When two balls collide with each other (not head on), will linear momentum still be conserved if there were friction between the balls? 
As shown in this figure, where the surfaces of the two balls A and B are not smooth. Therefore, when they collide there will be frictions.
My question is that since we derived the conservation of momentum using $F=ma$ which is valid for situations where forces are acting on a point mass, will the linear momentum be conserved in this case? I have doubts about it because the direction of friction is not along the centre of mass, and it causes both translational and rotational motion.
 A: Realize that friction is not a external force for the system of two bodies.
Say friction imparts a impulse J on one ball then it will impart -J impulse on the other ball yes the momentum will be conserved
A: Yes, the linear momentum will be conserved, the angular momentum (zero) and the overall energy. All are conservative properties.
However, the translational energy will not be conserved, only the total energy: On the collision energy will be transferred into different forms of energy. Initially the total energy is identical to the translational energy. After the collision it will be split into both, translational and rotational energy - but maintain the overall total.
A: Without external forces, the momentum of system must be conserved. $$F_{external}=0\iff \frac{dP_{system}}{dt}=0$$
this is not hard to prove.
Let's consider each mass point of the whole body, and name them with $1,2,3,...$
Let the force that $j$ gives $i$ be $F_{ij}$ , where $i,j=1,2,3,...,i\ne j$
Then the force that on $i$ would be (without external forces)
$$F_i=F_{i1}+F_{i2}+F_{i3}+\dots=\sum_{j}F_{ij}=\frac{dp_i}{dt}$$
Because $$F_{ij}=-F_{ji}$$
Then$$\sum_{i}\sum_{j}F_{ij}=0$$
We sum $F_i$ and have,$$LHS\qquad\sum_i F_i=\sum_{i}\sum_{j}F_{ij}=0$$
$$RHS\qquad \sum_i \frac{dp_i}{dt}=\frac{d\sum_i p_i}{dt}=\frac{dP_{system}}{dt}$$
Thus,$$\frac{dP_{system}}{dt}=0$$
A: At the time where the collision occurs, you have this situation.

writing the EOM's   you obtain
$$m_1\dot v_1 =f_\mu\\
m_2\dot v_2=-f_\mu\\
I_{o1}\,\dot \omega_1=f_\mu\,\rho_1\\
I_{o2}\,\dot \omega_2=-f_\mu\,\rho_2
$$
multiply with $~dt~$ and  integrating :
$$m_1\left(v_{1f}-v_{1i}\right) =\int f_\mu\,dt=dp\\
m_2\left(v_{2f}-v_{2i}\right) =-dp\\
I_{o1}\,\left(\omega_{1f}-\omega_{1i}\right)=\rho_1\,dp\\
I_{o2}\,\left(\omega_{2f}-\omega_{2i}\right)=-\rho_2\,dp\\
$$
addition equation is that the relative velocity at the contact point towards the normal vector is zero
$$v_{1f}+v_{2f}=0$$
you have  5 equations for 5 unknows $~v_{1f}~,\omega_{1f} ~,v_{2f}~,\omega_{2f}~,dp$
the conservation of the linear momentum means that
$$m_1\,v_{1i}+m_2\,v_{2i}=m_1\,v_{1f}+m_2\,v_{2f}\tag 1$$
with the solution of the above equations you obtain that:
$$v_{1f}=\frac{m_1\,v_{1i}+m_2\,v_{2i}}{m_1-m_2}\\
v_{2f}=-v_{1f}$$
those equation (1) is fulfilled and   the   linear momentum is conserved.

*

*$v~$ linear velocity at the center of mass $c_i$

*$\omega~$ angular velocity

*index f final state

*index i initial  state

*$I_o~$ moment of inertia about point o

*$f_\mu~$ friction force

