Elastic collision of two identical hard spheres 
Hello, I want to find the relation between impact parameter b (it is negative on the figure) and the scattering angle $\theta$ which is $\theta_1$ on the picture. The collision is elastic, and the spheres are identical: they have an equal radius R and an equal mass m. I have to solve this problem in laboratory frame. Since it is an elastic collision there is conversation of kinetic energy:
$\frac{m{v_1}^2}{2}+\frac{m{v_2}^2}{2}=\frac{m{v_0}^2}{2}$
In addition, there is conversation of momentum:
$mv_1\,\cos\theta_1+mv_2\,\cos\theta_2=mv_0$
$-mv_1\,\sin\theta_1+mv_2\,\sin\theta_2=0$
I assume I know the initial momentum (or $v_0$) and since I want to find the relation between b and $\theta_1$ I have three parameters that I can write as the function of $\theta_1$. But I still don't have any relation for b. How can I find it?
 A: There are some very neat results for an elastic collision conserving translational kinetic energy between a moving sphere and a stationary sphere of the same mass. I'd make use of these (having derived them), rather than continue your present approach.
(a) If the collision were head-on the sphere moving originally would pass all its momentum to the other sphere. Although your collision is not head-on you can consider momentum components along the line joining the centres. You must also assume that at right angles to this line no contact force component acts (otherwise some translational KE would be transferred to thermal  energy or to rotational KE).
You'll find that you need almost no algebra, and that very easy expressions emerge for the angles in terms of the impact parameter and the sphere diameter. For a nice check on your result...
(b) Draw a vector diagram (a triangle) to show conservation of momentum for the system without considering exactly how the momentum divides up between the two spheres. Impose the condition of conservation of translational kinetic energy. This shows the triangle to be a special one, implying a very simple value for $\theta_1 + \theta_2$ that is independent of impact parameter (as long as $0<b<2r$).
Good luck!
A: The standard approach is to move into the reference frame of the mass center, where the two bodies move toward each other with opposite horizontal velocity $v_0/2$. If $b$ is given, it is easy to deduce the deflection angles for the two bodies in this frame. The final step is to move back to the original reference frame and calculate $\theta$.
A: First of all notice that the force that will occur between the two spheres is normal to the surface (assuming no rotation/friction). This means that the change in momentum that both balls will experience will be along the vector $\mathbf r_1-\mathbf r_2$ where $\mathbf r_{1,2}$ are the positions of the centers of the two balls. This motivates us to decompose each velocity into two components. The component normal to $\mathbf r_1-\mathbf r_2$ will stay the same so we can focus on the component that's parallel. This reduces it to a 1D problem. Let's call the two velocities pre-collision $\mathbf u_1,\mathbf u_2$ and the velocities post-collision $\mathbf v_1,\mathbf v_2$ (forgive me for introducing notation).
Let's focus only on the magnitude of the parallel components pre- and post collision and call them $u_1,u_2$ and $v_1,v_2$ respectively. The velocities are given by
\begin{align}
v_1&=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\
v_2&=\frac{2m_1}{m_1+m_2}u_1+\frac{m_2-m_1}{m_1+m_2}u_2
\end{align}
You can derive these equations by transforming to the center of momentum frame and by noticing that in this frame the momenta just get reflected.
In your case ($m_1=m_2=m,u_2=0$) these equations get simplified to
\begin{align}
v_1&=0\\
v_2&=u_1
\end{align}
Now you can construct the post-collision velocities $\mathbf v_1,\mathbf v_2$ by
adding the parallel and normal components together.
To get your $b$-dependence in you will have to figure out how to decompose $\mathbf u_1,\mathbf u_2$ into parallel/normal components such that these depend on $b$. Then you would have to construct the final vectors and calculate the angle.
A: This is geometry problem

$$b=2\,R\sin(\alpha)$$
where R is the ball radius
how you can obtain $~\alpha~$ ?
first find the the collision   point B
A: Your diagram is not drawn quite correctly: the line between the centers of the two masses should point in the direction of $v_2$. If you correct this, you can see that
$$sin(\theta_2)=\frac{b}{2R}$$.
With the other equations you can then relate $\theta_1$ to $b$
