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Homework Help: Starting on elastic collisions

  1. Jan 19, 2005 #1
    Currently I'm stuck on this problem:

    A pair of bumper cars at an amusement park ride collide elastically as one approaches the other directly from the rear. One has a mass of 450 kg and the other 550 kg. If the lighter one approaches at 4.5 m/s and the other is moving at 3.7 m/s, calculate their velocities after the collision.

    Now, I'm sure that I understand the concept of using the conservation of momentum and relative velocity to get two equations to solve for the two unknowns, but it isn't working out. Here's what I did.

    Conservation of momentum:
    [tex]m_1v_1 + m_2v_2 = m_1v_1^{'} + m_2v_2^{'}[/tex]

    [tex]450(4.5) + 550(3.7) = 450v_1^{'} + 550v_2^{'}[/tex]

    [tex]4060 = 450v_1^{'} + 550v_2^{'}[/tex]

    Because of elastic collision, relative velocity is constant:
    [tex]v_1 + v_2 = v_1^{'} + v_2^{'}[/tex]

    [tex]4.5 + 3.7 = v_1^{'} + v_2^{'}[/tex]

    [tex]v_1^{'} = 8.2 - v_2^{'}[/tex]

    And now, subbing back into the conservation of momentum equation...

    [tex]4060 = 450(8.2 - v_2^{'}) + 550v_2^{'}[/tex]

    [tex]4060 = 450(8.2) - 450v_2^{'} + 550v_2^{'}[/tex]

    [tex]370 = 100v_2^{'}[/tex]

    BAM! The final velocity is equal to 3.7m/s! But wait, that's what it was in the BEGINNING! Grrr! I've had this problem with five different elastic collision questions, so I must be doing something consistently wrong! Could someone please point out where I'm veering off-course?

  2. jcsd
  3. Jan 20, 2005 #2
    You are making a fundamental mistake: setting the sum of velocities before and after collision equal to each other! Instead you should use energy conservation. Also, there is a ratio called the coefficient of restitution which relates relative velocites. For an elastic collison, this would be

    [tex]e = 1 = \frac{v'_{1}-v'_{2}}{v_{2}-v_{1}}[/tex]

    (check the signs...e should be positive).

    instead of your equation. Do you see the difference?

    I'll leave the task of reading about e to you (it should be mentioned in your general physics textbook). IN case you have a problem with my explanation, please feel free to ask and I'll clarify.

  4. Jan 20, 2005 #3

    Doc Al

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    Staff: Mentor

    Just to restate what vivek explained:
    In a perfectly elastic collision, the relative velocity is reversed. But the relative velocity is [itex]v_2 - v_1[/itex], not [itex]v_1 + v_2[/itex]. So that means [itex]v_2 - v_1 = v'_1 - v'_2[/itex]. (Note that this reversal of relative velocity is derived by combining conservation of momentum and conservation of energy.)
  5. Jan 20, 2005 #4
    To whoever moved this, I'm in grade 12 Physics!

    Anyways, thanks for the help, guys. Doc, I used your equation, but it's still not working! I'll show you what I did, taking it from the second equation:

    [tex]v_2 - v_1 = v'_1 - v'_2[/tex]

    [tex]3.7 - 4.5 = v'_1 - v'_2[/tex]

    [tex]-.8 - v'_2 = v'_1[/tex]

    [tex]4060 = 450(-.8 - v'_2) + 550v'_2[/tex]

    [tex]4060 = -360 - 450v'_2 + 550v'_2[/tex]

    [tex]3700 = 100v'_2[/tex]

    [tex]v'_2 = 37m/s[/tex]

    This is obviously incorrect... what am I doing wrong now?
  6. Jan 20, 2005 #5
    On the 2nd equation, it should be + v'_2 , shouldn't it?

    Btw, I don't know why this problem got moved to the college level forum :eek:
  7. Jan 20, 2005 #6
    Well, Doc gave that equation that way... I'll try adding it in a sec.
  8. Jan 20, 2005 #7
    Ok, that equation's not right either. Would somebody please just solve the equation for me, since evidently I'm completely unable to get the right answer?
  9. Jan 21, 2005 #8
    As I pointed out earlier, the coefficient of restitution (equal to 1 in your case) must be a positive quantity by definition. Make sure that the ratio you use is positive. Secondly, the cars are moving toward each other so if you take the direction of motion of one of the cars as positive, the other must be negative to maintain sign consistency (and even physics..because identical signs mean motion in the same direction). I do not know if you are introduced to vector algebra yet but I would suggest that you re-do the problem defining a positive direction of motion and prefixing signs to velocities (before and after collision) accordingly as they are parallel or antiparallel to the assumed +ve direction.

    Note that this problem arises only when terms which are linear in velocities are involved (such as the restitution ratio expression, momentum conservation, relative velocity...) but not if you used the energy conservation equation directly, i.e. kinetic energy before collision = kinetic energy after collision (of both cars).

    Hope that helps...

  10. Jan 21, 2005 #9

    Doc Al

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    Staff: Mentor

    As futb0l pointed out, you messed up on the last step. If you did it right, you'd get [itex]v'_2 = v'_1 + 0.8[/itex]. Now just combine that with the equation you had in your first post ([itex]4060 = 450v_1^{'} + 550v_2^{'}[/itex]), and you should be able to solve it easily.
  11. Jan 22, 2005 #10
    USE CONSERVATION OF ENERGY! That with conservation of momentum will give you your two equations. Since it is elastic, energy is conserved also.
  12. Jan 22, 2005 #11
    I at first tried using conservation of momentum with conservation of kinetic energy, but instead of simply plugging in the momentum equation to the kinetic energy one, I plugged in the kinetic energy equation to the momentum equation, giving me rooted values to deal with. I've got it now, thanks.
  13. Jan 22, 2005 #12
    glad you got it :smile:
  14. Jan 22, 2005 #13

    Doc Al

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    Staff: Mentor

    You are absolutely correct. That's the most straightforward way to solve the problem. But if you happen to know that relative velocity is reversed in an elastic collision, you can take advantage of that fact. The equations are slighty easier to work with, since they are linear.

    But just to be clear: By combining (1) conservation of momentum with (2) reversal of relative velocity, you are using conservation of energy! That's because (2) is derived by combining (1) with conservation of energy.
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