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Momentum question-just checking my method

  1. Oct 26, 2007 #1
    Momentum question--just checking my method

    Two men, of mass 100 kg each, stand on a cart of mass 300 kg. The cart can roll with negligible friction along a north-south track, and everything is initially at rest. One man runs towards the north and jumps off the cart at a speed of 5.0 m/s, relative to the cart. After he has jumped, the second man runs towards the south and jumps off the cart, again with a speed of 5.0 m/s relative to the cart. Calculate the speed and direction of the cart after both men have jumped off.

    Since there is no net force acting, the momentum is conseved
    p of system(s)= Ps
    ps will be 0
    hence Ps is 0.
    So now the final momentum Ps is equal to P of man 1+P of man 2+ P of the cart
    Hence Ps=500(N)-500(S)+P3
    0=500(N)-500(S)+P3
    500(S)=500(N)+P3
    Now by phythagorous theorem,
    P3= root of [(500^2)+(500^2)]
    P3= 702.106kgm/s
    And we know that m=300Kg
    so we can find v
    Does that seem right?
     
  2. jcsd
  3. Oct 26, 2007 #2

    Doc Al

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    I can't quite follow your reasoning. I'd do it in two steps. Look at the situation just before the first man jumps:
    The velocity of the "second man + cart" = v
    The velocity of the first man = v + 5
    Total momentum equals 0

    Use that to find the speed of "second man + cart" after the first man jumps; then repeat the reasoning for the second jump.
     
  4. Oct 26, 2007 #3
    i dint quite understand. Could you explain again?
     
  5. Oct 26, 2007 #4

    Doc Al

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    Pinpoint the part that you don't understand (otherwise I'll just repeat the same thing). Realize that as the first man runs, the cart and second man move at some speed in the opposite direction. If the speed of the cart is v, and the speed of the man with respect to the cart is 5 m/s, then the speed of the man with respect to the ground (which is what we want) is v + 5.
     
  6. Oct 26, 2007 #5
    Yes..ok..so then initial momentum=0kgm/s
    Final Momentum is 100*V+5+100*V-5+300V
    Yes..?
     
  7. Oct 26, 2007 #6

    Doc Al

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    Yes. The initial momentum of cart + 2 men = 0.
    I don't quite understand that expression. Let v = speed of the cart just before first man jumps; m = mass of man, M = mass of cart.

    So:
    momentum of "cart + second man" = (m + M)*v
    momentum of first man = m*(v + 5)

    Add them up to get total momentum and to solve for the speed of the cart after the first man jumps off. (Note that v will be negative.)
     
  8. Oct 26, 2007 #7
    so then final momentum will be (m+M)V+m(V+5)
    So i equate this to 0, and find V. Does that sound right? What about the angle?
     
  9. Oct 26, 2007 #8

    Doc Al

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    Yes. That will give you the speed of the cart after the first man jumps off. Then you have to repeat the analysis for the second man jumping.

    What angle? (The cart is on a track.)
     
  10. Oct 26, 2007 #9
    So when the 2nd man jumps, the initial momentum will be 1000kgm/s
    and the final momentum will be MV+m(V+5)..?
     
  11. Oct 26, 2007 #10
    the cart is at an angle of 180 degree with the track
     
  12. Oct 26, 2007 #11

    Doc Al

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    Just before the 2nd man jumps, the total initial momentum will be (m + M)*v, where v is the speed you found in step one.
     
  13. Oct 26, 2007 #12

    Doc Al

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    When they ask for the direction of the cart, they are looking for an answer like north or south, not an angle.
     
  14. Oct 26, 2007 #13
    ooh..ok..then it is south..
    And just to make sure for the momentum part, after the 2nd man jumps,
    400=MV+m(V+5)
    And we solve for V.
    I got one question though, since the direction is different, arent we supposed to use the phythagorean theorem?
     
  15. Oct 26, 2007 #14
    The way i did it, i got v as 2.35m/s(south)..Is that right?
     
  16. Oct 26, 2007 #15

    Doc Al

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    That's not the answer I get.
     
  17. Oct 27, 2007 #16
    I think the problem would be slightly simpler to tackle if you were to attach a coordinate system to the problem. For example, take north as the positive direction and the south as negative.

    Now the final direction of motion can be determined by the sign attached to the velocity. For instance if the final velocity comes out to be positive in your coordinate system then it goes north, else it goes south. (Try this!)

    Thus all your velocities in the moemntum equations should include their sign, since it gives you the direction of motion.

    Another simplification would be to have variables, say M for the mass of the cart, m for the masses of the men, u for the relative velocities of the men with respect to (w.r.t) the cart, v1 for the velocity of the 'man + cart' w.r.t the ground after the first man jumps, and v2 for the velocity of 'cart' w.r.t the ground after the second man has also jumped out. Now the problem is only about finding v2.

    One fact which you might find useful is:
    Velocity of man w.r.t ground = Velocity of man w.r.t cart + Velocity of cart w.r.t ground. Again include the signs corresponding to your coordinate system.

    If you can use all of the above correctly in the conservation of momentum equations, you'll get the right answer.

    Hint: v2 should be negative.
     
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