Rocket accelerating thru a dust cloud

[Gregie666]

hi...
a rocket of mass M is flying through a dust cloud the cloud has a density of P. the rocket's cross section is A. every dusticle the rocket colides with becomes permenantly attached.
the rocket is ejecting material (as a propellant) at the same rate that it assimilates it. the speed of the ejection relative to the rocket is
$$V_g$$
find the rockets acceleration as a function of its speed V

Homework Equations

$$F = M{{dv} \over {dt}} + (u - v){{dm} \over {dt}}$$

The Attempt at a Solution

i'v found dm, i think:
$$dm = APdx = APVdt$$

so presumably, to find
the acceleration i did this:
$$\eqalign & dV = adt - V_0 = V_1 - V_0 \cr$$
$$& - V_0 dm = - V_g dm + MV_1 \cr$$
but now I am totally stuck...
any help? please?

 

[OlderDan]

hi...
a rocket of mass M is flying through a dust cloud the cloud has a density of P. the rocket's cross section is A. every dusticle the rocket colides with becomes permenantly attached.
the rocket is ejecting material (as a propellant) at the same rate that it assimilates it. the speed of the ejection relative to the rocket is
$$V_g$$
find the rockets acceleration as a function of its speed V

Homework Equations

$$F = M{{dv} \over {dt}} + (u - v){{dm} \over {dt}}$$

The Attempt at a Solution

i'v found dm, i think:
$$dm = APdx = APVdt$$

so presumably, to find
the acceleration i did this:
$$\eqalign & dV = adt - V_0 = V_1 - V_0 \cr$$
$$& - V_0 dm = - V_g dm + MV_1 \cr$$
but now I am totally stuck...
any help? please?

This does not look right

$$dV = adt - V_0 = V_1 - V_0 \cr$$

The definition of acceleration is just a = dV/dt

Since the mass of the rocket is not changing, you can treat the collision//expulsion of a mass dm using conservation of momentum. It does not matter that the gas expelled is not the same piece of matter as the dm that is captured. Think of a head-on collision between two objects of mass M and dm with M moving at an initial velocity V and dm at rest. The final velocity of dm is known in terms of V and the relative exhaust velocity. Solve for the final velocity of M and find the rate of change of velocity dV/dt. This will look a lot like your force equation, but I don't think you have that equation quite right. Maybe it's just a matter of interpretation. What force are you representing by that equation?

 

[Gregie666]

ok.. i think you are right. so using conservation of momentum i can approach it like this:
in an infinitesimally small ammonut of time, the following equation holds true:

$$VM = (V + dV)M + (V - V_g )dm$$
$$VM = (V + dV)M + (V - V_g )APdx$$
$$VM = VM + Mdv + (V - V_g )APdx$$
$${{(V - V_g )APdx} \over {dt}} = M{{dv} \over {dt}}$$
$${{(V - V_g )APV} \over M} = a$$

p.s. how do i make a newline in tex?? \newline doesn't seem to work