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Can somebody check my answer for momentum

  1. May 27, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball with mass M moving horizontally at a speed v, collides elastically with a block of mass 3m that is initially hanging at rest from a celing on the end of a wire of length L. Find the maximum angle through which the block swings after it is hit.

    2. Relevant equations

    F=dp/dt


    3. The attempt at a solution
    I split the problem into 2 parts. Part 1 is trying to find the initial velocity of the block and part 2 is trying to find the angle.

    PART 1:
    let M1 equal the mass of ball and M2 be the mass of the block. (M2=3M1) V1 be the velocity of the ball abd V2 the velcotiy of the block.

    [itex] F=\frac{dp}{dt} [/itex]

    No forces in the x direction:

    [itex] 0 = \Delta p [/itex]
    [itex] P_i = P_f [/itex]
    [itex] M_1V_1=-M_1V_1+M_2V_2 [/itex]
    [itex] V_2=\frac{2M_1V_1}{M_2} [/itex]

    PART 2:

    you can use energy because the contact is elastic.... so conservation of energy is valid:

    system mass, earth, and bloc

    W=deltaE
    [itex]0=-\frac{1}{2}mv_2^2+mgh[/itex]
    we found v2 in Part 1.... substitute it and simply:
    [itex] h = \frac{2M_1^2V_1^2}{gM_2^2} [/itex]
    [itex] (L-\frac{2M_1^2V_1^2}{gM_2^2})/L [/itex]
    [itex] \theta = cos^{-1}((L-\frac{2M_1^2V_1^2}{gM_2^2})/L) [/itex]
    M2 = 3M1....
    [itex] \theta = cos^{-1}((L-\frac{2M_1^2V_1^2}{g3M_1^2})/L) [/itex]

    is this correct?
     
  2. jcsd
  3. May 27, 2015 #2

    nrqed

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    This step is incorrect. If that was true, you would have more kinetic energy at the end then you had at the beginning, which is not possible in a collision. The ball does not recoil at the same speed it was coming in. Elastic means that kinetic energy is conserved, not that ball 1 rebounds at the same speed (a ball rebounds at the same speed in an elastic collision only when it hits a stationary object)
     
  4. May 27, 2015 #3
    alright, i understand. but [itex] V_{12} = V_{i1E} [/itex] correct? V_(12) is the velocity of 1 respect to 2 and V_(i1E) is the initial velocity of 1 respect to the earth (or ground). my teacher then gave us the following mnemonic:

    [itex] V_{1E} = V_{12}+V_{2E} [/itex] where E stands for earth (or the ground).

    then because V_12 is the same as V_i1E, we can say

    [itex] M_1V_{i1G} = -M_1V_{f1G}+M_2V_2 [/itex]
    [itex] M_1V_{i1G} = -M_1V_{i1G}-M_1V_{2G} +M_2V_{2G} [/itex] (through the mnemonic i mentioned 3 lines above)
    [itex] V_{2G} = \frac{2M_1V_{i1G}}{M_2-M_1} [/itex]

    is that correct?
     
  5. May 28, 2015 #4
    Bump... Please see comments
     
  6. May 28, 2015 #5
    Please, convert all your data to the same reference frame. You cannot write an equation with variables in different frames.
     
  7. May 28, 2015 #6
    arent all my velocity's in reference to the ground?
     
  8. May 28, 2015 #7
    Your answer on first post is ok except a 2 factor that must be over.
     
  9. May 28, 2015 #8
    i thought my original post was completely incorrect? i said the velocity in reference to the ground were the same before and AFTER the contact.... tahts not true.... the velocity in reference to M2 thats the same before and after the impact. i made the correction in my comment.
     
  10. May 28, 2015 #9

    haruspex

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    No. How did you lose Vf1G?
    It was.
    Since it's elastic, you can use conservation of KE. That gets a bit messy because of the quadratics, but combining with conservation of momentum, when the smoke clears, you end up with Newton's Experimental Law for the case R=1, namely, that the relative velocities reverse: V1i-V2i=-(V1f-V2f). It's well worth remembering this. But note: it only applies for the components of motion perpendicular to the contact plane. No difficulty in the present case, but has to be borne in mind for oblique contacts.
     
  11. May 28, 2015 #10
    ok so using kinetic energy:

    lets say the system is the ball and the block.... ball as mass1 and block as mass2.
    there are no forces in the x direction so:

    [itex] W=\Delta E [/itex]
    [itex] 0 = \Delta KE_1 + \Delta KE_2 [/itex]
    [itex] 0=\frac{1}{2}m_1v_{i1G}^2-\frac{1}{2}m_1v_{f1G}^2+\frac{1}{2}m_2v_{i2G}^2-\frac{1}{2}m_2v_{f2G}^2 [/itex]
    now how do i bring in the conservation of momentum into this?
     
  12. May 28, 2015 #11

    haruspex

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    You need to distinguish the two mass variables.
    Write out the conservation of momentum equation using the same variables.
    Note that you can factorise the two parts of your KE equation.
     
  13. May 28, 2015 #12
    alright, the conservation of momentum (before and after the collision for m1) using the same variables is:

    [itex] m_1v_{i1G}=-m_1v_{f12} [/itex] tell me if this step is correct..
    sorry ignore this. i misunderstood. ill come up with anotehr reply in a few minutes after i work it out,
     
  14. May 28, 2015 #13
    ok so the conservation of momentum gives me the following right?

    [itex] 0=\Delta p [/itex]
    [itex] P_i = P_f [/itex]
    [itex] m_1v_{i1G} = -m_1v_{f1g}+m_2v_{f2g} [/itex]
    right?

    also factoring the KE equation gives me:
    [itex] m_1(v_{i1g}^2-v_{f1g}^2)+m_2(v_{i2g}^2-v_{f2g}^2) [/itex] right?
     
  15. May 28, 2015 #14

    haruspex

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    We can drop the g subscripts, yes? And to keep it general for now, introduce vi2 into the momentum equation, and for clarity, measure all velocities as positive in the same direction:
    [itex] m_1v_{i1}+m_2v_{i2} = m_1v_{f1}+m_2v_{f2} [/itex]
    [itex] m_1(v_{i1}^2-v_{f1}^2)+m_2(v_{i2}^2-v_{f2}^2)=0[/itex]
    Rearrange each equation to have m1 on one side and m2 on the other.
    You didn't yet factorise the energy equation. How do you factorise ##a^2-b^2##?
     
  16. May 28, 2015 #15
    ok so its:
    [itex] m_1(v_{i1}-v{f1})(v_{i1}+v{f1}) + m_2(v_{i2}-v{f2})(v_{i2}+v{f2}) =0[/itex]
    then
    [itex] m_1(v_{i1}-v{f1})(v_{i1}+v{f1}) = -m_2(v_{i2}-v{f2})(v_{i2}+v{f2}) [/itex]
     
  17. May 28, 2015 #16

    haruspex

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    Yes, and rearranging the momentum equation?
     
  18. May 28, 2015 #17
    v_{f2}=\frac{m_1(V_{i1}+v_{f1})}{m2} [/itex]
    not you can plug in vf2 into the KE equation and vi2 is 0 so you can just throw vi2 away right? then the only unknown you have is vf1 and then you have it solved correct?
     
  19. May 28, 2015 #18

    haruspex

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    Sure, but what I'm trying to guide you to is the elastic form of Newton's Experimental Law, which is a very handy equation to know.
    You have ##m_1(v_{i1}-v_{f1})(v_{i1}+v_{f1}) = -m_2(v_{i2}-v_{f2})(v_{i2}+v_{f2})## and ##m_1v_{i1}+m_2v_{i2} = m_1v_{f1}+m_2v_{f2}##.
    Rewriting the latter as ##m_1v_{i1}-m_1v_{f1} = m_2v_{f2}-m_2v_{i2}##, can you see how to combine these two equations to get one that does not involve any masses?
     
  20. May 28, 2015 #19
    uhm can you divide them?

    also vi2 is 0 in this case right?
     
  21. May 28, 2015 #20

    haruspex

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    Yes, and yes.
     
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