How Much Mass Must a Rocket Expel to Alter Its Course by 35 Degrees?

In summary, a rocket of 3180 kg traveling at 115 m/s towards the sun wishes to change its course by 35.0 degrees by firing its rockets perpendicular to its original motion. The rocket gases are expelled at a speed of 1750 m/s. Using the equation P=mv, it can be determined that to conserve momentum, the mass of the rocket (mri) multiplied by its initial velocity (vri) must equal the mass of the gases (mg) multiplied by their velocity (vg) plus the mass of the rocket after the gases have been expelled (mrf) multiplied by its new velocity (vrf). Therefore, the mass of the gases expelled must be 140 kg.
  • #1
pinkerpikachu
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Homework Statement


A rocket of total mass 3180 kg is traveling in outer space with a velocity of 115 m/s toward the sun. It wishes to alter its course by 35.0 degrees, and can do this by firing its rockets briefly in a direction perpendicular to its original motion. If the rocket gases are expelled at a speed of 1750 m/s, how much mass must be expelled?
Answer is 140kg

Homework Equations



P=mv

The Attempt at a Solution


http://uploader.neoextreme.com/files/1274/haba/physicsproblem13ch7.jpg

First off, I want to know if that picture is correct, otherwise I'm heading in the wrong direction with this.

of course, momentum is conserved, so whatever is on one side has to be on the other. The rocket has an initial (r for rocket (mri*vri))
then the gases expelled have a certain (g for gas NOT gravity(mg * vg))
and the rocket will have a new (mrf * vrf)

so
mri*vri = mg*vg + mrf*vrf

The vectors in the picture have to be spilt up into their x and y components

I seem to be going wrong here. I don't get a correct answer
 
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  • #2
Hi pinkerpikachu! :smile:
pinkerpikachu said:
First off, I want to know if that picture is correct …

Yes! :smile:
The vectors in the picture have to be spilt up into their x and y components

Yes … and the x component isn't changing, so you can forget that. :wink:

But I can't really follow what you've done :redface:

can you write it out in full for us, with the numbers?

(and you are remembering that the mass of the gas equals the mass lost by the rocket, aren't you? :wink:)
 
  • #3
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Last edited by a moderator: May 4, 2017

- • Mathematicians solve age-old spaghetti mystery

Hi and welcome to PF.

Your approach is on the right track. You are correct in stating that momentum is conserved, and that the initial momentum of the rocket must equal the final momentum after the gases are expelled.

However, I think you may be overcomplicating the problem a bit. Remember, momentum is a vector quantity and must be conserved in both the x and y directions. So instead of breaking up the vectors in the picture into their x and y components, try looking at the problem in terms of the total momentum in each direction.

So for the x direction, the initial momentum is mri*vri, and the final momentum is mrf*vrf*cos(35). (Do you see why? Draw a picture of the rocket's new velocity vector after the gases are expelled and you should be able to see why the cosine is involved.)

Similarly, in the y direction, the initial momentum is 0 (since the rocket is traveling in the x direction), and the final momentum is mrf*vrf*sin(35).

Now set up your conservation of momentum equation for each direction and solve for mrf. Remember, you have two equations and two unknowns (mrf and vrf), so you should be able to solve for both.

Hope that helps get you on the right track!

Last edited by a moderator: May 4, 2017

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FAQ: How Much Mass Must a Rocket Expel to Alter Its Course by 35 Degrees?

What is linear momentum conservation?

Linear momentum conservation is a fundamental law of physics that states that the total momentum of a closed system remains constant over time. This means that in a system where there are no external forces acting on it, the total momentum of all the objects in that system will remain the same.

How is linear momentum conserved?

Linear momentum is conserved because of Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. This means that when two objects interact, their individual momentum may change, but the total momentum of the system will remain constant.

What is an example of linear momentum conservation?

A classic example of linear momentum conservation is a collision between two objects. When two objects collide, their individual momentums may change, but the total momentum of the system will remain the same before and after the collision.

Is linear momentum conservation applicable in all situations?

No, linear momentum conservation is only applicable in closed systems where there are no external forces acting on it. If there are external forces present, the total momentum of the system will not remain constant over time.

How is linear momentum conservation related to energy conservation?

Linear momentum conservation is related to energy conservation through the concept of kinetic energy. In a closed system, where there are no external forces acting on it, the total momentum and the total kinetic energy of the system will remain constant over time. This is because energy cannot be created or destroyed, it can only be transferred between objects in the system.

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