Rocket moving through a cloud of dust, variable mass problem

[etotheipi]

Homework Statement

The rocket, cross sectional area A and mass M, travels through a cloud of dust of density p. Determine the distance the rocket travels until its velocity is half of its initial velocity.

Relevant Equations

CLM

The first way to solve this is to just say, by conservation of momentum, that M_{0}v_{0}=(M_{0}+Apx)\frac{v_0}{2} where Apx is the mass of dust the rocket comes into contact with in a distance x.

For the second method, by considering the change of momentum of the dust in 1 second, we know the force applied to the rocket is -Apv^{2}. So if we write down the Newton's second law relation, we get $$-Apv^{2} = (M_{0}+Apx) v \frac{dv}{dx}$$ which when solved does happen to give the same answer as in the first method. However, I thought Newton's second law is only applicable to closed systems of constant mass; so how is this equation giving the same answer?

Is it just a coincidence that method 2 works in this case, i.e. do we actually need to use the variable-mass equation of motion to do it properly?

 

[haruspex]

Newton's second law is only applicable to closed systems of constant mass

You are only applying the law over an infinitesimal time, dt. During that time the rocket mass is effectively constant. Having obtained a valid equation you are then free to solve it.

 

[etotheipi]

You are only applying the law over an infinitesimal time, dt. During that time the rocket mass is effectively constant. Having obtained a valid equation you are then free to solve it.

That's interesting. What's the difference between then just applying NII vs using the following variable mass equation, if it is even relevant at all?

 

[haruspex]

That's interesting. What's the difference between then just applying NII vs using the following variable mass equation, if it is even relevant at all?

View attachment 253586

In the above equation, both terms on the left represent applied forces. In your equation in post #1, F_ext is zero and the term on the left there is the other force in the equation above.

 

[etotheipi]

In the above equation, both terms on the left represent applied forces. In your equation in post #1, F_ext is zero and the term on the left there is the other force in the equation above.

Actually, this works nicely, thanks! I'm just a little confused as to when we then actually need to use the second equation instead of just using F=ma all of the time?

 

[haruspex]

When I solve this I get a weird factor of ln(2)2ln⁡(2)2\frac{\ln{(2)}}{2} popping up.

Edit I think the mistake I made just now was writing the mass as just M, when it should instead be some function of t/x etc.?

Are you saying you treated M as constant in solving the equation? Yes, that would be wrong.

Your first equation in post #5 is often deduced from $0=\frac{dp}{dt}=\frac{d(mv)}{dt}=m\dot v+v\dot m$. But this is invalid because in a closed system mass is constant. The only reason it works in problems such as in this thread is that the added mass had no momentum initially.