New Problem: Leaky, Rainy, and Slow

Here’s a classic physics problem I asked my MCAT students yesterday. They unanimously chose the wrong answer.

A cart runs along a frictionless track on a rainy day. The rain falls straight down, and some of it lands in the open cart. As the cart accumulates rain, does it slow down, speed up, or keep going the same rate? (Do not worry about the cart running into raindrops ahead of it. We imagine that the raindrops fall in such a way that they either land in the cart or don’t hit it. Also, there’s no wind resistance.)

Rain falls straight down into the cart, which is coasting to the right.

Next, the rain stops, but the cart gets a leak. Water pours out a hole in the bottom of the cart. Does the cart get faster, slower, or stay the same speed? How does its final speed, when all the water has leaked out, compare to its original speed before the rain? (Again, ignore friction.)

Water leaks out a hole in the bottom of the cart as it slides to the right.

The answer is now up here.

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Tags: cart, mechanics, physics, physics problems, water

This entry was posted on February 22, 2010 at 10:09 pm and is filed under problems and solutions. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

7 Responses to “New Problem: Leaky, Rainy, and Slow”

Ian Says:
February 22, 2010 at 11:41 pm
The cart slows down as it fills with rain, then maintains that slower speed as the rain leaks out the hole. The final speed is equal to the original speed times the ratio of the total full weight to the original empty weight.

Am I right? If I am I should get some bonus points for being a little drunk right now.
Brendan Says:
February 23, 2010 at 1:18 am
I’m going to guess that the cart’s speed doesn’t change at any point.
Mark Eichenlaub Says:
February 23, 2010 at 9:40 am
Yes. Now, is the slow-down bigger, smaller, or the same if you account for special relativity?
Nik Says:
February 23, 2010 at 7:13 pm
The cart slows down with the added weight of the rainwater since momentum remains constant.
The cart remains at the same speed as the leaks out. This is because the rainwater is, at this point, moving with the cart. So rather than velocity changing, momentum is lost with the loss in rainwater.
Resultantly the final speed is lesser than the initial speed. The ratio of the 2 speeds would be equal to the ratio of the initial and final weights.

If you count for special relativity, I think the slow-down would be bigger. If I’m wrong here, it’d be really sad.
Answer: Leaky, Rainy, and Slow « Arcsecond Says:
February 23, 2010 at 9:45 pm
[...] Arcsecond Playing on the Sea-Shore, Rough Pebbles Welcome « New Problem: Leaky, Rainy, and Slow [...]
Mark Eichenlaub Says:
February 23, 2010 at 10:03 pm
Hi Nikita,

In normal Newtonian mechanics, momentum is just

$p = mv$ .

If we want the momentum to be conserved, then

$dp = mdv + vdm = 0$

meaning in Newtonian mechanics

$dv = \frac{-vdm}{m}$ .

In relativity, momentum is

$p = \gamma m v$

with $\gamma$ something that increases as $v$ increases. So

$dp = m v d\gamma + \gamma v dm + \gamma m dv = 0$ .

This gives

$dv = \frac{-mvd\gamma - \gamma vdm}{\gamma m}$

$d\gamma$ is negative because the cart slows down, so this term decreases the slowdown compared to Newtonian mechanics. The second term is actually the same as the Newtonian term when you cancel the factors of $\gamma$ in the numerator and denominator, so overall the relativistic slowdown is less.
Leakier, Slower, and No Rain « Arcsecond Says:
December 7, 2010 at 2:50 pm
[...] while ago, I asked a standard freshman physics problem about a cart that has rain fall into it, then opens a hole and rain leaks out. Then I gave an [...]