How do you calculate the force exerted on the smooth wall with an elastic collision of a bullet of known mass travelling at a constant velocity? If a bullet elastically collides with a smooth wall (assume the wall wont break under the force) to rebound (in a vacuum, considering all momentum is conserved), and we know its mass and the constant velocity with which it is travelling, can we determine the force exerted on the wall during the contact (due to the deceleration of the bullet to zero velocity, then a velocity equal in magnitude to the initial velocity but opposite in direction)?
It seemed to me that $$F = \frac{mv - \mu}t$$ is a good start, but then we need to know the contact time then. Is that anyhow directly proportional to the mass (with a certain inertia) in any other way?
 A: 
can we determine the force exerted on the wall during the contact

There is no unequivocal answer to this question.
The answer is: it depends on the 'hardness' of the bullet.
Please note that momentum is always conserved (in the absence of external forces), regardles of whether the collision is elastic or not.
Re. my first statement, let's consider the wall to be of infinite hardness, so that it undergoes no deformation during the collision whatsoever.
So only the bullet undegoes reversible deformation. Reversible because the collision is elastic. If it was inelastic, permanent deformation of the bullet would constitute energy loss, characteristic of an inelastic collision.
As a very simple model and thought experiment, let's consider the bullet to be a Hookean spring, where $F$ is the restoring force:
$$F=-kx$$
and $x$ represents the deformation of the bullet during the collision.
Assume the bullet, pre-collision, has a velocity $v$ and mass $m$, then its Kinetic energy $K$ was:
$$K=\frac12 mv^2$$
As the bullet slows down during the collision, it loses $K$, while the   bullet gains elestic energy. Using Energy Conservation we can write:
$$\frac12 mv^2-\frac12 k x_{max}^2=0$$
$$\Rightarrow x_{max}=\sqrt{\frac{m}{k}}v$$
where $x_{max}$ is the maximum deformation during the collision.
We can then also write:
$$F_{max}=-k\sqrt{\frac{m}{k}}v=-v\sqrt{km}$$
where $F_{max}$ is the maximum deceleration force the bullet exerts on the wall (and vice versa of course, with $\text{N3L}$)
Because $k$ represents the 'hardness' of the bullet, it's clear $F_{max}$ depends on this hardness, as well as depending on $v$.
