# Bucket launched into cloud of star dust.

1. Nov 20, 2007

### deccard

1. The problem statement, all variables and given/known data

An empty bucket (mass M, area A) is launched with velocity v0 from a space station into a cloud of dust (density ρ). As the bucket moves through the dust it will collect dust in it until the bucket comes to a stop. Solve the place of the bucket x=x(t).

2. The attempt at a solution

Okay. Here is my attempt to solve this.

momentum of the bucket dp = v*dm + dv*m.

As a particle of dust with mass dm collides with the bucket its momentum changes by u*dm, where u=-v is the change of the velocity of the particle. On the other hand the change in momentum of the particle has to be the same as the change in momentum of the bucket. So...

dp = u*dm = -v*dm
2*v*dm+m*dv=0
1/v*dv = -2/m*dm
Let's integrate v=v0...v, m=M..m
ln(v0/v) = -2*ln(m/M)
--->
v = v0*(M/m)

On the other hand mass of the bucket M will change by
m(x) = M + A*ρ*x
where x is the length of travelled distance.

and this coupled with the equation of velocity we get.
dx/dt = v0*(1+ρ*A*x/M)^-2
v0*dt = (1+ρ*A*x/M)^2 * dx
Lets integrate again t=0..t x=0..x and we get

v0*t = x + k*x^2 + 1/3*k^2*x^3 where k=ρ*A/M

Now by solving x, which I'm not going to do here, we get x=x(t)
Have I done this right? Is there simplier way to do this?

2. Nov 20, 2007

### azatkgz

Let's say

total momentum of the whole system is P(t) at some time t.

so

$$P(t)=mv$$

$$P(t+dt)=(m+dm)(v+dv)$$

$$dP=P(t+dt)-P(t)=(m+dm)(v+dv)-mv=mv+mdv+dmv+dmdv-mv=mv+dmv$$

dmdv is very small,so we can neglect it

$$dP=Fdt=mdv+dmv$$ becase there's no external force on this system we put F=0

$$dm=\rho Avdt\Rightarrow m(t)=M+\rho Ax$$

$$mdv=-dmv$$

$$\frac{v(t)}{v_0}=\frac{M_0}{m(t)}$$

$$v(t)(M+\rho Ax)dt=M_0v_0dt$$

$$\int(M+\rho Ax)dx=\int M_0v_0dt$$

$$Mx(t)+\frac{\rho Ax(t)^2}{2}=M_0v_0t$$

3. Nov 20, 2007

### deccard

So let's solve x(t):
$$x(t)=\frac{-M_0+\sqrt{M_0^2+2\rho AM_0v_0t}}{\rho A}$$

and v(t):
$$v(t)=\frac{dx}{dt}=\frac{M_0v_0}{\sqrt{M_0^2+2\rho AM_0v_0t}}$$

$$v(t)\to0$$ when $$t\to\infty$$ as it should. But

$$x(t)\to\infty$$ when $$t\to\infty$$ doesn't make sense. It is obvious that the bucket will come to a stop at a some finite point x.

4. Nov 22, 2007

### deccard

Is it sure we can say that there is no external forces on the bucket? What about the particles hitting the bucket? Couldn't the hitting be considered as an external force?

5. Nov 23, 2007

### azatkgz

If particles are hitting him ,then bucket also is hitting them.So total force in bucket-particle system is zero.

6. Nov 24, 2007

### deccard

Okay. What about conservation of energy?
Shouldn't $$E=\frac{1}{2}M_0v_0^2=\frac{1}{2}(M_0+\rho Ax(t))(v(t))^2$$ with all values of t.

But when x(t) and v(t) solved from $$Mx(t)+\frac{\rho Ax(t)^2}{2}=M_0v_0t$$ are put into
$$E=\frac{1}{2}(M_0+\rho Ax(t))(v(t))^2$$

We get
$$E=\frac{1}{2}M_0v_0^2(m_0^2+2\rho AM_0v_0t)^{-\frac{1}{2}}$$

So something is not right with the solution.

Well on the other hand there are the same problems with solution
$$v_0t = x + kx^2 + /frac{1}{3}k^2x^3$$ where $$k=/rho A/M$$.

7. Dec 3, 2007

### deccard

Ah! Of course it shouldn't! The collision is inelastic so kinetic energy is not preserved. But still it bothers me that
$$x(t)\to\infty$$ when $$t\to\infty$$. I don't understant it!