Collision of elastic spheres : final velocities 
I was able to make the sketch of this, but I wanted to find the formulas for the x,y velocity components for each ball after the collision. I let $v_{x1}, v_{y1}$ be the velocities of the first ball after the collision, and similarly $v_{x2}, v_{y2}$ for the second ball. This is 4 unknown variables. Then I wrote the conservation of momentum along x, y axis, and conservation of energy:
$mv - (2m)v = mv_{x1} + 2mv_{x2}$ along x axis
$0 = mv_{y1} +2mv_{y2}$ along y axis
$\frac{mv^2}{2} + \frac{2mv^2}{2} = \frac{m(v_{x1}^2 + v_{y1}^2)}{2} + \frac{2m(v_{x2}^2 + v_{y2}^2)}{2}$ for energy.
This is only 3 formulas, and 4 unknowns, so I can not solve for the velocity components without more equations. I think there is some information involved with the geometry of the question (that one of the balls' center aligns with the other's top or bottom), but I do not know how to write this out mathematically, especially since the masses of the two objects are different.
 A: Shift your perspective into the rest frame of one of the masses. Then, you can use the geometry of the setup to determine which direction the at-rest ball goes after the collision, since it has to move in the direction of the force applied by the other ball. This will give you the fourth equation.
A: There is a hidden assumption that the spheres are smooth. This implies that they retain their initial components of velocity parallel to the plane surface of contact. Then there are only 2 unknown variables, the components perpendicular to this plane, which can be found using the 2 equations for conservation of kinetic energy and conservation of linear momentum perpendicular to the plane of contact.
If the spheres were rough you would need additional information : the coefficient of friction and the moments of inertia of the spheres. The impact would provide a torque on each sphere, causing rotational as well as translational motion.
A: I think the graphical nature of this problem is being missed. One: do not think about velocity. Think about momentum, in units of $p_0 = mv$. You identified that the initial (and hence final) momentum is $-p_0\hat x$. Moreover, the initial (and hence final) energy is:
$$ E_0 = \frac 3 {2m} p_0^2 $$
Now consider the final state, there are 2 momenta: $\vec p_1$ and $\vec p_2$, and their sum is fixed:
$$ \vec p_1 + \vec p_2 = -p_0\hat x$$
Draw this, per the problem's request. Note how the tail of $\vec p_1$ and the head of $\vec p_2$ look like the foci of an ellipse.
The foci are at:
$(0,0)$ and $(-p_0, 0)$.
Now add the energy constraint. Suppose $\vec p_1 = (x, y)$, then:
$$ \frac{(x^2 + y^2)}{2m} + \frac{(x+p_0)^2+y^2}{4m}=\frac 3 2 \frac{p_0^2}m$$
Set $p_0=1$ so the foci are at the origin and $(-1,0)$:
and the equation is:
$$ 3x^2+2x+3y^2=5$$
Complete that square and the ellipse give you the allowed value of $\vec p_1=(x, y)$ for a scattering angle, $\theta$ of
$$ \tan \theta = y/x $$.
From there you can compute $\vec p_2$ directly.
A: Let u = velocity of object  A and -u = velocity of object B 
Allow me to assume that the co efficient of restitution p = 1/4
 At the moment of impact , the line joining the two centers makes an angle z with the original direction of motion and 
$sin z = r/2r = 1/2 $
so $z = 30^0$ 
Let the line making 30 degrees be the new frame of reference then the principle of linear momentum  will  conserve velocity in the j direction as the normal reaction to the collision will be in the i direction
   ... velocity before impact       ....mass ....    velocity  after impact
[A]..  $ u cos 30 i + u sin 30 j  $   .... $m$   .......$x i + u sin 30 j$ [pclm]
[B].. $ -u cos 30 i - u sin 30 j  $....$m$ .......$y i - u sin 30 j$
In the i direction the ratios of the relative velocities before and after impact =p [ Newtons law of restitution] ,so
$[x-y]/[u cos 30 - (-u cos 30)] = -[1/4]$
$[x-y]/[2 u cos 30] =1/4$
$x-y = - [\sqrt 3]/4 $.......   Eq 1
Next use pclm in the i direction
mx + my = mu cos 30 +m[- u cos 30] = 0
x+y=0 ......  Eq  2
Solve for Eq  1 and Eq 2 then x= -[ sq rt 3]u/8 and y = [sq rt 3]u/8
So the new velocity v for A after impact is
v[A] = x i + u sin 30 j = -[sq rt 3] [u/8]i + u sin 30 j 
and the new velocity v for B after impact is 
v[B] = y i - u sin 30 j = +[sq rt 3][u/8]i - u sin 30 j 
