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Change in speed of the bullet after striking an object

  1. Dec 17, 2012 #1
    1. The problem statement, all variables and given/known data
    A gangster fires a bullet (m1 = 30 g, v1 = 190 m/s) at Stupendous Man (m2 = 31 kg), but it simply bounces away elastically. If Stupendous man was standing on a frictionless surface, what is the change in speed of the bullet after striking stupendous man? Be careful with signs and units!


    2. Relevant equations conservation of momentum, mv1+mv2=mv1+mv2



    3. The attempt at a solution the man is not moving initially, and final, i got 0.183 for the final bullet speed but thats not right, Iam confused about how to find the final for the bullet?
     
  2. jcsd
  3. Dec 17, 2012 #2

    SammyS

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    You haven't explained how you got that result, so it's hard to say where you went wrong.

    The problem doesn't state the direction in which the bullet bounces away, so I would assume that the bullet goes back toward the location it was fired from.

    Did you use the fact that it's an elastic collision ?

    Such a problem can often be solved most easily in the center of mass reference frame. If you do, don't round-off anything until your final answer.
     
  4. Dec 17, 2012 #3
    i thought i was using the formula for elastic collision, .03*190=.03*V2+31*V2, but I though the man did not move so I am confused about that, how would I apply the cm to this?
     
  5. Dec 17, 2012 #4

    SammyS

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    That is for an inelastic collision. That's evident because you have both the bullet and the man traveling at the same velocity.
     
  6. Dec 17, 2012 #5
    but i dont know either final velocity...
     
  7. Dec 17, 2012 #6
    Is the answer 0.367ms-1.

    Soln:
    We have two equations (one from momentum conservation and the other from kinetic
    energy conservation) with two variables (final velocities of the bullet and the man).Solve these
    two equations and get the final velocity of the man in terms of initial velocity of both and their masses.

    The final equation will be:

    V=M2U2/(M1+M2)

    where M1 is the mass of the man,
    M2 is the mass of the bullet,
    U2 is the velocity of the bullet.
     
  8. Dec 18, 2012 #7
    yes, that is the answer, why is the mass combined because the bullet bounces off.
     
  9. Dec 18, 2012 #8

    SammyS

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    In your above equation (from your Original Post), you have two unknowns. You need another equation in order to get a solution.

    Let's call the final velocity of the man, V2, and the final velocity of the bullet, u2.

    The initial velocities being V1 = 0 m/s and u1 = 190 m/s .

    Similarly, let M be the mass of the man, M = 31 kg. Let m be the mass of the bullet, m = 30 grams = 0.03 kg.

    (I took it upon myself to change some variable names.)

    Conservation of momentum gives [itex]\ mu_1+MV_1=mu_2+MV_2\ [/itex] where V1=0, for the (initially) stationary man.
    [itex]\ mu_1=mu_2+MV_2\ [/itex] ​

    The other equation you need is from conservation of kinetic energy (It's an elastic collision.)
    [itex]\ (1/2)m{u_1}^2=(1/2)m{u_2}^2+(1/2)M{V_2}^2\ [/itex]​
     
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