Collisions and loss of kinetic energy

[Carla1985]

Could someone please check my answers so far for this question as its getting very messy so I'm not sure if I've made a mistake :/

Two particles of masses 2m and m are moving along the x−axis with constant velocities u₁i and u₂i respectively, where u₁ > u₂. They collide ‘head on’, and the coefficient of restitution of the collision is e. Find the velocities of each of the particles after the collision, and show that the loss in the kinetic energy of the system of two particles is:

$\frac{(1-e^2)m(u_1-u_2)^2}{3}$

so far I have:
the velocities of the particles after impact are v₁=v₁i and v₂=v₂i

so 2mu₁i-mu₂i=2mv₁+mv₂
ie 2mu₁-mu₂=2mv₁+mv₂ (which is my first equation)

then from the law of restitution:
e(u₁-u₂)=v₁-v₂ (second equation)

solving the 2 equations I get
v₁=1/3((e+2)u₁-(e+1)u₂)
v₂=1/3((2-2e)u₁+(2e-1)u₂)

the to get the loss of kinetic energy:
(1/2*2m*(v₁)²+1/2*m*(v₂)²)-(m*(u₁)²+1/2*m*(u₂)²)

but obviously (v₁)² and (v₂)² gets very messy

Thanks x

 

[I like Serena]

Hi Carla1985!

Messy or not, it seems to me you're going the right way.
You did seem to make a sign mistake in your first equation (it should be $2mu_1+mu_2$).
You may want to correct that before getting into the messy stuff. ;)

 

[Carla1985]

Thats great, I'l keep at it then. Thank you :)

 

[Carla1985]

I cannot for the life of me get this into the form i need it in :/

 

[Ackbach]

Have you tried working backwards from the form you're supposed to prove? Not an ideal method of solution, but it can sometimes get the job done.

 

[Carla1985]

thats what I'm attempting atm :)

 

[Carla1985]

are the first two equations then:
2mu₁-mu₂=2mv₁+mv₂
e(u₁-u₂)=v₁-v₂​as I seem to lose the 1/3 when I do that?

 

[Ackbach]

Hmm. So you've got
$$ \frac{(1-e^{2})m(u_{1}-u_{2})^{2}}{3}
= \frac{\left(1- \left( \frac{v_{1}-v_{2}}{u_{1}-u_{2}} \right)^{2}\right)m(u_{1}-u_{2})^{2}}{3}
= \frac{\left(\frac{(u_{1}-u_{2})^{2} - (v_{1}-v_{2})^{2}}{(u_{1}-u_{2})^{2}}\right)m(u_{1}-u_{2})^{2}}{3}$$
$$= \frac{\left((u_{1}-u_{2})^{2} - (v_{1}-v_{2})^{2}\right)m}{3}
= \frac{m(u_{1}^{2}- 2u_{1}u_{2}+u_{2}^{2} - v_{1}^{2}+ 2 v_{1}v_{2} - v_{2}^{2})}{3}.$$

I have to run right now. Is this what you get?

 

[Carla1985]

no, but I went right back and I think I've managed to get the answer :D thanks for the help x