Answer to Question #8895 in Mechanics | Relativity for AB
Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s and is aimed at the side of a car which stops it. Assuming no splashing back, what is the force exerted by the water?
The rate of change of momentum is equal to the net force:
F = dp/dt.
This can be shown using Newton’s second law. So,
F = ΔP/Δt = (P_final - P_initial)/Δt = (0 - 30kg·m/s) / 1.0s = -30N.