Variable mass system : water sprayed into a moving container

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SUMMARY

The discussion centers on the dynamics of a variable mass system, specifically a water cannon ejecting water into a moving container. Key equations derived include the total mass of the system, represented as M(t) = M_C + m(t), and the relationship between force and mass flow rate, expressed as F = u (dm/dt) = ρ A u². Participants clarify the importance of considering the system's momentum and external forces, concluding that the correct application of Newton's second law for variable mass systems is crucial for accurate results. The final derived velocity function is v(t) = (ρ A u² t) / (M_C + ρ A u t).

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  • Knowledge of differential equations and their applications
  • Basic fluid dynamics concepts, particularly related to flow rates
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Students and professionals in physics, particularly those focusing on fluid dynamics, mechanical engineering, and systems involving variable mass, such as rocket propulsion and water cannons.

Su6had1p
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Homework Statement
A water cannon starts shooting a jet of water horizontally, at t = 0, into a heavy trolley
of mass M placed on a horizontal ground. The nozzle diameter of the water cannon is d,
the density of water is ρ, and the speed of water coming out of the nozzle is u. Find the
speed of the trolley as a function of time. Assume that all the water from the jet is
collected in the trolley. Neglect all frictional losses.
Screenshot 2025-09-26 210049.webp


Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system
$$M(t) = M_{C} + m(t)$$
$$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$
$$P_i = Mv + u \, dm$$
$$P_f = (M + dm)(v + dv)$$
$$\Delta P = M \, dv + (v - u) \, dm$$
$$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$
$$F = u \frac{dm}{dt} = \rho A u^2$$
from conservation of momentum , the cannon recoils with the same force which it applies.
$$\quad \frac{dm}{dt} = \rho A u$$
$$\therefore M \frac{dv}{dt} + (v - u) \frac{dm}{dt} = \rho A u^2$$
$$M \frac{dv}{dt} + (v - u) \rho A u = \rho A u^2$$
$$\frac{dv}{dt} = \frac{\rho A u [2u - v]}{M}$$
$$\int \frac{dv}{[2u - v]} = \int \frac{\rho A u}{M} \, dt$$
$$-\ln (2u - v) = \frac{\rho A u}{M} t + c$$ \\
given that at $t = 0, v = 0$
$$\therefore C = -\ln (2u)$$
$$ \frac{dM}{dt} = \rho A u \Rightarrow M(t) = M_C + \rho A u t$$
$$\ln (2u) - \ln (2u - v) = \frac{\rho A u t}{M}$$
$$\ln \left( \frac{2u}{2u - v} \right) = \frac{M - M_C}{M}$$
$$\exp \left[ \ln \left( \frac{2u}{2u - v} \right) \right] = \exp \left[ \frac{M - M_C}{M} \right]$$
$$\frac{2u}{2u - v} = \exp \left[ \frac{M - M_C}{M} \right]$$
$$2u \exp \left[ \frac{M_C - M}{M} \right] = 2u - v$$
$$v = 2u \left[ 1 - \exp \left( \frac{M_C - M}{M} \right) \right]$$
this doesnt match with the answer, am i doing anything wrong ? please help
 
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A is the area of the water cannon, forget to mention
 
Su6had1p said:
A is the area of the water cannon, forget to mention
Did you take into account that the water takes an increasing finite time to reach the container?
 
$$m\frac{d v}{dt}=S\rho (u-v)^2,\quad m=M+\rho S(u-v)t$$
UPD
 
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PeroK said:
Did you take into account that the water takes an increasing finite time to reach the container?
No, I'm not sure about that. I didn't think about it.
 
wrobel said:
$$\frac{d}{dt}(mv)=S\rho u(u-v),\quad m=M+\rho S(u-v)t$$
It's not rendered
 
Su6had1p said:
No, I'm not sure about that. I didn't think about it.
The simplest interpretation is that the cannon follows the container and effectively all the water ejected is in the container at time ##t##.

In that case, the cannon does not recoil.

Does that give the expected answer?
 
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To @Su6had1p :
I have a solution under certain assumptions. Please post the answer that you are told is correct to compare.
Su6had1p said:
this doesnt match with the answer, am i doing anything wrong ? please help
Note that the problem is asking you to find the velocity as a function of time. Your solution has no explicit time dependence on the right-hand side.
 
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kuruman said:
To @Su6had1p :
I have a solution under certain assumptions. Please post the answer that you are told is correct to compare.

Note that the problem is asking you to find the velocity as a function of time. Your solution has no explicit time dependence on the right-hand side.
in my solution there is ##M_{c}## which the the mass of container and ##M## which is shorthand for ##M(t)##. There could be a confusion between those two since the question mentions only ##M##. However now that i have mentioned my notations, i think it would be clear. There is time dependence, if you look at steps above without using the masses i can also write the same answer, that would give explicit time dependence.
since ##M(t) = M_{c} + \rho A u t##
$$v = 2u \left[ 1 - \exp \left( \frac{-\rho Aut}{M} \right) \right]$$
 
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  • #10
kuruman said:
I have a solution under certain assumptions. Please post the answer that you are told is correct to compare.
Unfortunately the Examiner hasn't provided any solution, neither any standard textbooks have a similar problem as far as i know. so my only go to was ChatGPT. It gave the solution :
$$v(t) = \frac{\rho Au^{2}t}{M_{C}+\rho Aut}$$
 
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  • #11
Su6had1p said:
. . . so my only go to was ChatGPT.
What about us at Physics Forums?

The ChatGPT solution matches mine. I recommend that you read this article which includes a step-by-step method for solving variable mass problems. Start with equation (2).
 
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  • #12
kuruman said:
What about us at Physics Forums?

The ChatGPT solution matches mine. I recommend that you read this article which includes a step-by-step method for solving variable mass problems. Start with equation (2).
Definitely, physics forum is a go to ! Especially when chatGPT doesn't correct my approach instead just gives solution. I expect Physics forum to help me identify what mistake in making so that I can correct my thought process.
 
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  • #13
Su6had1p said:
Unfortunately the Examiner hasn't provided any solution, neither any standard textbooks have a similar problem as far as i know. so my only go to was ChatGPT. It gave the solution :
$$v(t) = \frac{\rho Au^{2}t}{M_{C}+\rho Aut}$$
That's what I get with the assumption that the jet is never a significant distance from the container.
 
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  • #14
Here's my starting point. In the rest frame of the container, a small mass of water ##\Delta m## is ejected in time ##\Delta t## at speed ##u -v##.

Conservation of momentum gives:
$$\Delta m(u-v) = (M(t) + \Delta m)\Delta v \approx M(t)\Delta v$$
 
  • #15
PeroK said:
Here's my starting point. In the rest frame of the container, a small mass of water ##\Delta m## is ejected in time ##\Delta t## at speed ##u -v##.

Conservation of momentum gives:
$$\Delta m(u-v) = (M(t) + \Delta m)\Delta v \approx M(t)\Delta v$$
Isn't that the 5th step from my solution above?
 
  • #16
Su6had1p said:
Isn't that the 5th step from my solution above?
Okay, but that was my first step. That leads directly to a differential equation:
$$\frac{d v}{dt} = \frac 1 {M(t)}k(u-v)$$where ##k = \frac{dm}{dt}## is the constant rate of mass being expelled.
 
  • #17
Su6had1p said:
Definitely, physics forum is a go to ! Especially when chatGPT doesn't correct my approach instead just gives solution. I expect Physics forum to help me identify what mistake in making so that I can correct my thought process.
As you have discovered, ChatGPT doesn't give a hoot whether you learn or not because it is constitutionally incapable of diagnosing where you went wrong and why. Here is my diagnosis.

It looks like your mistake is in not having in your head a clear picture of what the "system" is in this case where the mass is variable. That is important for sorting out which force(s) are internal and which are not in the framework of Newton's second law. This task is especially tricky with variable mass systems. The correct equation for Newton's second law to use in variable mass systems is $$ M~\frac{dv}{dt}+(v-u)\frac{dm}{dt}=F_{\text{ext}}$$ where
##M = ~## the variable mass of the system of interest, here the trolley with the accumulated water.
##v=~## the instantaneous velocity of the system of interest relative to the ground.
##u=~## the velocity of the element ##dm## relative to the ground just before it hits the vertical wall.
##F_{\text{ext}}=~## any external force, e.g. gravity, acting on the total variable mass system in this case the system of interest plus all the moving water that could conceivably accumulate in the trolley.

All this is explained in the article I suggested that you to read. So you came up with the equation $$M \frac{dv}{dt} + (v - u) \rho A u = \rho A u^2$$ which is the above equation with an additional term on the right-hand side where the external force belongs. However, in this problem there is no external horizontal force acting on the moving water plus trolley system so the right-hand side should have been zero.
 
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  • #18
... I don't see why analysing forces was necessary, if you are given everything you could possibly know about momentum!
 
  • #19
kuruman said:
As you have discovered, ChatGPT doesn't give a hoot whether you learn or not because it is constitutionally incapable of diagnosing where you went wrong and why. Here is my diagnosis.

It looks like your mistake is in not having in your head a clear picture of what the "system" is in this case where the mass is variable. That is important for sorting out which force(s) are internal and which are not in the framework of Newton's second law. This task is especially tricky with variable mass systems. The correct equation for Newton's second law to use in variable mass systems is $$ M~\frac{dv}{dt}+(v-u)\frac{dm}{dt}=F_{\text{ext}}$$ where
##M = ~## the variable mass of the system of interest, here the trolley with the accumulated water.
##v=~## the instantaneous velocity of the system of interest relative to the ground.
##u=~## the velocity of the element ##dm## relative to the ground just before it hits the vertical wall.
##F_{\text{ext}}=~## any external force, e.g. gravity, acting on the total variable mass system in this case the system of interest plus all the moving water that could conceivably accumulate in the trolley.

All this is explained in the article I suggested that you to read. So you came up with the equation $$M \frac{dv}{dt} + (v - u) \rho A u = \rho A u^2$$ which is the above equation with an additional term on the right-hand side where the external force belongs. However, in this problem there is no external horizontal force acting on the moving water plus trolley system so the right-hand side should have been zero.
I see, F(ext) = 0
 
  • #20
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  • #21
Su6had1p said:
I see, F(ext) = 0
Here's another way to look at it. There are no external forces on the large mass M. A small mass, m, is fired at a known speed (the external force on m is already accounted for in its specified momentum). The mass m collides inelastically with the mass M. There are no external forces in the collision.
 
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  • #22
My approach was:
When the cart has moved s from the cannon, how much water has the cannon ejected? How much of that has reached the cart? What momentum did it supply?
In terms of ds/dt, how much momentum does the cart and contents have now?
From that I got ##\dot s=\frac{\lambda(ut-s)u}{1+\lambda(ut-s)}##, where ##\lambda=A\rho/M##.
I wound up with ##v=u(1-(1+2\lambda ut)^{-\frac 12})##.
 
  • #23
I'm trying to work this out from a Fluid Mechanics perspective with the Momentum Equation and Mass Conservation:

$$ \sum \boldsymbol{F} = \frac{ d}{dt} \int_{cv} \rho \boldsymbol{v} d V\llap{-} + \int_{cs} \boldsymbol{v}\rho ( \boldsymbol {V}\cdot d \boldsymbol{A} ) \tag{Momentum}$$

1758999156617.webp


## \boldsymbol {v} = v \text{i} ## ( the velocity of the cart and its liquid contents rel. to fixed inertial frame )
## \boldsymbol {u} = u \text{i}## the jet velocity ( rel. fixed inertial frame )
## \boldsymbol {V} = ( u-v )\text{i} ##
## \sum \boldsymbol F = 0 ##


With uniform flow properties I'm getting an equation that appears somewhat ugly:

$$ 0 = \frac{d}{dt} \left[ (M + m ) v \right] + \rho A u( u-v ) $$

$$ 0 = \frac{d}{dt} \left[ (M + \rho A ( u-v)) v \right] + \rho A u( u-v ) $$

The term on the left is the rate of momentum accumulating in the control volume ( body plus liquid mass).

is this result in alignment with the rest of the group?
 
  • #24
haruspex said:
1758999006190.webp
I get the same result. A nice approach is to use momentum conservation in an inertial frame moving with the stream of water.

1758999398849.webp

After finding v(t) in this frame, it's easy to transform back to the earth frame.
 
  • #25
erobz said:
Fluid Mechanics perspective
Yes and that is a proper argument:
an article
but nobody knows how to marry this argument with results of very informal speculations from textbooks.
 
  • #26
wrobel said:
Yes and that is a proper argument:
an article
but nobody knows how to marry this argument with very informal speculations from textbooks.
Well the liquid mass in the buggy must be accelerated from u to v inside the buggy? So it feels like an inelastic collision should be relevant like @PeroK mentioned, but when I try to apply the equations I'm getting some uglier...but not unsolvable ODE. Is what I'm doing the common mis application the article talks about? I don't have $40 to spend for a read I probably won't understand! It would be nice to try and talk it out.
 
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  • #28
You know what. Never mind. When I try to solve problems anymore, I get an almost nauseated with trembling anxiety, hyperactivity... It's very off-putting emotional response. Strange, perhaps unbelievable...I know. So just forget I asked. I tried problem solving after a good break from this it immediately returns.
 

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