special relativity - How would one compute the angle of deflection, in a relativistic collision - underspecified system?

Consider the simplistic case of two identical mass particles colliding elastically with the second particle initially stationary and the first particle travelling with energy $E$. By conservation of 4-momentum we have:

$$p_{1}^{\mu}+p_{2}^{\mu}=p_{1}'^{\mu}+p_{2}'^{\mu}$$

Taking the inner product of this with itself:

$$\left\langle p_{1}^{\mu} \middle| p_{1}^{\mu}\right\rangle + 2\left\langle p_{1}^{\mu} \middle| p_{2}^{\mu} \right\rangle+\left\langle p_{2}^{\mu} \middle| p_{2}^{\mu} \right\rangle = \left\langle p_{1}'^{\mu} \middle| p_{1}'^{\mu}\right\rangle + 2\left\langle p_{1}'^{\mu} \middle| p_{2}'^{\mu}\right\rangle +\left\langle p_{2}'^{\mu} \middle| p_{2}'^{\mu}\right\rangle$$

Using the fact that $\left\langle p_{i}^{\mu} \middle| p_{i}^{\mu}\right\rangle = m_{0}^{2}c^{2}$, we can simplify this:

$$2m_{0}c^{2}+2m_{0}E= 2m_{0}c^{2}+2\left(\frac{E_{1}'E_{2}'}{c^{2}}-\vec{p}_{1}'\cdot\vec{p}_{2}'\right) \implies m_{0}E=\frac{E_{1}'E_{2}'}{c^{2}}-\vec{p}_{1}'\cdot\vec{p}_{2}'$$

We note that $\vec{p}_{1}'\cdot\vec{p}_{2}'=\|\vec{p}_{1}'\|\|\vec{p}_{2}'\|\cos(\theta)$, where $\theta$ is the inner angle between $\vec{p}_{1}'$ and $\vec{p}_{2}'$.

However, this results in a system with $\|p_{1}'\|$,$\|p_{2}'\|$ and $E_{1,2}$ unspecified and I cannot see how we could thus extract $\theta$ from the initial conditions; what have I misunderstood or misapplied here?