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Intuitive explanation of momentum conservation problem

  1. Sep 28, 2012 #1
    Hi,

    So I recently worked out a problem in my mechanics class about two people jumping off a frictionless railroad cart at speed u. The result is that the cart will move faster if they each jump off separately than it would have if they both jumped at the same time.

    I've been trying to understand intuitively why this is so, but I actually reach the opposite conclusion. My explanation is that when one person jumps they exert force f. So when they jump together 2f is exerted on the cart. When they jump separately f is exerted twice, but the first time its exerted is on a heavier cart (because cart plus the mass of the person still on) so it will accelerate to a slower speed. This implies the opposite of the actual result.
     
  2. jcsd
  3. Sep 28, 2012 #2

    A.T.

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    How do they jump off? Back, side or forward?
     
  4. Sep 28, 2012 #3
    Everything happens on the x-axis. They jump to the right completely horizontally with speed u
     
  5. Sep 28, 2012 #4
    Is the speed u with reference to

    a) the ground
    b) the rest frame of the cart before the jump.
    c) the rest frame of the cart after the jump.
     
  6. Sep 28, 2012 #5
    When the second person jumps the cart is in motion so some of the force he is exterting is only moving him forward without actually leaving the cart. Well, actually the force is moving him backward to undo the motion the cart gave to him.
     
  7. Sep 29, 2012 #6

    Here's the actual problem:

    Two hobos, each of mass mh, are standing at one end of a stationary railroad flatcar with frictionless wheels and mass mfc. Either hobo can run to the other end of the flatcar and jump off with some speed u (relative to the car). (a) Use conservation of momentum to find the speed of the recoiling car if the two men run and jump simultaneously. (b) What is it if the second man starts running only after the first has already jumped? Which procedure gives the greater speed to the car?
     
  8. Sep 29, 2012 #7

    A.T.

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    For an intuitive grasp, I would not go into forces here. What you know is the separation speed of u. How it is achieved is irrelevant. It is given as a constraint, that each separation creates a velocity difference of u between the separating objects. So the more such separations, the more velocity difference you create in total.

    Consider the extreme case where the wagon mass is negligible compared to hobo mass:

    a) If they jump together the velocities will be:
    v_hobos ~= 0
    v_wagon ~= -u

    b) If they jump separately the velocities will be:
    v_hobo_1 ~= u/2
    v_wagon_with_hobo_2 ~= -u/2
    then:
    v_hobo_2 ~= -u/2
    v_wagon ~= -u3/2

    Note that in a) you waste two hobos to add the same velocity as you do with a single hobo in b) in the 2nd jump.
     
    Last edited: Sep 29, 2012
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