Answer
0.056 N.
Work Step by Step
By Newton’s third law, the force on the pan due to the impacting rain is equal in magnitude to the force on the rain due to the pan.
Calculate the force on the rain from its change in momentum. $$F_{average}=|\frac{\Delta p}{\Delta t}|=|\frac{m(v_f-v_o)}{\Delta t}|$$
The rain does not rebound, so its final speed is 0.
$$F_{average}=\frac{m(v_o)}{\Delta t}$$
The mass striking the pan is (volume)(density).
$$F_{average}=\frac{\rho V(v_o)}{\Delta t}=\frac{\rho Ah(v_o)}{\Delta t}$$
$$F_{average}=\frac{(1000kg/m^3)(1.0m^2)(0.025m) (8.0m/s)}{3600s}=0.056N$$
