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What is the force imparted by the bullet on the block

  1. Dec 5, 2016 #1
    1. The problem statement, all variables and given/known data
    A bullet with a mass of 0.045 kg is shot at 560m/s fwd, and it gets dislodged in a stationary block that is 0.5kg.
    The collision lasts 0.35 seconds. What is the force imparted by the bullet on the block?

    2. Relevant equations
    F=(m(vf-vi))/t

    3. The attempt at a solution
    F=(0.045kg(0-560m/s))/0.35s
    F=-72N [FWD]

    It doesn't make sense that the force the bullet has on the block is backwards 72N[BWD], so I'm guessing thats the force of the block on the bullet, therefore using newton's 3rd law which if F Action= -F reaction i get the force of the bullet on the block is 72N [FWD]

    Is this right, shouldn't the initial calculation be the final answer for the question? Help is appreciated
     
  2. jcsd
  3. Dec 5, 2016 #2

    collinsmark

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    Are you sure the "stationary block" remains stationary after the collision?

    To me, the fact that the problem statement mentions the 0.5 kg mass of the block is a hint that the bullet+block combination has a non-zero velocity after the collision.
     
  4. Dec 5, 2016 #3
    I get 46.2m/s (fwd) for the velocity of the block and bullet using conservation of momentum, and with that i still get -66N fwd. But I think you use the velocity of the bullet relative to the block, which is still 0 because it's inside of it.
     
  5. Dec 5, 2016 #4

    collinsmark

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    That sounds about right. :smile: The force applied to the bullet.

    You'll need to ensure that your frame of reference is the same before and after the collision.

    You can't use a stationary frame before the collision, and then switch to a moving frame after, when calculating velocities and momentum values.
     
    Last edited: Dec 5, 2016
  6. Dec 5, 2016 #5

    haruspex

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    As an aside, the wording of the question could be better. It should make clear that the block is free to move, and it should not ask for "the force imparted". You can only calculate an average force, and forces are not imparted. Imparting implies transfer of some state, such as energy or momentum. Forces are applied or exerted.
     
  7. Dec 5, 2016 #6
    Okay I see what your saying, so using conservation of momentum i get that both the block and bullet are going 46.2m/s FWD. If i use that final velocity to find force I get -66 N [FWD], which is still a negative.
     
  8. Dec 5, 2016 #7

    haruspex

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    You calculated the average force that slowed the bullet. That is therefore the force exerted on the bullet, but the question asks for the force the bullet exerted on the block.
     
  9. Dec 5, 2016 #8
    So according to newton's 3rd law the average force the bullet exerted on the block should be 66N [FWD], correct?
     
  10. Dec 5, 2016 #9

    haruspex

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    Yes.
     
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