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Momentum head on collision problem

  1. Jun 25, 2005 #1
    Code (Text):
    A                B
    -->           <--
    24kgm/s      10kgm/s
    Two trucks A and B are about to collide head on; their values of linear momentum are as shown in the figure above. After the collision the two trucks separate and move away from each other, at which time truck A has a linear momentum of 8.0 kgm/s.

    My answer was 22 kgm/s because I assumed in my calculations that A and B will move in opposite directions after the collision, as this is what I infer from the statement 'after the collision the two trucks separate and move away from each other'.

    However, this is wrong and the answer given in the book is 6 kgm/s.

    I'm not sure what to make of this. :confused: Can someone please explain?
  2. jcsd
  3. Jun 25, 2005 #2
    I'm guessing it's looking for truck B's momentum? Conserve energy and momentum is the trick here, there are no external forces. Also, post some of your work (equations).
  4. Jun 25, 2005 #3
    Could you show your working please? I also got 6 kg m s^-1.
  5. Jun 25, 2005 #4
    Oops, sorry! I forgot to include the question asked.

    Calculate the momentum of truck B after collision and state its direction of travel.

    I was thinking that

    Total original momentum = Total current momentum
    14 kgm/s = -8 kgm/s + Current momentum of B
    Current momentum of B = 22 kgm/s

    There is nothing said in the question about energy being conserved. It does not say anywhere that it is an elastic collision.
  6. Jun 25, 2005 #5
    If you notice, the net momentum vector of the system (obtained by adding the initial momentum vectors of each component) is 14Ns to the right. Since momentum is conserved (no outside forces) this will not change. Taking the right to be positive, you have:

    Total momentum = A + B = +14Ns
    A Momentum = 8Ns
    B's Momentum must be such that the total momentum is 14Ns to the right, so it must be 6Ns.
  7. Jun 25, 2005 #6
    I really thought A would move to the left after collision. It's this part of the question 'After the collision the two trucks separate and move away from each other' that confuses me.
  8. Jun 25, 2005 #7


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    Well, as long as the "forward" truck has higher speed, they can move in the same direction and still separate.
  9. Jun 25, 2005 #8
    Yes, I see that now. However, I still feel that the question lacks some information. We don't know the masses of A and B for one.

    Is the question ambiguous?
  10. Jun 25, 2005 #9
    You don't need them. The conservation of linear momentum is

    [tex]p_{A1} + p_{B1} = p_{A2} + p_{B2}[/tex], where the letters and numbers specify the momenta of each particle before and after the collision (eg. [itex]p_{A1}[/itex] is the momentum of A before the collision, etc).
  11. Jun 25, 2005 #10
    I just found out from the internet that 'head-on' collisions are in fact elastic collisions. We weren't taught this at school.

    Right now, I would like to apply the principle of conservation of energy to the problem but I can't seem to find a starting point.
  12. Jun 25, 2005 #11


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    Recon, you are right. Out of the two choices for the final momentum of A, it does look like you made the right choice. That last sentence must be interpreted in the original rest frame, and so I too would have made the same choice.

    PS : "Head on" collisions do not have to be elastic, and even if this one was, there's not enough data to apply CoE.

    This is just an ambiguously worded question and the solution has probably assumed p_A (final) = 8 kgm/s with the positive x-axis pointing to the right. (This wasn't drawn in the picture, was it ?)
    Last edited: Jun 25, 2005
  13. Jun 25, 2005 #12


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    I agree. The question seems ambiguous to me also. You cannot know before solving the problem which way truck A is moving after the collision unless you assume the 8kgm/s is to be taken as positive, continuing in the original direction. I thought perhaps you could eliminate one possibility based on energy (assuming no explosion to increase the energy of the system), but in fact you cannot.

    You can show (below) that either solution demands the trucks be of equal mass. If energy were conserved, both trucks would recoil with an exchange of velocity. Since the -8 and +22 solution has the trucks recoiling with less than inital velocities, no kinetic energy is created, but not much is lost. The +8 and +6 solution corresponds to a much higher, and no doubt more realistic energy loss. However, without specifying energy loss or final direction, both solutions are valid

    [tex] m_A v_{Ai} + m_B v_{Bi} = m_A v_{Af} + m_B v_{Bf} [/tex]

    [tex] m_A \left( {v_{Ai} - v_{Af} } \right) = m_B \left( {v_{Bf} - v_{Bi} } \right) [/tex]

    [tex] \frac{{m_A }}{{m_B }} = \frac{{\left( {v_{Bf} - v_{Bi} } \right)}}{{\left( {v_{Ai} - v_{Af} } \right)}} [/tex]

    [tex] m_A v_{Ai} + m_B v_{Bi} = 24Ns - 10Ns = 14Ns [/tex]

    [tex] K.E._i = \frac{{\left( {24Ns} \right)^2 }}{{2m_A }} + \frac{{\left( {10Ns} \right)^2 }}{{2m_B }} [/tex]

    Case 1: Mass A moves to the left

    [tex] m_A v_{Af} + m_B v_{Bf} = - 8Ns + 22Ns = 14Ns [/tex]

    [tex] \frac{{m_A }}{{m_B }} = \frac{{\left( {v_{Bf} - v_{Bi} } \right)}}{{\left( {v_{Ai} - v_{Af} } \right)}} = \frac{{22Ns - \left( { - 10Ns} \right)}}{{24Ns - \left( { - 8Ns} \right)}} = \frac{{32}}{{32}} = 1 \Rightarrow m_A = m_B = m [/tex]

    [tex] \Delta K.E. = \frac{{\left( {8Ns} \right)^2 + \left( {22Ns} \right)^2 }}{{2m}} - \frac{{\left( {24Ns} \right)^2 + \left( {10Ns} \right)^2 }}{{2m}} = \frac{{\left( {548 - 676} \right)N^2 s^2 }}{{2m}} = - \frac{{64N^2 s^2 }}{m} [/tex]

    Case 2: Mass A moves to the right

    [tex] m_A v_{Af} + m_B v_{Bf} = 8Ns + 6Ns = 14Ns [/tex]

    [tex] \frac{{m_A }}{{m_B }} = \frac{{\left( {v_{Bf} - v_{Bi} } \right)}}{{\left( {v_{Ai} - v_{Af} } \right)}} = \frac{{6Ns - \left( { - 10Ns} \right)}}{{24Ns - \left( {8Ns} \right)}} = \frac{{16}}{{16}} = 1 \Rightarrow m_A = m_B = m [/tex]

    [tex] \Delta K.E. = \frac{{\left( {8Ns} \right)^2 + \left( {6Ns} \right)^2 }}{{2m}} - \frac{{\left( {24Ns} \right)^2 + \left( {10Ns} \right)^2 }}{{2m}} = \frac{{\left( {100 - 676} \right)N^2 s^2 }}{{2m}} = - \frac{{288N^2 s^2 }}{m} [/tex]

    Case 2 does seem more representative of an actual collision, but not the only possible answer.
    Last edited: Jun 25, 2005
  14. Jun 25, 2005 #13

    *looks at olderdan's post*

    I wish it had, though. :biggrin:

    But, seriously, I'm going to look through that in the morning (Brunei time). Need to catch some sleep now. Thanks for all the help guys!
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