Solution 13
Consider $N$ particles moving randomly in a 2D square frame of side $L$. The force exerted on a wall is the rate of change of momentum.
For a single particle hitting a vertical wall with x-velocity $v_x$: $$ \Delta p = 2mv_x $$ Time between collisions with the same wall: $$ \Delta t = \frac{2L}{v_x} $$ Force by one particle: $$ f = \frac{\Delta p}{\Delta t} = \frac{2mv_x}{2L/v_x} = \frac{m v_x^2}{L} $$
Total force by $N$ particles: $$ F = \sum_{i=1}^{N} \frac{m (v_{x,i})^2}{L} = \frac{N m}{L} \langle v_x^2 \rangle $$ Since the motion is random in 2D ($v_x^2 + v_y^2 = v^2$), we assume isotropy: $$ \langle v_x^2 \rangle = \langle v_y^2 \rangle = \frac{v^2}{2} $$
Thus, the average force is: $$ F = \frac{N m}{L} \left( \frac{v^2}{2} \right) = \frac{N m v^2}{2L} $$
$N = 50$, $m = 5.0\text{ g} = 0.005\text{ kg}$, $v = 2.0\text{ m/s}$, $L = 5.0\text{ m}$. $$ F = \frac{50 \times 0.005 \times (2.0)^2}{2 \times 5.0} $$ $$ F = \frac{0.25 \times 4}{10} = \frac{1.0}{10} = 0.1 \text{ N} $$