Calculating Post-Collision Speeds of Billiard Balls

[pupatel]

I was wondering if someone can help me with this problem:

Two biilard balls of equal mass undergo a perfectly elastic head-on collison. If the speed of one ball was initially 2.00 m/s and of the other 3.00 m/s in the opposite direction, what will be their speeds after the collision??

 

[Chen]

See this:
http://hyperphysics.phy-astr.gsu.edu/hbase/colsta.html#c2
In short - the speed of the first ball would be 3 m/s and the speed of the second ball would be 2 m/s.

Conservation of momentum:
m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2

m_1 = m_2

v_1 + v_2 = u_1 + u_2

(v_1 + v_2)^2 = (u_1 + u_2)^2

v_1^2 + 2v_1v_2 + v_2^2 = u_1^2 + 2u_1u_2 + u_2^2

Conservation of energy:
\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2

m_1 = m_2

v_1^2 + v_2^2 = u_1^2 + u_2^2

So we have:
v_1^2 + 2v_1v_2 + v_2^2 = u_1^2 + 2u_1u_2 + u_2^2

v_1^2 + v_2^2 = u_1^2 + u_2^2

Substract them:
2v_1v_2 = 2u_1u_2

u_1 = \frac{v_1v_2}{u_2} = v_1 + v_2 - u_2

u_1 = v_1v_2 = v_1u_2 + v_2u_2 - u_2^2

u_2^2 - (v_1 + v_2)u_2 + v_1v_2 = 0

A quick inspection would reveal that the solutions to that are v1 and v2. v2 is the speed before collision, so we are left with v1. Hence u1 = v2 and u2 = v1. There are probably much easier ways to come to this but it's late. :zzz:

 

[pupatel]

I am still confused...I use these equations but it doesn't give me the right answer...

 

[krab]

I was wondering if someone can help me with this problem:

Two biilard balls of equal mass undergo a perfectly elastic head-on collison. If the speed of one ball was initially 2.00 m/s and of the other 3.00 m/s in the opposite direction, what will be their speeds after the collision??

They will interchange speeds. Ball 1 was at 2m/s, after collision is -3m/s. Ball 2 was at -3m/s, after collision is 2m/s.

Edit: I notice Chen already gave this answer. Here's another derivation:
v_1+v_2=u_1+u_2\mbox{ and }v_1^2+v_2^2=u_1^2+u_2^2
Move them around a bit:
v_1-u_1=u_2-v_2\mbox{ and }v_1^2-u_1^2=u_2^2-v_2^2
Divide these last two equations and get:
v_1+u_1=v_2+u_2
Subtract this from the first equation and also add it to the first equation, get:
v_2=u_1\mbox{ and }v_1=u_2

 

[Chen]

I knew there was an easier way to get to that.