Suppose 2 sphere of mass 1Kg move with velocity 5m/s and 10m/s in same direction. Let us call the ball moving with 5m/s blue and other one red with blue ahead of red from origin.

Initial KE of Blue = 12.5 ; Initial KE of Red = 50

Now, assuming elastic head on collision and by conservation of momentum,

$1×5 + 1×10 = 1×u₁ - 1×u₂$ (the one which moves ahead is u₁ and the one which turns back is u₂)

also by conserving KE,

Blue: 0.5×(25 −u₁²)---(a)

Red: 0.5×(100+u₂²)---(b)

and Blue+Red = 0.5×(25+100) or $u₂² = u₁²$ I could solve this by using the above 2 equation.

Now since I have not explicitly mentioned that u₁ is new velocity of blue and u₂ of red, I could easily replace u₁ and u₂ in (a) and (b)

Which won't change the numerical value of u₁ and u₂ but suppose if the balls were indistinguishable i.e point size of same colour, how would I know which one picked u₁ and which one u₂

  • $\begingroup$ u₂ does NOT go negative. Velocity does not have direction. Momentum does not go negative. And if one turned back it would be the liter. u₂ correct or incorrect sign you cannot cannot conclude u₂²=u₁² from conservation of KE. So you cannot distinguish - that is not a physics problem - you start with two identical balls and you end with two identical balls. $\endgroup$ – paparazzo Aug 1 '15 at 15:54
  • $\begingroup$ @Frisbee Sorry, not true. Velocity does indeed have direction. It is a vector after all. Same for momentum. $\endgroup$ – Involutius Aug 1 '15 at 17:21
  • $\begingroup$ @YourAverageMechEng OK velocity changes direction. I mean the - on U2 was wrong. Wrong equation for both. No - (minus) in momentum and no red + (plus) in KE. Even with proper or in-proper sign on KE would not get u₂²=u₁². $\endgroup$ – paparazzo Aug 1 '15 at 17:45

Solving this particular set of equations would yield two values for $u1$ and as such two values for $u2$ - 10 and 5 ,that is, if the value of $u1$ were 5 the value of $u2$ would be 10 and vice-versa.You may arrive at this by solving a quadratic equation in $u1$ (or for that matter,$u2$)

Now,you may have, while solving quadratics seen we,sometimes acquire values of roots that sometimes do not satisfy the physical situation of the problem.These are extraneous solutions.

Here,if the velocity of blue were 5 and that of red were 10, what would that indicate?

It would indicate that even though an external force(an impulse due to the collision) has acted on each of the bodies it has failed to produce any change in their momentum,which would clearly violate newtons laws.This is therefore the unacceptable solution yielded by the quadratic..the other root-the one that says blue moves with a velocity of 10 and red with 5 is completely acceptable.

Another way to look at it would be from the frame of.. say, blue... In the frame of blue, after collision red and blue would exchange their velocities(since bodies with identical masses exchange their velocities in a head on collision- a result which is very easy to prove).Since in this frame, blue is at rest and red is moving with a velocity of 5,after collision, red would be at rest and blue would move with a velocity of 5 in its initial direction.In the lab frame this translates to the result already achieved earlier.


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