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Conservation of Momentum

  1. May 2, 2016 #1
    1. The problem statement, all variables and given/known data
    After shooting a 28g arrow with an initial velocity of 92m/s[forward], an archer standing on a frictionless surface travels in the opposite direction at a speed of 0.039m/s. Calculate the combined mass of the archer and the bow.

    Given: **Subscript of 1 indicates values for the arrow and subscript of 2 indicates values for the archer

    m1=0.028kg
    vi1=92m/s[forward]
    vf1=0m/s

    vi1=0.039m/s[backward]
    vf2=0m/s

    2. Relevant equations
    m1vi1+m2vi2=m1vf1+m2vf2

    3. The attempt at a solution
    The only way this problem works is if you treat the archer and the bow as one body/one object and using the above formula and values, solve for m2:

    (0.029)(92)+m2(0.039)=0
    2.576=-m2(0.039)
    -66.05128205kg=m2

    I got the correct answer aside from the negative sign in front of my answer. Why is my answer negative? Does that mean I did something wrong?

    Also, why do you treat the archer and the bow as one collective object? I thought the law of conservation of momentum, which gives us the equation I used, applied to situations when two objects collide in an isolated system. If so, then what collision occurs in this problem; how are both the bow and the archer colliding with the arrow?
     
  2. jcsd
  3. May 2, 2016 #2

    BvU

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    With your initial and final you describe a fully inelastic collision of an arrow and an archer plus bow. Rather fatal for the archer ...

    Is it strange you get a negative mass ? The recoil from shooting the arrow makes the archer go backwards (that's why it's called recoil) wrt the arrow ! Yet you have the same sign for both speeds !
     
  4. May 2, 2016 #3
    I thought that it was okay not to plug in negative signs indicating direction for vector values when doing calculations to find a scalar value, like mass.
     
  5. May 2, 2016 #4

    BvU

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    Well, apparently you thought wrong: a negative mass is unphysical !
     
  6. May 2, 2016 #5
    Good to know
     
  7. May 2, 2016 #6
    But why are the bow and archer one mass?
     
  8. May 2, 2016 #7

    haruspex

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    Because the archer is assumed to hold the bow rigidly. In a different model, the archer could let go of bow and arrow at the same instant. The arrow would travel a smallish distance one way, and the bow a smaller distance the other.
    Note that with a firearm that second model is closer to what happens. The impulse is so great that it is not possible (and maybe not a good idea to try) to hold the weapon rigidly.
     
  9. May 2, 2016 #8
    What is the collision that happens, the bow and the arrow are not colliding, they were already in contact.
     
  10. May 3, 2016 #9

    haruspex

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    Conservation of momentum is not limited to collisions.
     
  11. May 3, 2016 #10

    BvU

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    Physically the time reversal of a collision happens: first they are together (sum of momenta = 0) and then they separate -- without any external forces acting. So the center of mass does not accelerate and sum of momenta remains zero.

    Newton 3: action = - reaction makes that afterwards the individual momenta are equal and opposite.
     
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