# Momentum and changed a changing mass

let there be a spaceship which is floating in space with a surface size of A. The spaceship floats in a constant speed of $$v_0$$, dust with density of $$\rho$$ gets stuck on the (only) surface of the spaceship. The starting mass of the spaceship is $$m_0$$

Find the velocity as a function of time

So we now that the momentum is preserved as there is no outer force acting on the spaceship:

$$\frac{d(mv)}{dt}=0$$

Because the mass is a function of time the derivative is:

$$\dot{m}v+m\dot{v}=0$$

At an instant of time when the spaceship is in the dust the change in the mass is:

$$dm=\rho Avdt$$

Using again the fact the the momentum is preserved we get:

$$mv=m_0v_0$$

So:

$$\frac{m_0v_0}{v}\dot{v}+v^2\rho A=0$$

How did we arrive to the last equation?

If we take:

$$dm=\rho Avdt\iff \frac{dm}{dt}=\rho Av$$

And:

$$mv=m_0v_0\iff m=\frac{m_0v_0}{v}$$

taking the derivative:

$$\frac{dm}{dt}=\frac{d}{dt}\frac{m_0v_0}{v}\Rightarrow \rho Av=-\frac{m_0v_0\dot{v}}{v^2}$$?

• You want to get to the final answer as a function of time? Commented Apr 8, 2019 at 13:01
• @KV18 correct a function of time Commented Apr 8, 2019 at 13:22

I am not entirely sure about this one. But I used the control volume approach and ended up with a first order differential equation which I tried to solve. I will be posting it below, and suggestions are welcome. I, probably, might have got it wrong.

Here,

  v = Instanteneous velocity of spaceship
v_{d} = Absolute Velocity of dust striking the spaceship