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How do I figure out the final position of the block?

  1. Jan 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Three blocks of identical mass are placed on a frictionless table as shown. The center block is at rest, whereas the other two blocks are moving directly towards it at identical speeds v. The center block is initially closer to the left block than the right one. All motion takes place along a single horizontal line.

    Problem #2 on this website: https://www.aapt.org/physicsteam/2010/upload/2009_F-ma.pdf

    Suppose that all collisions are instantaneous and perfectly elastic. After a long time, which of the following is true?
    (A) The center block is moving to the left.
    (B) The center block is moving to the right.
    (C) The center block is at rest somewhere to the left of its initial position.
    (D) The center block is at rest at its initial position.
    (E) The center block is at rest somewhere to the right of its initial position.

    (Correct Answer: D)
    2. Relevant equations
    p
    = mv
    mv1 + mv2 = mv3 + mv4

    KE = (0.5)m(v^2)
    (0.5)m(v1^2) + (0.5)m(v2^2) = (0.5)m(v3^2) + (0.5)m(v4^2)
    3. The attempt at a solution
    I observed that there was no friction, meaning that the velocity of each mass should, in theory, remain constant until it experienced a force from another mass after colliding with it. This, in my mind, would mean that the blocks should continue colliding forever due to Newton's first law of inertia... but that thinking is obviously flawed.

    I tried writing equations using conservation of kinetic energy and conservation of linear momentum, but I seemed to be getting nowhere in terms of finding the position after a long period of time. Honestly, I don't even know how I'm supposed to attempt such a problem. Should I be considering velocity of the center of mass or change in position of the center of mass?
     
  2. jcsd
  3. Jan 14, 2016 #2

    ehild

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    What do v1, v2, v3, v4 mean?
    Follow the events. What is the first collision, that between the left block and the central one, or that between the right block and the central one?
    What happens if two blocks of the same mass collide elastically, and one of them was in rest before the collision?
     
  4. Jan 14, 2016 #3
    Ohhhh Since the collision is perfectly elastic, the speed of the block on the left will become zero, and the speed of the block on the left will be v. Then, after the other two blocks collide, the center block will collide again with the left block, and the speed of the center block will become zero. I get it.
     
  5. Jan 14, 2016 #4

    ehild

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    Well done!
    Does it get to the initial position?
     
  6. Jan 14, 2016 #5
    Yes. It does get to its initial position since the point at which the center block collides with the left most block for the second time is at the initial position. Since the center block comes to rest at the instant of collision due to conservation of momentum at the point of collision, that must mean that the center block comes to rest at its initial position.
     
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