At what angle do billiard balls scatter if they collide off center? The angle defined by joining a line from the centers of the balls must be important. But do they follow this angle when viewed in the rest frame of one of the balls or in the CM frame?
The spheres are hard and friction is ignored. I've tried googling a lot. All of the examples include the final angle as given.
Please if give an argument for the answer as well. Thanks in advance.
 A: In any practical application, if the balls collide off-center, they will separate at an angle of 90 degrees. This is implied by momentum conservation:
Let p1i be the initial momentum of ball 1, p1 the final momentum of ball 1, and p2 the final momentum of ball 2. Then by conservation of momentum we have
p1i = p1 + p2, and therefore, (p1i ^ 2) = (p1 + p2) ^ 2 = p1 ^ 2 + p2 ^ 2 + 2p1(dot)p2
Where (dot) symbolizes the dot product, since our momenta are definitely vectors. Now, assuming the balls are rolling on a flat surface, all they exchange is kinetic energy, which is conserved without friction. Therefore we can write (given the mass of each ball is the same):
v1i ^ 2 = v1 ^ 2 + v2 ^ 2, where v are velocities, and thus, p1i ^ 2 = p1 ^ 2 + p2 ^ 2
So therefore, p1i ^ 2 = p1 ^ 2 + p2 ^ 2 + 2p1(dot)p2 = p1 ^ 2 + p2 ^ 2
And thus we see that p1(dot)p2 = 0, implying p1 and p2 (and therefore v1 and v2) are orthogonal. Therefore, the balls will scatter at 90 degrees.
A: If you assume idealised conditions you can calculate this using very little assumptions. For the following I assume elastic collisions (no energy lost) and no rotations. The balls have masses $m_1$ and $m_2$ and they initially have velocities $u_1$ and $u_2$. After the collision they have velocities $v_1$ and $v_2$. Start with conservation of momentum and conservation of energy:
$$m_1\vec u_1+m_2\vec u_2=m_1\vec v_1+m_2\vec v_2\\
\tfrac 1 2m_1u_1^2+\tfrac 1 2m_2u_2^2=\tfrac 1 2m_1v_1^2+\tfrac 1 2m_2v_2^2$$
Let's transform to the Center Of Momentum frame to make the calculations easier. The COM frame is defined such that the total momentum is zero. If $\vec P=m_1\vec u_1+m_2\vec u_2$ then in the COM frame we have
$$\vec u_1'=\vec u_1-P/(m_1+m_2)\\
\vec u_2'=\vec u_2-P/(m_1+m_2)$$
and the same for the $\vec v_1'$ and $\vec v'_2$. I'll leave it as an exercise to show the new total momentum is zero. By transforming to this frame we see that the velocities of the two particles are always opposite to each other:
$$m_1\vec u'_1=-m_2\vec u_2'\\m_1\vec v'_1=-m_2\vec v_2'$$
Using conservation of energy we can further constrain the after-collision velocities. If we fill in $\vec u_2'=-m_1/m_2\vec u_1'$ we get
$$\tfrac 1 2m_1u_1'^2+\tfrac 1 2m_2u_2'^2=\tfrac 1 2\left(m_1+\frac{m_1^2}{m_2}\right)u_1'^2$$
Using the same reasoning for $v$ and repeating for particle 2 we get 
$$\tfrac 1 2\left(m_1+\frac{m_1^2}{m_2}\right)u_1'^2=\tfrac 1 2\left(m_1+\frac{m_1^2}{m_2}\right)v_1'^2\\
\implies \begin{cases}u_1'^2=v_1'^2\\u_2'^2=v_2'^2\end{cases}$$
So from conservation of momentum and energy we can conclude two things. In the COM frame we have a) the velocities are in opposite directions before and after the collision and b) the magnitude of the velocities stays the same after the collision ($|\vec u_1|=|\vec v_1|$). I won't discuss a non-COM frame because you can easily go back to the normal frame and outside the COM frame things become messy. If we want to know more about the angles we have to use something else besides conservation of momentum/energy. During the collision the particles interact using a normal force. If we assume perfectly spherical particles we have that the normal force points in the direction between the centers of the spheres. If sphere's centers are positioned at $r_1, r_2$ then the momentum can only change in the direction of $\vec r_2-\vec r_1$. In the picture below I have taken $m_1=m_2$ to see what this condition implies for the angles:

If we look at the velocity of a sphere pre- and post collision we see that the component perpendicular to $\vec r_2-\vec r_1$ is reflected but the parallel component remained the same. From this picture we also see that the point of contact matters. If the collision is head on we have that $\alpha=0$: the particles bounce back in the direction they came from. Generally they get reflected by an angle $2\alpha$. The variables $\alpha$ and $d$ are also called scattering angle and impact parameter in high energy physics (although they are usually called $\theta$ and $b$ and defined slightly different, yeah I messed that up). Using some trig you can derive that $2r\sin \alpha=d$ with $r$ the radius of the spheres.
A: Collision of the balls in billiards is under the influence of a wide variety of factors. This includes spin, static friction, rolling friction, air resistance, angular velocity of the balls etc. Hence the expected observations given below will not be observed in practise.
From the information you provided, the collision would be elastic in nature. This gives us two constraints- kinetic energy is conserved, momentum is conserved. Under idealistic conditions, the two balls will separate at an angle of ${\pi}\over 2$ with respect to each other. The object ball will go in the direction along the line connecting the center's of the two balls. The cue ball will go perpendicular to the object ball abiding with the constraints. In practice, the angle between the balls can be literally anything based on the spin applied.
