A freight car, open at the top, weighing 1.80 tons, is coasting along a level track with negligible friction at 6.80 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 0.100 tons of rain?

Been dreading this one so I saved it for last. If I could just get a hint on where to start that would be really appreciated!

Question

A freight car, open at the top, weighing 1.80 tons, is coasting along a level track with negligible friction at 6.80 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 0.100 tons of rain?

Been dreading this one so I saved it for last. If I could just get a hint on where to start that would be really appreciated!

matt101 · Answer

It seems tricky because the mass is continuously changing...but it's actually really easy!

Consider conservation of momentum :)

anonymous · Answer

So then its just$$m1 \Delta v1 = -m2 \Delta v2$$
$$1.80(6.80) = -(1.80 + 0.100) v2$$
$$v2 = \frac{ 1.90 * 6.80 }{ 1.9 } = 6.44$$

matt101 · Answer

That's the right answer! I just want to tweak your equation a bit.

If you use delta v, you're solving for impulse (change in momentum), not momentum itself. Also no need for a negative on the other side like you might do for thermochemistry.

So you have m1v1=m2v2  -->  (1.8)(6.8)=(1.9)v2  -->  v2=6.44

anonymous · Answer

Okay, that makes more sense. I just took what we had on our formula sheets. I'll be sure to study this equation versus the one I used. Thanks a lot!!