We can use a qubit as a particle detector that doesn't change the particle's energy. This can be implemented as follows. We start out with a qubit initialized in the state $\left|0\right>$ and we apply the Hadamard gate $U$ that acts as follows:
$$
\begin{split}
U\left|0\right> &= \frac{1}{\sqrt{2}}\left[\left|0\right\rangle + \left|1\right\rangle\right]\\
U\left|1\right> &= \frac{1}{\sqrt{2}}\left[\left|0\right\rangle - \left|1\right\rangle\right]
\end{split}
$$
Note that $U$ is its own inverse, so applying $U$ again will bring the qubit back to the state $\left|0\right\rangle$ we started out with. But now consider what happens if during the time the qubit spends being a superposition of $\left|0\right\rangle$ and $\left|1\right\rangle$ a particle collides with it, but such that no energy is exchanged. Then the qubit will get entangled with the particle, so the qubit-particle system will be in a state of the form:
$$\left|\psi\right\rangle = \frac{1}{\sqrt{2}}\left[\left|0\right\rangle \left|D_0\right\rangle + \left|1\right\rangle\left|D_1\right\rangle\right]$$
where the states $\left|D_{i}\right\rangle$ are the particle states after scattering off the qubit in state $\left|i\right\rangle$. You might think that because the qubit was not affected by the interaction at all, we cannot perform a measurement on the qubit's state to find out that it has interacted with a particle. But watch what happens if we apply the Hadamard gate again to the qubit:
$$U\left|\psi\right\rangle =\left|0\right\rangle\left|D^{+}\right\rangle+\left|1\right\rangle \left|D^{-}\right\rangle$$
where $D^{\pm} = \frac{1}{2}\left[\left|D_0\right\rangle\pm\left|D_1\right\rangle\right]$
So, had there been no interaction, the qubit would have returned to the initial state $\left|0\right\rangle$ but now we end up with an entangled state of the qubit and the particle such that there is now a finite probability of finding the qubit in the state $\left|1\right\rangle$, despite the fact that the collision with the particle happened in a purely elastic way such that it did not affect the physical state of the qubit in any way at the time of the collision. The probability of finding the qubit in the state $\left|1\right\rangle$ is $\frac{1}{2}\left[1-\operatorname{Re}\left\langle D_0\right|D_1\left.\right\rangle\right]$, so it depends on the overlap between the two particle states corresponding to scattering off the qubit in the two states of the superposition.
If the states $\left|D_i\right\rangle$ are orthogonal, then you have 50% probability of finding the qubit in the states $\left|0\right\rangle$ and $\left|1\right\rangle$; the density matrix after tracing out the particle state is $\frac{1}{2}\left[\left|0\right\rangle\langle 0| + \left|1\right\rangle\langle 1|\right]$.
