A steady beam of argon atoms (molar mass 0.040 kg/mol) moves in a vacuum chamber…

The mass of an atom. $m = \dfrac{0.040\ \text{kg/mol}}{6.022 \times 10^{23}\ \text{mol}^{-1}} = 6.642 \times 10^{-26}\ \text{kg}$. A useful product: $n m u^2 = 2.0 \times 10^{20}\ \text{m}^{-3} \times 6.642 \times 10^{-26}\ \text{kg} \times (500\ \text{m/s})^2 = 3.32\ \text{Pa}$.

Step 1. Elastic bounce (a). Atoms hitting one square metre per second: $nu\cos\theta = 8.660 \times 10^{22}\ \text{s}^{-1}$. Each reverses its normal velocity and gives the plate $2mu\cos\theta = 5.752 \times 10^{-23}\ \text{kg m/s}$. So

The sideways parts of the momentum do not change, so the force is along the normal.

Step 2. Atoms stick (b). Now each atom gives up all its momentum, $mu$, along the beam. The same number arrives per second, so the force on a square metre is $nmu^2\cos\theta = 3.32\ \text{Pa} \times 0.866 = 2.88\ \text{N}$, along the beam. Its parts are $2.49\ \text{N}$ along the normal (a pressure) and $1.44\ \text{N}$ along the plate (a drag).

Step 3. Moving plate (c). In the plate's frame the atoms come in at $u + w = 600\ \text{m/s}$, and the density is the same. They hit one square metre at the rate $n(u + w)$ and each gives $2m(u + w)$:

Step 4. Speed after the bounce (d). In the plate's frame an atom leaves at $600\ \text{m/s}$. Add the plate's own $100\ \text{m/s}$ to return to the chamber: $u + 2w = 700\ \text{m/s}$.