Conservation of momentum — Water dropped from a firefighting airplane

[Amik]

Homework Statement

It is in the attempt section

Relevant Equations

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I could not unserstand part d.Does problem ask about plane's speed immediately after it starts to drop water or after the water is dropped completely?

 

[Cutter Ketch]

I believe it means after all of the water has left the plane.

 

[Amik]

Can you explain the answer?

 

[Cutter Ketch]

Hmm ... in the first image you posted you have a set of questions. The second image appears to be from some solution key and each solution includes the answer and an explanation. The explanation for part D seems pretty thorough to me. I’m not sure I could improve upon it.

Perhaps you could tell me what about the explanation seems wrong or confusing?

 

[Amik]

I just can not understand why velocity does not change.

 

[Amik]

let me think about where i think It is wrong

 

[PeroK]

I just can not understand why velocity does not change.

It takes a force on the plane to change the velocity of the plane.

 

[Cutter Ketch]

Suppose you are traveling at a constant speed with a ball in your hand. At some point you open your hand. You don’t put any force on the ball just opening your hand. You don’t accelerate the ball and the ball doesn’t accelerate you. If it weren’t for gravity and wind resistance the ball would continue traveling with you at zero relative speed as if you were still holding it.

 

[Amik]

But vertical velocity increases?

 

[kuruman]

I do not mean to derail the discussion, but it seems to me that the answer provided in (b) is flawed which makes the answer in (c) also flawed. Here is why. The plane's mass increases over a 12 s interval, not instantaneously. If we assume that the mass increases linearly over the 12 s interval, we have $m(t)=m_0+\mu t$ where $\mu=(6000/12) ~\mathrm{kg\cdot s^{-1}}$. Then one needs to solve Newton's second Law $$\frac{dp}{dt}=m(t)\frac{dv}{dt}+v(t)\frac{dm}{dt}=0$$to find $v(t)$, then obtain $a(t)$ by differentiation and finally the average force (over time) from $$\langle F \rangle_T =\dfrac{{\int_0^{T}m(t)a(t)~dt}}{{\int_0^Tdt}}$$where $T=12~\mathrm{s}$. This method gives the same answer $v(12~\mathrm{s})=24~ \mathrm{m/s}$ as it should for this "inelastic collision" but a larger value for the average force, also as it should. Perhaps this treatment is beyond OP's level in which case I apologize for the interjection.

 

[Amik]

The knowledge I learned is not enough for me to understand your solution.Most likely you are right.However, this is a AP physics 1 problem

 

[kuruman]

Thanks for the clarification. Just so that you understand the the solution given to you is an approximation.

 

[haruspex]

I do not mean to derail the discussion, but it seems to me that the answer provided in (b) is flawed which makes the answer in (c) also flawed. Here is why. The plane's mass increases over a 12 s interval, not instantaneously. If we assume that the mass increases linearly over the 12 s interval, we have $m(t)=m_0+\mu t$ where $\mu=(6000/12) ~\mathrm{kg\cdot s^{-1}}$. Then one needs to solve Newton's second Law $$\frac{dp}{dt}=m(t)\frac{dv}{dt}+v(t)\frac{dm}{dt}=0$$to find $v(t)$, then obtain $a(t)$ by differentiation and finally the average force (over time) from $$\langle F \rangle_T =\dfrac{{\int_0^{T}m(t)a(t)~dt}}{{\int_0^Tdt}}$$where $T=12~\mathrm{s}$. This method gives the same answer $v(12~\mathrm{s})=24~ \mathrm{m/s}$ as it should for this "inelastic collision" but a larger value for the average force, also as it should. Perhaps this treatment is beyond OP's level in which case I apologize for the interjection.

$\dot p=m\dot v+\dot mv$ is popular but not really valid. That equation would only make sense if a force acting on an object could change its mass. Note that applied to the later part of the question, where it drops the water, it yields the wrong result. It only works for the first part because the water being scooped has no momentum of its own.

I do not understand your objection to $F_{avg}=\frac{\Delta p}{\Delta t}$ for finding the average force exerted on the plane. Your method seems to be answering a different question, the average force exerted by water being scooped up on the plane+water already scooped.

 

[kuruman]

$\dot p=m\dot v+\dot mv$ is popular but not really valid.

Can you explain why not? We have a plane + some water of total mass $m$ moving with speed $v$. At time $\Delta t$ later the mass is $m+\Delta m$ and the speed $v+\Delta v$. Momentum conservation requires that $mv=(m+\Delta m)(v+\Delta v)$ which, to first order, becomes $0=m \Delta v+v \Delta m$. Is this not equivalent to $\dot p=m\dot v+\dot mv$ in the limit $\Delta t \rightarrow 0$?

Your method seems to be answering a different question, the average force exerted by water being scooped up on the plane+water already scooped.

Perhaps. I imagined myself on this plane with instruments that independently record the acceleration and the mass of the plane at regular time intervals. If someone asked me "what is the force on the plane at any given time?", I would say "the product of the two measurements." The time-averaged product would be the average force. Since water is the only entity exerting a force on the plane, this would be the "average force of the water on the plane" that the problem asks for. However, considering OP's post #11, the intended calculation of the average force is a simplification of this.

 

[PeroK]

Can you explain why not? We have a plane + some water of total mass $m$ moving with speed $v$. At time $\Delta t$ later the mass is $m+\Delta m$ and the speed $v+\Delta v$. Momentum conservation requires that $mv=(m+\Delta m)(v+\Delta v)$

Conservation of momentum only applies if you account for all the mass before and afterwards. In your equation you have $\Delta m$ appearing on the RHS (final momentum). But, there is no term on the LHS for the initial mometum of the $\Delta m$.

If an object "gains mass", then it will speed up or slow down depending on the collision with the additional mass.

Likewise, if an object loses mass, then it may speed up (if it fires the mass in one direction), slow down (if it fires the mass in another direction) or remain at constant velocity (if it simply releases the mass).

 

[kuruman]

Conservation of momentum only applies if you account for all the mass before and afterwards. In your equation you have $\Delta m$ appearing on the RHS (final momentum). But, there is no term on the LHS for the initial mometum of the $\Delta m$.

I agree. Perhaps I was not clear about $\Delta m$. It is the mass of the water that is scooped up in time $\Delta t$. It is assumed to have zero velocity before being scooped up. If it were to be formally included in the LHS side, the equation would be $mv+\Delta m*0=(m+\Delta m)(v+\Delta v)$.

 

[haruspex]

It is the mass of the water that is scooped up in time Δt. It is assumed to have zero velocity before being scooped up.

Yes, that is the flaw in the equation.
It has the look of some nice general law, but if we define p(t) as the momentum of the plane plus the water it is carrying at whatever instant then dp/dt only represents the net force acting on it if the incoming mass brings no momentum with it.

 

[haruspex]

the average force is a simplification of this.

No, it is not a simplification. It answers the question intended, the average force acting on the plane, as an entity distinct from its cargo. (Since when is $\frac{\Delta p}{\Delta t}$ not the definition of average force on an object, objects being of fixed mass?)

 

[kuruman]

No, it is not a simplification. It answers the question intended, the average force acting on the plane, as an entity distinct from its cargo. (Since when is $\frac{\Delta p}{\Delta t}$ not the definition of average force on an object, objects being of fixed mass?)

Yes I see that now. I still don't see why $\dot mv+m\dot v=0$ is invalid. Here is a derivation that sets $\dot P(t)=0$ aside. Let
$M_0$ = mass of empty plane; $m_0$ = total mass of water that is scooped up.
Consider splitting $m_0$ into $N$ discrete pieces each of mass $m_0/N$. Let
$v_k$ = velocity of the airplane plus water after scoop of the $k$th piece. We write the momentum conservation equation for each scoop
$M_0 v_0=\left(M_0 + \dfrac{1*m_0}{N}\right)v_1$
$\left(M_0 + \dfrac{1*m_0}{N}\right)v_1=\left(M_0 + \dfrac{2*m_0}{N}\right)v_2$
$\left(M_0 + \dfrac{2*m_0}{N}\right)v_2=\left(M_0 + \dfrac{3*m_0}{N}\right)v_3$
$\dots \dots \dots$
$\left((M_0 + \dfrac{(k-1)*m_0}{N}\right)v_{k-1}=\left(M_0 + \dfrac{k*m_0}{N}\right)v_k$
$\left(M_0 + \dfrac{k*m_0}{N}\right)v_k=\left(M_0 + \dfrac{(k+1)*m_0}{N}\right)v_{k+1}$
$\dots \dots \dots$
$\left(M_0 + \dfrac{(N-1)*m_0}{N}\right)v_{N-1}=\left(M_0 + \dfrac{N*m_0}{N}\right)v_N$
It is immediately obvious that if all the equations are added together, we end up with
$M_0v_0=(M_0+m_0)v_N$ as expected. By adding up to each of the two intermediate equations betwen the dots, It should also be obvious that
$M_0v_0=\left(M_0 + \dfrac{k*m_0}{N}\right)v_k$
$M_0v_0=\left(M_0 + \dfrac{(k+1)*m_0}{N}\right)v_{k+1}=\left(M_0 + \dfrac{k*m_0}{N}\right)v_k$
Rearranging the last equation,
$M_0(v_{k+1}-v_k)+\dfrac{m_0}{N}[(k+1)v_{k+1}-k v_k]=0$
Now define $(\Delta v)_k \equiv v_{k+1}-v_{k}$ as the increment required to change the velocity from $v_k$ to $v_{k+1}$ and then substitute $v_k=v_{k+1}-(\Delta v)_k$ into the last term on the LHS.
$M_0(v_{k+1}-v_k)+\dfrac{m_0}{N}[(k+1)v_{k+1}-k (v_{k+1}-(\Delta v)_k)]=0$
which becomes
$$\left[M_0+\dfrac{k*m_0}{N}\right](\Delta v)_k +\dfrac{m_0}{N}v_{k+1}=0$$
The mass of the plane plus the mass of the scooped water after the $k$th scoop is, $M_{k}=M_0+\dfrac{k*m_0}{N}$. The first term on the LHS is the momentum change over the $(k+1)$th scoop of the plane plus the water inside it. The second term on the LHS side can be viewed as the momentum change over the $(k+1)$th scoop of the water that is not in the plane. Momentum conservation requires that the two changes add to zero. In other words the plane + water of mass $M_k$ has its speed reduced from $v_k$ to $v_{k+1}$ while the water that is scooped up has its speed increased from zero to $v_k$. Now we can write the change in mass of plane plus water as $(\Delta M)_k=m_0/N$. With this, the momentum conservation equation becomes
$M_k(\Delta v)_k +v_{k+1}(\Delta M)_k=0$.
As $N \rightarrow \infty$, it becomes $Mdv+vdM=0$ which is implied in $\dot Mv+M\dot v=0$.

If I did or assumed something incorrectly, please let me point it out because I cannot see what or where it is.

Happy New Year y'all.

 

[haruspex]

Yes I see that now. I still don't see why $\dot mv+m\dot v=0$ is invalid.

It is true from first principles that $\dot p=m\dot v+\dot mv$, but we need to be clear what p is the momentum of. If it is the momentum of plane+current cargo then m varies, but we cannot blithely apply conservation of momentum to something which is gaining mass from or losing mass to the outside. That mass transfer might very well involve momentum transfer too. When the plane scoops the water there is no such incoming momentum, but when it drops the water momentum is lost with it.

 

[kuruman]

It is true from first principles that ˙p=m˙v+˙mv\dot p=m\dot v+\dot mv, but we need to be clear what p is the momentum of.

Yes! We are in complete agreement here. Unambiguously identifying the system, its components and its boundary are absolutely important. Going back to my posts on this thread, I see that I omitted identifying the system. Belatedly, the system here is the empty plane plus the 6000 kg that will eventually end in its tank at t = 12 s. The momentum of this particular system is conserved from t = 0 to t = 12 s, but not from t = 12 s to t = 14 s if the plane's pilot decides to scoop water for an additional 2 s.

 

[haruspex]

Belatedly, the system here is the empty plane plus the 6000 kg that will eventually end in its tank at t = 12 s.

So for that period $\dot m=0$, yes? And we cannot say what $\dot v$ is because different parts of m have different velocities.

 

[kuruman]

So for that period $\dot m=0$, yes? And we cannot say what $\dot v$ is because different parts of m have different velocities.

Yes to both statements. The way to do it is to define as the system the plane and what's in it and have an external force $F=\text{-}v\frac{dm_{water}}{dt}$ act on it.