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How much energy is lost in an inelastic collision?

  1. Nov 19, 2016 #1
    1. The problem statement, all variables and given/known data
    "Suppose a ##1000 kg## car slides into a stationary ##500 kg## moose on a very slippery road, with the moose being thrown through the windshield. (a) What percent of the original kinetic energy is lost in the collision to other forms of energy? (b) What percent of the original kinetic energy is lost if the car hits a ##300 kg## camel? (c) Generally, does the percentage loss increase or decrease if the animal mass decreases?"

    2. Relevant equations
    ##m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}##
    ##K=\frac{1}{2}mv^2##

    ##m_0=1000kg##
    ##m_1=500kg##
    ##m_2=300kg##

    ##K_0=\frac{1}{2}m_0v_0^2##

    3. The attempt at a solution
    Okay, so I figured out pretty much (a) and (b), but will need someone to check how I got the values for those problems. I also made a derivative proof for (c) that I would like someone to go over. Let ##m_0## be the car, ##m_1## be the moose, and ##m_2## be the camel.

    (a)

    ##(m_0)(v_0)=(m_0+m_1)(v_1)##
    ##v_1=(\frac{m_0}{m_0+m_1})(v_0)=\frac{2}{3}v_0##

    I set this equal to the new velocity for the kinetic energy calculation needed to calculate the loss in ##K_0##.

    ##K_0'=\frac{1}{2}m_0v_1^2=\frac{1}{2}m_0(\frac{2}{3}v_0)^2=\frac{2}{9}m_0v_0^2##
    ##K_1'=\frac{2}{9}m_1v_0^2##

    Now here, I took the sum of all the kinetic energy remaining (in the moose and in the car).

    ##∑K'=\frac{2}{9}v_0^2(m_1+m_0)##

    Then I divided it by ##K_0## to yield the kinetic energy remaining out of kinetic energy before the collision.

    ##\frac{∑K'}{K_0}=(\frac{\frac{2}{9}v_0^2(m_1+m_0)}{\frac{1}{2}m_0v_0^2})=\frac{4}{9}(\frac{m_1+m_0}{m_0})=\frac{2}{3}##

    The total percentage of energy lost is given by: ##1-\frac{∑K'}{K_0}=\frac{1}{3}##

    (b)

    ##(m_0)(v_0)=(m_0+m_2)(v_2)##
    ##v_2=(\frac{m_0}{m_0+m_2})(v_0)=\frac{10}{13}v_0##
    ##K_0'=\frac{1}{2}m_0v_2^2=\frac{1}{2}m_0(\frac{10}{13}v_0)^2=\frac{50}{169}m_0v_0^2##
    ##K_2'=\frac{50}{169}m_2v_0^2##
    ##∑K''=\frac{50}{169}v_0^2(m_2+m_0)##

    ##\frac{∑K''}{K_0}=\frac{\frac{50}{169}v_0^2(m_2+m_0)}{\frac{1}{2}m_0v_0^2}=\frac{100}{169}(\frac{m_2+m_0}{m_0})=(\frac{100}{169})(\frac{1300}{1000})=\frac{130}{169}##

    The percentage of energy lost is: ##1-\frac{∑K''}{K_0}=\frac{39}{169}##

    (c) Here is the part that I wish for someone to check.

    I'm looking for the percentage of kinetic energy loss when an object of fixed mass is forced into an inelastic collision with a body of variable mass. This value is given by:

    ##g(m)=1-\frac{∑K}{K_0}=1-2(\frac{m_0}{m+m_0})^2(\frac{m_0+m}{m})=1-2(\frac{m_0^2}{m+m_0})(\frac{1}{m})##

    So: ##g(m)=1-2(\frac{m_0^2}{m^2+m_0m})##

    I differentiated this with respect to ##m## and got:

    ##g'(m)=\frac{2(m_0^2)(2m+m_0)}{(m^2+m_0m)^2} ≥ 0## for all ##m##, since mass can never be negative or zero.

    Therefore, as mass increases, the percentage of energy loss will increase. Conversely, as mass decreases (as is asked by the problem), the percentage of energy loss will decrease. Can anyone check my derivation of part (c)? Additionally, can anyone check if my process in (a) and (b) are correct? I'm particularly worried about if ##m## goes to zero, in part (c).
     
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  3. Nov 19, 2016 #2

    haruspex

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    If you take the equation you start with in c) and plug in the values for the masses you do not get the answers you found in a) and b).
    The answer to b) can be simplified.
     
  4. Nov 19, 2016 #3

    Simon Bridge

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    You are free to pick variables to be anything you like - and you have been clear. It is best practice to make the variables easy to read though, so m_m and m_c for the mass of the moose and the camel respectively and, maybe M for the car since it is a big mass? You can see how it makes the equations later easier to read and thus troubleshoot if needed? (You can also use u and v for initial and final velocities etc.) It's a niggle only because different people feel differently about this, and you were clear enough in your definitions.

    You appear to have assumed that the animal sticks to the car in each case - reasonable but, ewww.

    loss of KE would be ##-\Delta K## (check: because a negative loss would be a "gain".)
    The percentage loss would be ##-100\Delta K/K## right?
    If ##\Delta K = K'-K## (final minus initial) and we take ##g## as the "percentage loss", then:
    ##g = 100(K-K')/K = 100K(1-K'/K)/K = 100(1-K'/K)## ... aaand this is where you lost me, because my equation now looks different from yours.

    ... I am not following the reasoning, so I cannot show you where it went wrong.
    Suggest recheck.
     
  5. Nov 19, 2016 #4

    haruspex

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    It says the animal joins the driver in the vehicle. Passenger seat, one hopes.
     
  6. Nov 19, 2016 #5

    Simon Bridge

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    Yeah, "through the windsheild"... ikk. And the driver keeps going straight through the whole thing.
     
  7. Nov 19, 2016 #6
    Let me try again: ##g(m)=1-\frac{m_0}{m+m_0}##. Differentiate to get: ##g'(m)=\frac{m_0}{(m+m_0)^2}≥0## for all ##m##. So as ##m## increases, percentage loss in ##K_0## increases; conversely, as ##m## decreases, percentage loss in ##K_0## decreases. My only gripe with this, now, is that if ##m=0##, ##g'(0)=\frac{1}{m_0}##. How would I interpret this if it were correct?

    ##g(m_1)=\frac{3}{13}##

    I'm sorry. I should have labelled my variables better. I also took into account the kinetic energy that was transferred to the stationary animal when calculating percentage of kinetic energy loss. That was what the ##K_2'## and ##K_1'## were supposed to represent. I added the kinetic energy of the animals and the car, and divided them over the initial kinetic energy of the car before the collision.
     
  8. Nov 19, 2016 #7

    haruspex

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    Why is that a problem? g(0)=0.
     
  9. Nov 19, 2016 #8
    You're right. I'm just wondering about how I should interpret ##g'(0)=\frac{1}{m_0}##.
     
  10. Nov 19, 2016 #9

    haruspex

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    It tells you how the energy loss grows, from zero, as the mass of the victim increases, from zero. Hit a fly mass mf and the fraction of energy lost is mf/m0.
     
  11. Nov 19, 2016 #10
    Mm, okay then. Thanks for your help, everyone.
     
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