Rocket ejecting mass in a direction perpendicular to its velocity

[LCSphysicist]

Homework Statement

Find the relation between the mass of the rocket (initial mass = M_o) and the angle between the velocity and the initial velocity. The gas is ejected with velocity u with respect to the rocket in a direction perpendicular to rocket's velocity v.

Relevant Equations

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That was my approach:
$$P_f - P_i = [(m-dm)(v + dv) + dm(u+v+dv)] - [m(v)]$$
$$= mdv - dmv + dmu + dmv = mdv + dmu = 0$$

Since the variation of the rocket's velocity is perpendicular to itself, $$ dv = v d \theta => m v d \theta + dm u = 0$$

So we have $$\frac{dm}{m} = \frac{-v d \theta}{u}$$

And so, $$M = M_o e^\frac{-v \theta}{u}$$

Is that right? I am afraid that it can be wrong, but i am not sure why.

 

[kuruman]

Please fix the LaTeX. We cannot read the bottom line.

 

[LCSphysicist]

Please fix the LaTeX. We cannot read the bottom line.

I am not sure where should i fix. In my pc everything is right:

 

[Delta2]

I am not sure at all in what I am going to say but I think the rocket's initial component of velocity (say $\vec{V_{0y}}$will remain unchanged, and all you have to do is calculate the other perpendicular component $\vec{V_x}$ (setting as $V_{0x}=0$) and then vector add the two components $V_x,V_{0y}$ as being perpendicular to each other to get the final velocity $\vec{V}=\vec{V_x}+\vec{V_{0y}}$.

 

[Orodruin]

Homework Statement:: Find the relation between the mass of the rocket (initial mass = M_o) and the angle between the velocity and the initial velocity. The gas is ejected with velocity u with respect to the rocket in a direction perpendicular to rocket's velocity v.
Relevant Equations:: .

That was my approach:
$$P_f - P_i = [(m-dm)(v + dv) + dm(u+v+dv)] - [m(v)]$$
$$= mdv - dmv + dmu + dmv = mdv + dmu = 0$$

Since the variation of the rocket's velocity is perpendicular to itself, $$ dv = v d \theta => m v d \theta + dm u = 0$$

So we have $$\frac{dm}{m} = \frac{-v d \theta}{u}$$

And so, $$M = M_o e^\frac{-v \theta}{u}$$

Is that right? I am afraid that it can be wrong, but i am not sure why.

You could motivate your treatment in a better way. For example, don’t use $dv$ because the change in the speed $v$ is zero so $dv=0$ by construction. Instead, use reasoning to say that you are taking the component of $d\vec v$ perpendicular to $\vec v$.

Apart from that, it looks reasonable.

I am not sure at all in what I am going to say but I think the rocket's initial component of velocity (say $\vec{V_{0y}}$will remain unchanged, and all you have to do is calculate the other perpendicular component $\vec{V_x}$ (setting as $V_{0x}=0$) and then vector add the two components $V_x,V_{0y}$ as being perpendicular to each other to get the final velocity $\vec{V}=\vec{V_x}+\vec{V_{0y}}$.

If this were the case, the exhaust would need to be perpendicular to the original direction of motion, not the current direction of motion (they are the same only at the beginning).

 

[Delta2]

If this were the case, the exhaust would need to be perpendicular to the original direction of motion, not the current direction of motion (they are the same only at the beginning).

Thanks, I now understand my faulty reasoning.

 

[kuruman]

I am not sure where should i fix. In my pc everything is right:

It is fixed now, thanks. I could see all the LaTeX lines except for the last, but now I see them all.