Loss of energy in the system as a result of the collision

1. Jun 5, 2010

thereddevils

1. The problem statement, all variables and given/known data

Particle A with mass 2m and particle B with mass 5m move towards each other with speeds of 8u and 2u respectively on a smooth horizontal plane . After the collision , particle B moces with speed u in a direction opposite to that of the original direction . Find in terms of m and u , the loss of enerygy in the system as a result of the collision .

2. Relevant equations

3. The attempt at a solution

0.5(2m)(8u)^2-0.5(5m)(2u)^2=0.5(5m)(u)^2+0.5(2m)(v')^2

simplify and

v'^2=(107u^2)/2

this is the velocity of particle A after collision.

I am not sure how to relate this to the loss of energy , i thought of doing

Loss of enerygy=(0.5)(m)((107u^2)/2) but it doesn't make sense

2. Jun 5, 2010

Fightfish

Re: momentum

The question already tells you explicitly that energy is not conserved in the collision. How can you then use conservation of energy (and what's with the minus sign in your equation?) to tackle the problem?
There's something else that is definitely conserved - and that is encapsulated in the title of your post itself.

3. Jun 6, 2010

thereddevils

Re: momentum

ok attempt no2 :

Yes momentum is conserved -

2m(8u)-(2u)(5m)=2mv'+5mu

v'=u/2 ---1

KE before collision = (0.5)(2m)(8u)^2+(0.5)(5m)(2u)^2=74mu^2

KE after collision=(0.5)(m)(u/2)^2+(0.5)(m)(u)^2=(5mu^2)/(8)

take the difference , KE loss = 73(3/8)mu^2

correct ?