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Combining conservation of momentum and kinetic energy

  1. Dec 7, 2017 #1
    Hi everyone, hopefully someone will be able to point me in the right direction with this problem, I get as far as combining the two equations together and no matter how I rearrange them they don't seem to cancel nicely and leave me with an awful quadratic, I can find vx, but the question specificity states to find m first.

    Question:
    Consider two railway trucks that can travel along a straight railway track which defines the x-axis. A laden truck of mass M =6 .6×103 kg and an empty truck of unknown mass m approach each other at constant velocity. The laden truck has an initial velocity component of ux = +2 .0ms−1 which is half that of the initial velocity component of the empty truck (but in the opposite direction). After an elastic collision, the velocity component of the laden truck is halved, but it continues in the same direction. The unknown post-collision velocity component of the empty truck is vx.
    (b) Rearrange the momentum equation into the form: vx = ...,and rearrange the energy equation into the form m = ....Then combine these simultaneous equations to determine values, first for m and then for vx, using the known values of M and ux.

    The attempt at a solution

    Rearranging equations:
    Momentum:
    $$ m_1u_1+m_2u_2=m_1v_1+m_2v_2 $$
    $$ m_1u_1+m_2u_2-m_1v_1=m_2v_2 $$
    $$ v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2} $$
    $$ v_2=\frac{m_1u_1-m_1v_1}{m_2}+u_2 $$
    $$ v_2=\frac{m_1\left(u_1-v_1\right)}{m_2}+u_2 $$

    Kinetic energy:
    $$ \frac{1}{2}m_1u_1^2\ +\ \frac{1}{2}m_2u_2^2\ =\ \frac{1}{2}m_1v_1^2\ +\ \frac{1}{2}m_2v_2^2 $$
    $$ m_1u_1^2+m_2u_2^2=m_1v_1^2+m_2v_2^2 $$
    $$ m_1u_1^2=m_1v_1^2+m_2v_2^2-m_2u_2^2 $$
    $$ m_1u_1^2-m_1v_1^2=m_2v_2^2-m_2u_2^2 $$
    $$ m_1\left(u_1^2-v_1^2\right)=m_2\left(v_2^2-u_2^2\right) $$
    $$ m_2=\frac{m_1\left(u_1^2-v_1^2\right)}{\left(v_2^2-u_2^2\right)} $$


    This is where It starts getting messy, I cant seem to find a way to get the square terms to vanish which makes this problem a mess, can anyone point me in the right direction?
     
  2. jcsd
  3. Dec 7, 2017 #2

    kuruman

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    It will not be as messy if you put in what the problem is giving you including the symbols. For example, the laden truck has initial velocity ##u_x## and mass ##M## while the empty truck has mass ##m## and twice the velocity of the laden truck in the opposite direction. I would write the initial momentum as ##P_{initial}=Mu_x+m(-2u_x)##. How would you write ##P_{final}##?
     
  4. Dec 7, 2017 #3
    So it would look something like:

    $$ P_{final} = M\frac{u_x}{2} + mv_x $$
     
  5. Dec 7, 2017 #4

    kuruman

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    Yes. Now do the initial and final energy along these lines.
     
  6. Dec 7, 2017 #5
    So we have;

    $$E_{k initial} = \frac{1}{2}Mu_x^2 + \frac{1}{2}m(-2u_x)^2 $$
    $$ E_{k final} = \frac{1}{2}M(\frac{u_x}{2})^2 + \frac{1}{2}mv_x^2 $$
     
  7. Dec 7, 2017 #6

    kuruman

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    Correct. Can you proceed and finish the problem on your own?
     
  8. Dec 7, 2017 #7
    Sorry if i'm being a bit slow, but i'm not seeing how this makes it simplify out, unless i'm going to far these new equations give us:

    $$ v_x=\frac{M\left(u_x-\left(\frac{u_x}{2}\right)\right)}{m}+\left(-2u_x\right) $$
    and
    $$ m=\frac{M\left(u_x^2-\left(\frac{u_x}{2}\right)^2\right)}{\left(v_x^2-\left(-2u_x\right)^2\right)} $$

    and i'm not sure how this helps.
     
  9. Dec 7, 2017 #8

    kuruman

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    Take a good look at your expressions. Can you simplify them?

    On edit: Add the fractions.
     
  10. Dec 7, 2017 #9
    This is the part where I got stuck when I tried it this way before. I'm not sure where I am meant to be looking, the only other suggestion I have found suggests finding the ratio of M/m as a first step, which would give:

    $$ \frac{M}{m} = \frac{( v_x^2 - (-2u_x)^2)}{u_x^2 - (\frac{u_x}{2})^2)} $$

    Which can be expanded;

    $$ \frac{M}{m} = \frac{(v_x - (-2u_x)) (v_x + (-2u_x))}{(u_x - (\frac{u_x}{2})(u_x + (\frac{u_x}{2})} $$
     
  11. Dec 7, 2017 #10

    kuruman

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    You are digressing. You are asked to find an expression vx = ... as your first task. You already have $$v_x=\frac{M\left(u_x-\left(\frac{u_x}{2}\right)\right)}{m}+\left(-2u_x\right)$$
    Can you simplify this as ##v_x=(something)\times u_x##? I will get you started. Begin by noting that ##u_x-\frac{u_x}{2}=\frac{u_x}{2}##.
     
  12. Dec 7, 2017 #11
    Embarrassing how I missed that, would:
    $$ v_x = (\frac{M}{2m} - 2) u_x $$
    be the correct simplification?
     
  13. Dec 7, 2017 #12

    haruspex

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    You are well on the way to solving this one anyway, but for future reference:
    Newton's Experimental Law gives a general relationship between relative velocities before and after collision that does not involve masses. In the special case of KE conservation, it simplifies to: relative velocity after = -(relative velocity before). This can be deduced from conservation of energy and conservation of momentum, but it is worth remembering in itself because it involves no masses and no quadratics.
     
  14. Dec 7, 2017 #13

    kuruman

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    It would be. The comment by @haruspex is good for future reference. The path that the problem suggests is also good for this particular problem. Now you need to complete the second task, write the mass of the empty truck as ##m = ...## Once you have done that, I suggest that you replace in it the expression for ##u_x## that you have already found and solve for ##m##. It should simplify to ##m = (some~number)\times M##.
     
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