Concept Question: Rainfall & Draining from Cart

[Izmad]

[SOLVED] Concept Question

Homework Statement

Q2: Suppose rain falls vertically into an open cart which is rolling along a straight track with negligible friction. The speed of the cart

a. increases
b. decreases
c. doesn't change


Q2: Now that the cart is full of water, you poke a hole in the bottom and the water drains out. What happens?

Homework Equations

change in momentum=mv(final)-mv(initial)

The Attempt at a Solution

For the first question I was able to reason out the the speed of the cart decreases.

For the second question I've hypothesized that the speed increases as the weight decreases.

Could someone please verify if I'm correct or not, and attempt to direct me in the correct direction?

thanks



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[Mindscrape]

You are correct. :)

Think about conservation of momentum. Your momentum stays constant, so if m goes up then v must go down. Similarly, if m goes down then v must go up.

 

[saket]

Assuming ideal and symmetric situations, your answer doesn't look good to me!
Suggestion: If you are trying to apply conservation of linear momentum, take care of what is your system. (Initial and final system should be same.) Not that, initially you are considering cart full of water and finally just the cart (without water!).

 

[saket]

I mean, if water is flowing out, take that also in account! (Note hole is at the bottom!)

 

[Dick]

Not correct (not you saket, the OP and mindscrape). Conservation of momentum again. m(initial)*v(initial)=m(leaked)*v(leaked)+m(final)*v(final). m(final)=m(initial)-m(leaked). v(initial)=v(leaked). What does this tell you about v(final)? I should know this one. I made an ass of myself by overly complicating it and giving a wrong answer a few days ago.

 

[Mindscrape]

If the hole is at the bottom the water's momentum isn't contributing to the momentum in the x-direction. So it would be $m_0 v_0 = m_f v_f$ where $m_f = m_0 - m_l$, so then $v_f = \frac{m_0 v_0}{m_0 - m_l}$. Since $m_0 - m_l$ is less than $m_0$ then the velocity will increase. Am I missing something?

 

[Dick]

Yes. You have three mass values and only two terms in your conservation of momentum equation. What's the relation between the initial velocity and the missing term? You are leaking momentum. I already confessed to breaking into an already resolved thread and insisting we have to use F=dp/dt. And doing it wrong. Ask learningphysics. Review my previous post.

 

[Mindscrape]

Hah, maybe I'm about to learn something, or be embarrassed, or both, but I really don't see why we care about the water's momentum if it is orthogonal to the carts momentum. I don't get it in the sense of vectors. Are you guys talking about the total momentum, the magnitude of momentum? If so, I am assuming (and maybe wrongly) that the problem wants to consider just the one component. If that still doesn't clear it up then I am severely humbled and I have to come with some excuse why I am an idiot.

 

[Dick]

Right. You DON'T care about the momentum component of the leaking water that's orthogonal to the carts momentum. You DO care about the momentum of the leaking water that parallel to the carts momentum. The carts momentum isn't constant as it leaks because the leaking water shares the forward velocity of the cart. For it not to do that, the cart would have to spray the water backwards with velocity equal to that of it's forward motion. But then it wouldn't be called a leak. It would be called a 'jet'.

 

[saket]

Just one thing: as the hole is at the bottom, even if it was a 'jet', it doesn't matter. (I am assuming, 'jet' is along the area vector of the hole.)
The direction of motion of the cart: call it X.
The water which is just about to be "leaked", has a velocity component equal to the cart's speed in X-direction. As it leaves through the hole, there is no impulse acting on it in X-direction (we are neglecting any air drag), so it's velocity component along X-direction would remain same.
I hope it helps.