Calculating Final Speed & Angle of a Rocket Motion

[tamref]

Homework Statement

A rocket is traveling with a speed 20 km/s in a non-gravity space. To fix the direction of motion, it turns on an engine, which pushes the gasses with constant speed 3 km/s w.r.t. the rocket perpendicularly to the direction of its motion. The engine is on till the mass of the rocket declines for 1/4 of the initial.

For what angle has the rocket changed the direction of motion and what is its final speed?

Homework Equations

x-axis: d( p(rocket) ) = - d( m(gasses) ) v(gasses) cos( alpha+d(alpha) )

dm/dt * v(rocket_final) + dv/dt m( rocket_new )= - dm/dt * v(gasses) (cos(alpha)cos(d(alpha)) - sin(alpha)sin(d(alpha)) )

dm/dt * v(rocket_final) + dv/dt ( m- dm/dt *t) = - dm/dt * v(gasses) *cos(alpha)cos(d(alpha))

The Attempt at a Solution

I tried to express the change in momentum of the rocket w.r.t. the angle of motion like above for both x- and y- axis, but I get unknown velocity and angle.

 

[dx]

Taking the direction of motion of the rocket to be the x-axis, write the momentum conservation equation in the y direction. This will give you the final velocity of the rocket in the y direction.

Now, you know both the x and y components of the final velocity, and you can calculate the angle that the velocity makes with the x-axis using trigonometry. (Hint: Draw a picture. What is tan(θ) in terms of V_x and V_y?)

 

[tamref]

Thank you, dx.

But the problem I see is, that the direction of gasses also changes with moving rocket. Therefore also Vx changes. The whole momentum does not go into Vy.

 

[dx]

The direction of motion changes, but the orientation of the rocket stays the same. So the direction in which the gas is ejected is the same.

 

[tamref]

What refers orientation of the rocket to, dx?

I thought of this problem again and maybe this is a solution.

Let us say, that the rocket would push the gasses with a given Vr=3 km/s in front of itself. Then its final Vx would be 19,2 km/s.

On the other hand, if the rocket would push the gasses with a given Vr perpendicularly to the initial direction of motion, its final Vy would be 0,86 km/s.

Therefore in any case 0<Vy<0,86 and 19,2<Vx<20. Therefore the angle for which it turns is
0< theta < 2,6 degrees.


Using the fact about the angle, we can see, that the change in Vx in x-axis can be neglected, since sin(2,6)~0,045. Also since cos(2,6)~0,998, we can take V=Vr for the relative velocity of gasses in y-axis during turning.

Therefore we can really assume that the gasses are ejected perpendicularly to the initial direction of rocket's motion.