# How is the final velocity in 2 body collision distributed between the 2 bodies?

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

• 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. – paparazzo Aug 1 '15 at 15:54
• @Frisbee Sorry, not true. Velocity does indeed have direction. It is a vector after all. Same for momentum. – Involutius Aug 1 '15 at 17:21
• @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. – 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,$u2$)