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Space rock collision problem

  1. Nov 28, 2008 #1
    1. The problem statement, all variables and given/known data
    In outer space rock 1, with mass 6 kg and velocity < 4100, -2900, 3200 > m/s, struck rock 2, which was at rest. After the collision, rock 1's velocity is < 3900, -2400, 3700 > m/s. What is the final momentum of rock 2 (p2f)?

    3. The attempt at a solution

    ok from my understanding of head on collision of non equal masses (honestly the problem deosnt tell me if they are equal masses or not so i assumed they are not. so i got p2f = 2p1i
    that is the final momentum of rock 2 is equal to twice the initial momentum of rock 1.
    so initial momentum of rock one = 6 * < 4100, -2900, 3200 > = <24600,-17400,19200> so the final momentum of rock 2 would be 2*<24600,-17400,19200> =<49200,-34800,38400>, but its telling me wrong answer. anything am doing wrong. Thanks.
     
  2. jcsd
  3. Nov 28, 2008 #2

    mgb_phys

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    Re: Collision

    Remember momentum is conserved.
     
  4. Nov 28, 2008 #3
    Re: Collision

    ok so the change of momentun in the system + the change of momentum in the surroundings = 0 here we are ignoring the surrounding so pf system - pisystem = 0 ?
     
  5. Nov 28, 2008 #4
    Re: Collision

    Hello Lebprince,

    How did you manage to achieve this result (i.e. p2f = 2p1i)? Wouldn't it be more accurate if you consider consider the conservation of momentum in each dimension separately, since it is (or, at least the question is) presented as a head-on collision?

    Regards,
    Horatio
     
  6. Nov 28, 2008 #5
    Re: Collision

    ok so would this make sense now? p1f + p2f = p1i + p2i ?

    so 6<3900,-2400,3700> + p2f = 6<4100,-2900,3200> + <0,0,0> and then i can find p2f?
     
  7. Nov 28, 2008 #6
    Re: Collision

    If we assume the collision is head-on, your solution would and should be right.

    Regards,
    Horatio
     
  8. Nov 28, 2008 #7
    Re: Collision

    Thanks i got that part right. can i get some help on this?

    Suppose the collision was elastic (that is, no change in kinetic energy and therefore no change in thermal or other internal energy of the rocks). In that case, after the collision, what is the kinetic energy of rock 2?

    i wanted to use K= p^2/2m i have p but i dont have m. so any other way around it? Thanks
     
  9. Nov 28, 2008 #8

    mgb_phys

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    Re: Collision

    You know the magnitude of the velocity change in rock1 and you know it's mass
    If no energy is lost all this ke must have all gone into rock 2. You would need the mass of rock2 to get it's speed but not it's ke.
     
  10. Nov 28, 2008 #9
    Re: Collision

    ok thanks i used Efsystem = Eisystem and figured it out. Thanks but when thermal energy gets involved it confuses me like this question suppose that in the collision some of the kinetic energy is converted into thermal energy of the two rocks, where Ethermal,1 + Ethermal,2 = 1.44106 J. What is the final kinetic energy of rock 2? i just dont get what to do here
     
  11. Nov 28, 2008 #10

    mgb_phys

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    Re: Collision

    If it isn't elastic, ie. if kinetic energy is lost, than you need more information so solve it.
     
  12. Nov 28, 2008 #11
    Re: Collision

    so if we have an inelastic collision what would i need to calculate DeltaEthermalA + DeltaEthermalB? Thanks
     
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