Actually, I use 13 equations
This should come as no surprise, since:
A rigid body has 6 degrees of freedom.
In addition, the force (or impulse) is unknown as well, giving 3 extra unknowns.
Hence, you should expect to need 6+6+3=15 unknowns.
The "normal impact" condition specifies the tangential impulses to be zero, hence you're left with 13 unknowns.
The equations of linear and angular momentum of both objects give you 12 equations; the last is given with equal normal velocity at the contact point at the end of the first impact period.
This uniquely determines the motion state at the end of the first impact period.
To bring the analysis to finish, we use a simple relation between the secondary impulse and the primary impulse in order to gain the final motion states.
Your professor is incorrect.
Let M objects make N normal impacts with each other (that is, there exist N points of collisions, distibuted among the M objects. The location of these N points are ASSUMED KNOWN)
We therefore have 6M+N unknowns to determine!
In the first impact period, 6M independent equations are gained from linear and angular momentum equations. The remaining N equations are the equal normal velocity demands at the N contact points.
Specifying the secondary impulses in terms of the primary impulses enable us to determin the final motion state.
Hope this clears up a bit; if you'd like a detailed discussion of your dumb-bell problem, just say so
EDIT:
I would, however stress that your professor has identified a very important point, although he had dismissed the possibility of solving it too soon:
In order to solve general impact problems, you have to use some assumptions regarding the basically unknown impulse. The normal impact assumption is by far the most useful one (it implies the "normal" two-collision equations).
To give one subtle example:
Assume that you have solved a general impact problem up to and including the first period of impact. In particular, we assume that we've solved an oblique impact problem, where the primary tangential impulse(s) have not been zero.
What now of modelling the secondary impulse?
Now, we know this is associated with the reversal of elastic deformations; but we cannot in general conclude that the coefficients of elasticity is equal in the normal and tangential directions!
Reversals in the normal directions arre typically responses to compression deformations, whereas reversals in the tangential directions are responses to shear deformations.
Most materials have completely different attitudes to these deformation types.
Hence, it is not generally true that the primary and secondary impulses are collinear..
2.EDIT:
Finally, I'll explain why your professor is correct in saying you'll need only 3 equations in your case.
This is not because my approach is wrong (it is not; it is the most general approach of impact analysis if we want to avoid a differential equations approach).
You'll strictly need only 3 equations because you've got lots of information in this particular case yet unused!
A) 2-dimensional normal impact problem:
This means each rigid body has only 3 degrees of freedom; hence, in general, you'll need only 3+3+1=7 equations (in the first period).
B) One of the objects is a sphere/circle.
Since we have normal/direct impact, this means that the sphere cannot gain rotational velocity as a result of the impact (i.e, it experience a centered impact)
This reduces our number of fundamental unknowns by one.
C)Unidirectional motion:
Impacts are along the horizontal, hence there will be generated no vertical C.M. velocities; our fundamental unknown velocities are therefore reduced to 3:
New velocity of the sphere, C.M velocity of dumb-bell, angular velocity of dumb-bell.
Under our assumptions, we see that for the system sphere+dumb-bell, we gain by conservation of linear and angular momentum, we have 2 of the necessary 3 equations.
(Note that the impulse itself falls out in the system!)
Do we need the impulse here (that is, a total of 4 unknowns), or can we find a 3rd equation relating the velocities?
This is given by Newton's condition, which for an elastic collision says that the final separation velocity at the contact point must be the negative of the initial collision velocity at the contact point.
(Collision velocity is one object's relative velocity "into" the other object)
Newton's condition is certainly implied in my approach; occasionally (as in your case), Newton's condition plus conservation of the system's angular and linear momenta are sufficient to solve the problem.
That is, we may by-step the admittedly cumbersome procedure I've sketched; just remember that for general problems, this remains the technique to solve them (in contrast to simplified versions which merely hold in certain cases).