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Pellet fired through one foam block, embeds in second block

  1. Dec 16, 2015 #1
    1. The problem statement, all variables and given/known data
    A 0.5 g pellet is fired horizontally at two foam squares resting on a frictionless tabletop. The pellet rips through the first block (mass = 1 kg) and embeds itself in the second block (mass = 0.5 kg). As the bullet interacts with the blocks, it causes them to move with speeds of speed = 0.43 m/s and speed = 1.1 m/s, respectively. What is the speed of the pellet as it emerges from the first block?

    2. Relevant equations
    P=mv; conservation of momentum

    3. The attempt at a solution
    We began by calculating the momentum of the system in its final state, by adding the mass of the second block to the mass of the pellet, then multiplying by the final velocity of the second block (with the pellet embedded):

    P3 = (0.0005kg + 0.5kg)(1.1m/s) = 0.55055 kg m/s

    We then reasoned that, by conservation of momentum, the momentum of the pellet just prior to entering the second block would be equal to the momentum of the block+pellet after the pellet had embedded in the second block:

    0.55055kg m/s = (0.0005kg) v

    , where v is the entry velocity of the pellet into the second block. Solving for v:

    v = (0.55055kg m/s) / 0.0005kg = 1101.1 m/s

    Assuming that the difference between the velocity of the pellet as it exits the first block, and its velocity as it enters the second block, is negligible, the final answer would therefore be 1101.1 m/s.

    However, the solution from the instructor took a different approach, and got a different answer. She states: "based on conservation of momentum, the second block and pellet should be equal to the combination of the first block and the pellet leaving the block."

    Based on this reasoning, the solution given by the instructor is:

    0.55055kg m/s = 0.0005v + 0.43
    v = 241 m/s

    What was wrong with our reasoning?
     
  2. jcsd
  3. Dec 16, 2015 #2
    There is something dubious about your instructor's reasoning, unless there are missing details that you have not supplied in your description of the problem. Your instructor's solution is only correct if the first block had at some point collided with the second block and come to rest in the end. If the two blocks do not interact with each other at all throughout, then certainly your reasoning is correct.
     
  4. Dec 16, 2015 #3

    CWatters

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    The first block will start moving/accelerating the moment the pellet hits it. However if the two blocks are in contact then the second one will also start moving while the pellet is still going through the first.

    At the instant the pellet emerges from the first block the first block stops accelerating so it will have reached it's final velocity of 0.43 m/s. That means at this point both blocks must be going at 0.43m/s.

    The pellet then enters the second block and accelerates that one from 0.43m/s to 1.1m/s.
     
  5. Dec 16, 2015 #4

    haruspex

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    It's worse than that. The instructor is saying that as the bullet emerges from the first block, not only does the first block stop but it somehow transfers its momentum back to the bullet. So the bullet is now moving at its original speed.

    As CWatters points out, it is not clear whether the blocks make contact before the bullet transits to the second block. Your calculation assumes it doesn't. If it does, the answer is between 600 and 700 m/s. Since that requires knowledge of the 0.43 m/s speed, and yours doesn't, I would suggest that is the intended interpretation of the problem.
     
  6. Dec 17, 2015 #5
    You guys are fantastic! Thanks for the quick and helpful replies ("like" given to all 3), which we discussed together yesterday, and used to work the problem again this morning.

    To address the case where the blocks make contact, we solved for the following: "If a 0.5kg block is moving at 0.43 m/s, and a 0.0005kg pellet hits it such that the block is now moving at 1.1m/s, how fast was the pellet moving when it entered the block?"

    Using P=mv and the conservation of momentum, m1v1 = m2v2 -> (0.5005)(0.67) = (0.0005)v2 -> v2 = 671 m/s.

    We'll go back and talk with the instructor! Thanks again for your great help. Since we're new: if there's another way we can thank or up-vote your help, beyond the "likes" we gave to each reply, please let us know!
     
  7. Dec 17, 2015 #6

    haruspex

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    That's what I got.
     
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