1. The problem statement, all variables and given/known data A rocket is travelling 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? 2. Relevant 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)) 3. 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.
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}?)
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.
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.
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.