
#1
Apr2609, 03:10 AM

P: 5

1. The problem statement, all variables and given/known data
A rocket is travelling with a speed 20 km/s in a nongravity 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 xaxis: 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. 



#2
Apr2609, 03:52 AM

HW Helper
PF Gold
P: 1,962

Taking the direction of motion of the rocket to be the xaxis, 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 xaxis using trigonometry. (Hint: Draw a picture. What is tan(θ) in terms of V_{x} and V_{y}?) 



#3
Apr2609, 04:55 AM

P: 5

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. 



#4
Apr2609, 06:38 AM

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PF Gold
P: 1,962

Rocket in space
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.




#5
Apr2609, 07:18 AM

P: 5

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


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