice versa.
    Momentum is equal to mass*velocity.
    Momentum before collision: momentum of a = (mass of block A)*(velocity of block A), momentum of block B = (mass of block B)*(velocity of block B)
    Momentum after collision: momentum of a = (mass of block A)*(velocity of block B), momentum of block B = (mass of block B)*(velocity of block A)
    We assume that the blocks will now rise until they have converted all kinetic energy into potential energy (height essentially), so we can once again set potential energy equal to kinetic energy to find the final height of each block.
    (mass of block A)*g*(final height) = (1/2)*(mass of block a)*sqrt(2g*initial height of B)^2
    after a little algebra, you'll find that the final height of block A = the initial height of block B (HB).

    Hopefully this makes sense! Please let me know if you have any questions!
     

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