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Conservation of linear momentum.

  1. Feb 7, 2014 #1
    say a bullet with mass mb and initial speed v0 strikes and becomes embedded in a block of mass mc, which is initially at rest. The coefficient of kinetic friction between the bock and the surface is uk. ( a situation I set up)

    my question is, because its an impact problem. Can I ignore the friction force over time and use the conservation of linear momentum. because I can assume it is a non impulsive force in this situation?
    Ive done some other problems similar to this where there is a constant force, and it is included. So I figure I can remove it when it is an impact.

    G1+[itex]\int^{t2}_{t1}[/itex][itex]\SigmaFdt[/itex]=G2

    here is the Problem
    [URL=http://s1341.photobucket.com/user/nebula-314/media/20140207_224649_zps5b1d47a5.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20140207_224649_zps5b1d47a5.jpg[/URL][/PLAIN]

    my work. I am unsure about how to go from here, If I solve for V. I cant really use vdv=ads because I cant assume constant acceleration. And the solution does not have time in it. but this is impulse/momentum equation. So I am not to sure.

    [URL=http://s1341.photobucket.com/user/nebula-314/media/20140207_225157_zpse4356e4f.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20140207_225157_zpse4356e4f.jpg[/URL][/PLAIN]
     
    Last edited: Feb 7, 2014
  2. jcsd
  3. Feb 7, 2014 #2
    Not quite sure what you mean by "and it is included", but you can't use conservation of momentum for the whole problem. You can to find the initial velocity of the block, but after that friction takes momentum out of the system, so you can't use it for that part.
     
  4. Feb 7, 2014 #3
    by included i mean that, I use the equation that is written with the friction force integrated with respect to time.
     
  5. Feb 7, 2014 #4
    Oh, I see, using impulse. Introducing time into the problem makes it too cumbersome -- you would have to create a second equation with time in order to eliminate it. And you can assume constant acceleration, since when the block is moving, the only force acting along the axis of motion is friction, which is constant. Since you have the initial velocity of the plug, try now finding the initial velocity of both masses.
     
  6. Feb 7, 2014 #5
    ok, so I take it that means IMMEDIATELY after impact thus.

    v=[itex]\frac{Ma\sqrt{2gr}}{(Ma+Mc)}[/itex]
     
  7. Feb 7, 2014 #6
    Yes, that looks right. Then it travels a certain distance with a constant frictional force, so you can use those kinematic equations for constant acceleration to solve the rest.
     
  8. Feb 7, 2014 #7
    Ahhh ok thank you, I thought I could only assume constant acceleration if the only force acting was gravity?
     
  9. Feb 7, 2014 #8
    No problem! And no, it just depends on the case. Gravity works because it is constant (at least roughly), and so does kinetic friction since it is also constant (roughly). It's always a good idea to draw out all your forces in a free-body diagram to see what the case is -- that is, constant or variable net force, for a particular problem.
     
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