2 Cars Colliding with same mass

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In an elastic collision between two cars of equal mass, the speeds of the cars after the collision will simply swap. For example, if Car1 is moving at +4 m/s and Car2 at -2 m/s before the collision, after the collision, Car1 will move at -2 m/s and Car2 at +4 m/s. This principle holds true for all elastic collisions involving equal masses. The discussion emphasizes the importance of using momentum and kinetic energy equations to derive these results. The conclusion is that the swapping of speeds is a consistent outcome in such scenarios.
YinHoNg
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Hiya. The problem I'm trying to solve is this:

  • I have two cars, each with the same mass.
  • Speed before Collision of Car1 = Xm/s
  • Speed before Collision of Car2 = Ym/s

To solve the problem of what speeds the two cars have after an ELASTIC collision, i used simultaneous equations.
My process was:

  1. make a K.E equation
  2. Make a momentum equation
  3. Solve for V1, and then thus V2

From my workings, i seem to have found that if the masses of the two vehicles are equal, then in an elastic collision; the speeds afterwards of the two vehicles will just swap.

e.g.

  1. Both cars' mass = 2Kg
  2. Speed of Car1 before collision = +4m/s
  3. Speed of Car2 before collision = -2m/s
  4. Speed of Car1 after collision = -2m/s = Speed of Car2 before collision
  5. Speed of Car2 after collision = +4m/s = Speed of Car1 before collision

Will that be the same for all collisions where the mass is the same and the collision is elastic?
 
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YinHoNg said:
Will that be the same for all collisions where the mass is the same and the collision is elastic?
Indeed it will. (Good thinking!) Why not try to prove this for yourself in general using the same method that you used above, but using symbols, not plugging in specific numbers. You'll need to solve these two equations simultaneously:
mv_1 + mv_2 = mv'_1 + mv'_2
mv_1^2 + mv_2^2 = mv'_1^2 + mv'_2^2
 
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