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Perfectly Inelastic Collision

  1. Dec 14, 2004 #1
    This is probably easy for most people, but I can't quiet seem to understand it.

    Two automobiles have a perfectly inelastic collision. The automobiles, which are identical models except for the color and which contain only drivers of identical masses, meet at an intersection. Each was moving with speed 12m/s, one coming from the south, the other from the east. What is the velocity of the final composite object?

    It was an old homework problem, just studying for my final exam and I am trying to understand it. The answer set from my prefessor says that the final speed is (6^2+6^2)^(1/2) = 8.4 m/s. I am not sure why its half the speed of the original... I know the formula is mv + Mv = (M+m)V. Any tips on why/how this problem is solved would be appreciated. Thanks.
  2. jcsd
  3. Dec 14, 2004 #2


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    You have this

    [tex] m \vec{v}_{1} + m \vec{v}_{2} = (m + m) \vec{v}_{both} [/tex]

    [tex] m \vec{v}_{1} + m \vec{v}_{2} = 2m \vec{v}_{both} [/tex]

    [tex] \frac{\vec{v}_{1}}{2} + \frac{\vec{v}_{2}}{2} = \vec{v}_{both} [/tex]

    then you have

    [tex] \frac{-12 \vec{i}}{2} + \frac{12 \vec{j}}{2} = \vec{v}_{both} [/tex]


    [tex] \sqrt{6^2 + 6^2} = | \vec{v}_{both} | [/tex]
    Last edited: Dec 14, 2004
  4. Dec 14, 2004 #3

    Andrew Mason

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    The momentum of the two cars is 12 m/s North and 12 m/s West. So the momentum is ((12m)^2 + (12m)^2)^1/2 = 2mv which can be written as ((6^2 x (2m)^2 + 6^2 x (2m)^2)^1/2 = 2mv

  5. Dec 14, 2004 #4


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    You can use the properties of momentum to calculate this, so you can use vectors for the individual components. So just look up :biggrin:
  6. Dec 14, 2004 #5
    Ah thanks, I see what I did wrong. Gosh, such a silly algebra mistake.
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