Question

A freight car, open at the top, weighing 2.80 tons, is coasting along a level track with negligible friction at 4.10 ft/s when it begins to rain hard. The raindrops fall vertically with respect to the ground. What is the speed (in feet/second) of the car when it has collected 0.200 tons of rain?

Answer 1

Momentum is conserved so P(i)=P(f). M(1)v(i)=(M(1)+M(2))v(f). Convert the tons to lbs (1 ton= 2000 lbs) and plug them in. (2.8*2000)*(4.10)=((2.8+.2)*2000)*v(f)= 3.8267 ft/s