Momentum and Its Relation to Force on a Rocket

[JPhillips]

Homework Statement

A 4200-kg rocket is traveling in outer space with a velocity of 120 m/s toward the Sun. It needs to alter its course by 23.0$$^{}0$$, which can be done by shooting its rockets briefly in a direction perpendicular to its original motion. If the rocket gases are expelled at a speed of 2200 m/s relative to the rocket, what mass of gas must be expelled?

Homework Equations

Momentum=mass * velocity

The Attempt at a Solution

I start by drawing a diagram

The Y$$\uparrow$$ positive axis and Positive Right Axis$$\rightarrow$$

Because there is movement in both the y and x directions I start by breaking them into components

Assuming no external forces are acting on the rocket (for simplicities sake) $$\sum$$Fext.=0 So total momentum should be conserved (constant).

Given:
Mo= 4200kg
Mf= (Mo-Mg)
Vox= 120m/s
Voy= 0
Vg= 2200m/s
$$\theta$$= 23 degrees

Unknown:
Mg
Vf

Unsure how to approach a situation with two unknowns I attempt to build a system of equations:
For momentum in the y I'm getting

0 + 0 = (Mo-Mg)Vfsin23 - MgVg
(1) Mgvg = (Mo-Mg)Vfsin23

and for momentum in the x I'm getting
(2) VoxMo=(Mo-Mg)Vfcos23


Now in order to get rid of the Vf unknown I divided equation (1) by equation (2)
which achieves the correct answer of 97.2 kg [just noticed I could do that while positing this, thanks Physics Forums!]

but I still have a question,
in another solution for this problem that I found on a popular homework help site, I saw a different way of doing this problem that didn't jive well with me.

in that solution, in their very first step they assume that Vf's component perpendicular to Vo is equivalent to voxtan23. which achieves 50.9m/s. This is not at all obvious to me, and has in fact caused me quite a bit of heart burn. What am I missing about this problem that allows the tangent function to be used in such a way as to solve for one of the two unknowns so easily?

Any consul would be appreciated!

 

[krausr79]

The gas released will cause motion in the y direction and it already has motion in the x direction. The gas won't affect x at all, so once the boost is over we want it to go 120 m/s forwards and y m/s up such that the x and y make a 23° angle. If the setup is a right triangle then the 'adjacent' is the x and the 'opposite' is the y, then tan(23)=opp/adj=y/120. That's where you should start is finding how big a boost gives it that velocity 'up'.

 

[JPhillips]

Thank you, that was very helpful, I didn't realize that the adjacent side was really just equivalent to initial velocity because the perpendicular velocity would have no effect on it. Thanks again!