Momentum Eqn. Homework Problem:

[UFeng]

I posted this in the advanced forum by mistake so I'm am posting it here as I think it may be more appropriate in this forum.

Homework Statement

A motor boat is speeding at velocity W0 when the motor is turned off and a scoop is lowered into a still lake. The scoop captures flow with a cross section A. If the initial mass of the boat is M0 and the wave and friction drag are negligible, what is the velocity of the boat as a funtion of time?

Homework Equations

the momentum equation. I'm not sure how to account for the "scoop" though

The Attempt at a Solution

I am using the momentum equation but am not sure how to incorporate the effect of the scoop into it.
Any help would be greatly appreciated!

 

[tiny-tim]

Welcome to PF!

Hi UFeng! Welcome to PF!

(try using the X₂ tag just above the Reply box )

A motor boat is speeding at velocity W0 when the motor is turned off and a scoop is lowered into a still lake. The scoop captures flow with a cross section A. If the initial mass of the boat is M0 and the wave and friction drag are negligible, what is the velocity of the boat as a funtion of time?
…
I am using the momentum equation but am not sure how to incorporate the effect of the scoop into it.

I was totally confused until I saw the words "initial mass" …

clearly, the scoop is collecting all the water that flows into it, and somehow sucks it onto the boat, making the boat heavier.

So you need two equations: one to tell you how much heavier the boat is getting each second, and a conservation of momentum equation (remember, the scooped water starts with zero velocity, and ends with the same velocity as the boat).

 

[UFeng]

Oh ok. I think I completely missed the point of the question. I was thinking that the scoop was just acting as a "brake," but it makes more sense that it is collecting water and adding weight to the boat.

I'm thinking I'll need the continuity equation to show the flow onto the boat as a function of time. Does this sound right? Then I will need to figure out how this effects the velocity of the boat as a function of time taking into account the added mass I'm not sure exactly how to go about this, but I will work with it and try to figure out how to relate these two equations.
Also I'm not sure what you mean by "remember, the scooped water starts with zero velocity, and ends with the same velocity as the boat." Wouldn't water flow onto the boat at the same velocity that the boat is traveling (starting with velocity W0 and slowing over time until the flow is zero)? Am I missing something?

Thanks very much!

 

[gamer_x_]

Wouldn't water flow onto the boat at the same velocity that the boat is traveling

If the lake is stationary (i.e. the water is undisturbed), you scoop it up and as soon as it is in your scoop it has the speed of the boat. While it is in the lake it is still stationary, which means that at some point when you began collecting it, it was accelerated to your speed. However, this isn't free acceleration, it's going to cost something for you to continuously accelerate the water from 0 -> speed of boat.

 

[tiny-tim]

Hi UFeng!

I'm thinking I'll need the continuity equation to show the flow onto the boat as a function of time. Does this sound right?

I'm not sure what yo mean by "continuity equation" … you just need the rate of flow of water across a pipe of cross-section A.

Also I'm not sure what you mean by "remember, the scooped water starts with zero velocity, and ends with the same velocity as the boat." Wouldn't water flow onto the boat at the same velocity that the boat is traveling (starting with velocity W0 and slowing over time until the flow is zero)? Am I missing something?

As gamer_x_ says, at some point, it accelerates to the speed of the boat.

Before it's scooped, it has zero velocity, and in the end it has the velocity of the boat … it doesn't matter exactly when or how fast the change happens.

 

[UFeng]

ok, I've come up with this to describe the mass of the boat plus the scooped up water as a function of time:

M(t) = d/dt (density of water)*Volume + Mo
M(t) = [(density of water)*(velocity)*(area of Scoop))* (time)] + M0

I'm still not quite sure what to do to find the velocity of the system as a function of time, W(t). I assume I will need M(t) to find this, but since M(t) has velocity in the equation it would seem to depend on W(t) and W(t) seems to depend on M(t). In other words, it seems like they depend on each other. I think this is where I am getting confused. Thanks again for the help!

Also, my book suggests using the integral momentum equation to find W(t). This makes sense to me but I'm still stuck with the above question. I may be making this much harder than it actually is.

 

[tiny-tim]

Hi UFeng!

M(t) = [(density of water)*(velocity)*(area of Scoop))* (time)] + M0

That's right.

… since M(t) has velocity in the equation it would seem to depend on W(t) and W(t) seems to depend on M(t). In other words, it seems like they depend on each other. I think this is where I am getting confused.

Well, they both depend on t.

Also, my book suggests using the integral momentum equation to find W(t). This makes sense to me but I'm still stuck with the above question. I may be making this much harder than it actually is.

Yes.

Just use conservation of momentum.

 

[UFeng]

ok, I'm still stuck.

Do I need to find an equation for the acceleration(which would be negative) and then use it to find velocity as a function of time.
i.e. dW/dt => then separate variables and integrate from 0 to W(t) => then solve for W(t)

Does this sound right? Any hints would be greatly appreciated!

 

[tiny-tim]

Just do the obvious …

what is the total mass at time t?

how much of that is boat, and how much is water?

what was the original momentum (before the scoop started operating) of all that mass?

and what is its momentum now (at time t)?

 

[UFeng]

the total mass at time t: M(t) = M0 + density * A * W(t) * t

the mass of the boat is M0 and the mass of the water is, density*A*W(t)*t

the original momentum before the scoop is lowered would just be: M0 * W0

the momentum now would be M(t) * W(t)

is this correct?

my problem is finding an expression for W(t) using the integral momentum equation.
There is something very basic I am completely missing here and I keep going in circles. For some reason I can 't seem to understand this easy problem. Thanks for your help!

 

[tiny-tim]

the total mass at time t: M(t) = M0 + density * A * W(t) * t

the mass of the boat is M0 and the mass of the water is, density*A*W(t)*t

the original momentum before the scoop is lowered would just be: M0 * W0

the momentum now would be M(t) * W(t)

What about the momentum of the water?

 

[UFeng]

What about the momentum of the water?

The momentum of the water would be:

density*[W(t)^2]*A*t => @t=0 the momentum of the water would be zero...correct?

so I think the equation would look like this:

M(t)*W(t) = M0*W(t) + density*[W(t)^2]*A*t => @t=0, W(t)=W0...I think. So @ t=0, M(t)*W(t) = M0*W0

does that look right?

 

[tiny-tim]

That's better, but you've calculated the mass of water at time t as if the speed was constant.

It isn't, so you'll need to integrate to get the mass of water.

 

[UFeng]

here is my understanding of the problem. I may be wrong.

The boat is orignally traveling at velocity, M0, and then the motor is shut off. At this point water is scooped onto the boat thus causing it to gain mass and slow down. So, at t = 0, the mass is equal to M0, but when the scoop gathers water the total mass as a function of time is M(t) = M0 + [Mass of water as a function of velocity which is a function of time]. So from what I understand the rate of mass increase is slowing because the velocity is also slowing. If my understanding is wrong this may be why I'm having such a hard time getting it.

 

[tiny-tim]

… So from what I understand the rate of mass increase is slowing because the velocity is also slowing.

Yes, that's correct … so you have to integrate …

total water scooped into the boat after time t = ∫ dt of … ?

 

[UFeng]

That's better, but you've calculated the mass of water at time t as if the speed was constant.

It isn't, so you'll need to integrate to get the mass of water.

ok I'll try this again.

Integrating the mass of water I come up with this:

I'll use "S" as the integration symbol and integrate from W0 to W(t)

Mass water as a function of time=density*A*S[from W0 to W(t)] dt = density*A*[W(t)-W0]

 

[UFeng]

oops, sorry. The last line would be the mass per time not the mass.
I think it would be [W0 - W(t)] instead of [W(t) - W0] also

 

[tiny-tim]

(have an integral: ∫ and a rho: ρ )

I'll use "S" as the integration symbol and integrate from W0 to W(t)

That doesn't make any sense …

how can you integrate over dt from W0 to W(t)?

If you're integrating over dt, it's from 0 to t.

 

[UFeng]

ok how about this

total mass of water from t = 0 to t:

rho * A * S(t=0 to t) W(t) dt

wouldn't this just leave me with rho*A*W(t)*t again...unless, the total mass of water is:

= rho*A*t*[W0 - W(t)] => would this be the mass of water assuming the velocity stayed at W0 minus the mass of water accumulated after, thus leaving you with the actual mass of water?

Thanks again for all your help. Hopefully I'll get this thing eventually.

 

[UFeng]

ok I have this now,

the mass of water added to the boat:
=rho*A*[W0-W(t)]*t

so the total mass of the boat + water scooped up, M(t);

M(t) = M0 + rho*A*[W0-W(t)]*t

again I'm stumped on how to find find velocity as a function of time, W(t) using the integral momentum equation. Any hints?

 

[UFeng]

anyone know what to do next?

 

[UFeng]

unfortunately I'm still stuck. I'd appreciate any help!