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Small cube Momentum Problem

  1. Nov 3, 2004 #1
    Say I was wondering if I could maybe get some help with this problem.


    In a physics lab, a small cude slides down a frictionless incline as shown in the figure below, and elastically strikes a cude at the bottom that is only one-half its mass. If the incline is 30 cm high and the table is 90 cm off the floor, where does each cube land?

    So basically they're asking how far does mass 1 and mass 2 land from the bottom of the table.

    But I don't really get it...

    Thank you in advance,
    Andrew J. Leer
     
  2. jcsd
  3. Nov 4, 2004 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Since you didn't include the "figure below", I assume that the second cube is at the edge of a 90 cm table.

    You are told the height of the incline so you can calculate the potential energy of the first cube relative to the bottom of the incline. That is the same as its kinetic energy at the bottom so you can find its speed and thus know both kinetic energy and momentum and, using conservation of each, can find the speed (which will be horizontal for both) of the both cubes after the collision. Knowing that, you can find the equations of motion of both cubes. Determine the time until they hit the ground and use that to find the distance they move from the table.
     
  4. Nov 5, 2004 #3
    I solved for the speed of the first block just before it hits the second block, and this came out to be:

    the square root of 2gh,

    Where h is the height of the incline.
    I assigned this to a varible called V_1.

    Next I tried to solve for the speed of the blocks after they collide,
    using the conservation of momentum equation (Detla P=0).

    When I'm done I get:

    V_2' = (m_1(v_1-v_1'))/(m_2)

    (for the second block)

    Next when I try to use V_2' (after solving for time in the air) in the equation which solves for Delta X_2:

    I.E.

    DX_2 = V_2' * t

    DX_2 = (m_1(v_1-v_1'))/(m_2) * t

    DX_2 = (m_1(v_1-v_1'))/(0.5*m_1) * t

    DX_2 = ((v_1-v_1'))/(0.5) * t


    But when I plug it into the equation and try to solve for it I'm still left with the unknown varible V_1' (as shown above).

    I also tried solving for V_1' and it's equation as the varible V_2 in it, so that doesn't work either.

    What do I do from here?

    Thanks again,
    Maskkkk
     
  5. Nov 5, 2004 #4

    Doc Al

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    Staff: Mentor

    first find both speeds

    Right.
    As HallsofIvy stated, you need to use both conservation of momentum and conservation of energy to find the speed of each block after the collision. When you're done you will know both v_1' and v_2' in terms of v_1.

    Then you'll be able to find how far they travel.
     
  6. Nov 5, 2004 #5
    Well actually I solved for them both, but I was too lazy to write the other one up here.

    But the problem I'm having is that both v_1 and v_2 are unknowns, and are within each other's equations when I solve for them.

    So when I plugin one for the other I wind up with a situation where I cannot solve for either.
     
  7. Nov 6, 2004 #6

    Doc Al

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    Staff: Mentor

    two equations, two unknowns: solve it!

    If they are still "unknown" after you solve for them, then you didn't really solve for them, right? :smile:

    You should have two equations with two unkowns. Once you solve them, you will know the speeds of both blocks. Start by writing both of those equations. (Don't waste time trying to solve for the distance until you've figured out how to solve for v_1' and v_2'.)
     
  8. Nov 6, 2004 #7
    Step 1: Just mass 1 sliding down the incline to the point
    just before it hits.

    [tex]&v_{1}=?$&[/tex]

    [tex]\Delta K_{E}+\Delta P_{E}=W_{NC}[/tex]

    [tex](K_{E1}-K_{E0})+(P_{E1}-P_{E0})=0[/tex]

    [tex](K_{E1}-P_{E0})=0[/tex]

    [tex]K_{E1}=P_{E0}[/tex]

    [tex]\frac{1}{2}m_{1}v_{1}^{2}=m_{1}gh[/tex]

    [tex]\frac{1}{2}v_{1}^{2}=gh[/tex]

    [tex]v_{1}=\sqrt{2gh}[/tex]



    Step 2:
    Figure out the momentum, by relating the before hitting
    and after hitting together.

    [tex](P_{2}^{~}+P_{1}^{~})-(P_{2}+P_{1})=0[/tex]

    [tex](m_{2}v_{2}^{~}+m_{1}v_{1}^{~})-(m_{2}v_{2}+m_{1}v_{1})=0[/tex]

    [tex](m_{2}v_{2}^{~}+m_{1}v_{1}^{~})-(m_{1}v_{1})=0[/tex]

    [tex]m_{2}v_{2}^{~}+m_{1}v_{1}^{~}-m_{1}v_{1}=0[/tex]

    [tex]m_{2}v_{2}^{~}=m_{1}v_{1}-m_{1}v_{1}^{~}[/tex]

    [tex]m_{2}v_{2}^{~}=m_{1}(v_{1}-v_{1}^{~})[/tex]

    [tex]v_{2}^{~}=\frac{m_{1}(v_{1}-v_{1}^{~})}{m_{2}}[/tex]




    Step 3: The blocks fall, and we need to determine how far
    in the x direction they will fall.

    Solve for: [tex]\Delta X_{1}[/tex]

    [tex]\Delta X_{1}=v_{1}^{~}t+\frac{1}{2}a_{x}t^{2}[/tex]

    [tex]\Delta X_{1}=v_{1}^{~}t[/tex]

    [tex]t=?[/tex]

    [tex]\Delta Y=v_{1}t+\frac{1}{2}a_{y}t^{2}[/tex]

    [tex]\Delta Y=\frac{1}{2}a_{y}t^{2}[/tex]

    [tex]-H=-\frac{1}{2}gt^{2}[/tex]

    [tex]H=\frac{1}{2}gt^{2}[/tex]

    [tex]\sqrt{\frac{2H}{g}}=t[/tex]

    [tex]\Delta X_{1}=v_{1}^{~}(\sqrt{\frac{2H}{g}})[/tex]

    [tex]\Delta X_{1}=(v_{1}-\frac{1}{2}v_{2}^{~})(\sqrt{\frac{2H}{g}})[/tex]

    [tex]\Delta X_{1}=(v_{1}-\frac{2(v_{1}-v_{1}^{~})}{2})(\sqrt{\frac{2H}{g}})[/tex]

    [tex]\Delta X_{1}=(v_{1}-(v_{1}-v_{1}^{~}))(\sqrt{\frac{2H}{g}})[/tex]

    And right about here is where it doesn't make any sense, because I keep getting the same varible I plugged in for back again.

    If what I wrote here doesn't make sense, please download what I did so far in lyx format, it's in a tar.gz archive on my space at http://home.pct.edu/~leeand00/ I just put it up there it's the only file there is on the entire page.

    Thank you,
    Andrew J. Leer
     
  9. Nov 6, 2004 #8

    Doc Al

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    Staff: Mentor


    Nothing wrong with step 1.

    One more time: you need to apply both conservation of momentum and conservation of energy in order to solve for the speeds after the collision:
    (1) [tex]m_1v_1 = m_1v_1' + m_2 v_2'[/tex]
    (2) [tex]1/2 m_1 v_1^2 = 1/2m_1v_1'^2 + 1/2m_2v_2'^2[/tex]

    You know v_1 from step one, and you know that [itex]m_2 = m_1/2[/itex]. So solve for v_1' and v_2'.
    Don't waste time with step 3 until you complete step 2.
     
  10. Nov 6, 2004 #9
    Okay now I tried to solve for Momentum and Potental Energy, just to get to the equations that you are at.

    Conservation of Momentum, no problem, came up to the same place that you are.

    Conservation of Energy on the other hand...I don't understand.

    Part 2: Conservation of Energy:

    Okay as I understand it, the energy before the collision is equal to the energy after the collision,
    thus:

    [tex]\Delta K_{E}+\Delta P_{E}=\Delta K_{E}^{'}+\Delta P_{E}^{'}[/tex]

    [tex](K_{E2}-K_{E1})+(P_{E2}-P_{E1})=(K_{E2}^{'}-K_{E1}^{'})+(P_{E2}^{~}-P_{E1}^{'})[/tex]

    The potental energy before the collision is equal to zero since the bottom of the incline is our 0 potental energy point. I'm not sure about the potental energy after the collision, but my guess would be that it too is set to zero since immediately after the collision both blocks only have a horizontal component of velocity. Also since block 2 is at rest before the collision it has not kinetic energy And so...
    [tex]K_{E1}=K_{E2}^{'}-K_{E1}^{'}[/tex]

    The only part I don't understand is, where does the plus in your (beginning of the equation) come from, since I get:
    [tex]\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{'2}-\frac{1}{2}m_{1}v_{1}^{'2}[/tex]

    and you get:

    [tex]\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{1}v_{1}^{'2}+\frac{1}{2}m_{2}v_{2}^{'2}[/tex]
     
  11. Nov 6, 2004 #10

    Doc Al

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    I'm not sure what you are trying to write. Conservation of energy says:
    [tex]\Delta {KE} + \Delta {PE} = 0[/tex]

    Since during the collision [itex]\Delta {PE} = 0[/itex], we can drop that term. (The collision takes place at a single height.) So:
    [tex]\Delta {KE} = 0[/tex]

    You messed up your statement of conservation of energy, which led you to an incorrect expression. Conservation of energy says that the kinetic energy before the collision (left side of my equation) must equal the kinetic energy after the collision (right side of my equation).

    Let me know if that makes sense to you.
     
  12. Nov 10, 2004 #11
    Thanks!

    That did the trick thanks Doc!
     
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