The experiments at LHC hit protons on each other at total energy of 7 TeV.
In comparison, a flying moscquito has a kinetic energy of
1 TeV: a trillion electronvolts, or 1.602×10−7 J, about the kinetic energy of a flying mosquito[12]
Could one collide two flying bees of 3.5TeV kinetic energy and get a proton proton collision? i.e. give the total energy of the bee on a proton of each bee's proboscis?
The answer is "no" because macroscopic objects as the bee's proboscis have a surface composed by a very large number of atoms and the contact of the two proboscises will disperse the energy among those atoms much before it can reach one proton.
So it is true that macroscopic objects have a lot of kinetic energy in TeV, but to get it on an elementary particle level is a problem that has not been solved. Actually, maybe, with nanotechnology something might be devised, with magnetic fields and electric fields and God knows what, but certainly it will not be with pistons.
The perfect head on collision is a special case where we don't need to worry about any relative angles. We can solve it using Physics 1 conservation of momentum and energy, all in the lab frame.
For equal mass particles
$$m u_1 = m v_1 + m v_2 \implies u_1 = v_1 + v_2,$$
and
$$\frac{1}{2} m {u_1}^2 = \frac{1}{2} m {v_1}^2 + \frac{1}{2} m {v_2}^2 \implies {u_1}^2 = {v_1}^2 + {v_2}^2 .$$
We can square the momentum equation and find
$${u_1}^2 = {v_1}^2 + {v_2}^2 + 2 v_1 v_2 = {v_1}^2 + {v_2}^2.$$
So
$$2 v_1 v_2 = 0.$$
This means one of the particles has zero velocity after the collision. Either particle 1 passed right through particle 2 with no effect (aphysical), or particle 1 stops and transfers all of its momentum and energy to particle 2.
In the center of mass frame this would be scattering at $180^\circ$. The fact that the first particle stops must break one of M&T's assumptions. In this case the scattering angle of particle 1 $\theta$ is ill defined. What is the direction of motion for a particle that is stopped?
In billiards you see a ball bounce backwards, because it was spinning backwards before the collision (while sliding along the table). It stops. Angular momentum is conserved so it keeps spinning. Friction catches the spin on the table and it starts to roll backwards. If the ball is rolling on the table it will continue to move forward after the collision for the same reason.
I don't have the reputation to comment, but setting $\theta=0$ is not the problem, as another answer suggests:
$$\tan\xi=\cot 0\rightarrow\infty \implies \xi=\frac{\pi}{2}$$
Best Answer
We do. The LHC accelerates two protons, each with 3.5 TeV of energy, giving a total of 7 TeV in the CoM frame (The energies are from the initial phase of the previous LHC run. Later in the run this was increased to 8 TeV and the combination of the two dataset was what discovered the Higgs boson. The energies are roughly doubling now for Run II, to 13 TeV).
The main reason for this is as you mentioned, the energy involved. In any frame, we have the following invariant quantity, $s = (p_1 + p_2)^2$ which its square root, $\sqrt{s}$, gives the Centre of Mass (CoM) energy for the experiment, and here $p_i$ represents the momentum four-vector for each particle i. In a collision where two particles are moving in opposite directions with equal energy we have the following:
$$ s \equiv (p_1 + p_2) \cdot (p_1 + p_2) = (E + E, \mathbf{p}_1 − \mathbf{p}_2 ) \cdot (E + E, \mathbf{p}_1 − \mathbf{p}_2 ) = (2E , 0 ) \cdot (2E, 0) = 4E^2$$
and now the CoM energy is given by the square root of this quantity, $$\to E_{CoM} = \sqrt{s} = 2E$$
In a an experiment where one of the particles is at rest (has mass $m_t$) and the other is travelling with momentum $\mathbf{p}$ (and has mass $m_b$) we have the following:
$$ s \equiv (p_1 + p_2) \cdot (p_1 + p_2) = (E_b + m_t, \mathbf{p}_b) \cdot (E_b + m_t, \mathbf{p}_b) = E_b^2 + m_t^2 + 2E_b m_t − p_b^2 = m_t^2 + m_b^2 + 2E_b m_t $$
Assuming the masses are negligible, we have the fixed target (FT) CoM energy,
$$\to E^{\text{FT}}_{CoM} = \sqrt{s} = \sqrt{2E_b m_t}$$
Thus we would need much more energy input in a fixed target experiment to achieve the same energies as in the case with two co-moving particles.
EDIT: Regarding a comment below which I think arises from confusion of what the CoM frame is. $\sqrt{s}$ gives the CoM energy in both cases. This is useful because we can now compare between a fixed target experiment and an experiment where both particles are accelerated at the same speed but in different directions.
So, say my collider has the capability to produce a magnetic field which at its maximum, can accelerate a charged particle so as to have energy of 3.5 TeV. Now in the case that we have two particles with the same energy going in opposite directions, we will give a total CoM energy of 7 Tev, following the result above. In the second case though, theres only one accelerating particle, hence $\sqrt{s} = \sqrt{2 \times 3.5 \times m_t}$ and since E $\ll$ m, this is always less than in the first case.
So be careful, because both experiments can be transformed into a CoM frame. In the CoM frame $|\mathbf{p_1}| = -|\mathbf{p_2}|$. Note this is true in both experiments, even in the second case where one of the particles is stationary. Well, the whole point is that we can use the above formulas so we can skip transforming to the CoM frame; we can compute this quantity directly.