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Collisions and loss of kinetic energy

Carla1985

Member
Feb 14, 2013
93
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
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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

Member
Feb 14, 2013
93
Thats great, I'l keep at it then. Thank you :)
 

Carla1985

Member
Feb 14, 2013
93
I cannot for the life of me get this into the form i need it in :/
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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

Member
Feb 14, 2013
93
thats what I'm attempting atm :)
 

Carla1985

Member
Feb 14, 2013
93
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

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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

Member
Feb 14, 2013
93
no, but I went right back and I think I've managed to get the answer :D thanks for the help x