Edexcel M2: Collisions and Kinetic Energy Question

v2=(11-2v1)/3=√((29-2(v1)^2)/3)
(1/3)(11-2v1)=(1/3)(√87-v1√6)

A particle, mass 2kg and velocity 4m/s, collides horizontally with an equal-sized particle, mass 3kg, which is moving in the same direction of the first particle, at a quarter of the speed.
The loss of kinetic energy in this collision equals 3J. Calculate the velocities of both particles which result from this collision.


My Working:

Conservation of momentum: m1u1+m2u2=m1v1+m2v2

2(4)+3(1)=11=2v1+3v2

Conservation of energy: 3=(1/2)[m1(u1)^2+m2(u2)^2-m1(v1)^2-m2(v2)^2]

3=(1/2)[2(4)^2+3(1)^2-2(v1)^2-3(v2)^2]
3=16+(3/2)-(v1)^2-(3/2)(v2)^2
3=17.5-(v1)^2-(3/2)(v2)^2
(v1)^2+(3/2)(v2)^2=14.5
2(v1)^2+3(v2)^2=29

Rearrange to v2

v2=(11-2v1)/3=√((29-2(v1)^2)/3)
(1/3)(11-2v1)=(1/3)(√87-v1√6)
v1(2-√6)=11-√87
v1=(11-√87)/(2-√6)=-3.72

v2=(1/3)(11-2(-3.72))=6.15

The markscheme says v1=1 and v2=3.

The second quoted line doesn't follow from the first quoted one.

Can you correct it, or do you need further input?

Help me out there. How should they follow?

$\displaystyle \frac{11-2v_1}{3}=\sqrt{\frac{29-2v_1^2}{3}}$

Now square both sides, so:

$\displaystyle \frac{121-44v_1+4v_1^2}{9}=\frac{29-2v_1^2}{3}$

Then rearrange into a quadratic - as brianeverit mentioned.

Two solutions. One you will be able to discard because it will give a physically impossible combination of v1,v2 - work them both out first to find out which.