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

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A rocket with a mass of 3180 kg traveling at 115 m/s aims to change its course by 35 degrees by expelling gas at 1750 m/s. The conservation of momentum is central to solving the problem, requiring the initial momentum of the rocket to equal the combined momentum of the expelled gas and the remaining rocket. The discussion emphasizes the need to break down vectors into x and y components, noting that the x component remains unchanged. The correct mass of gas that must be expelled to achieve the desired course alteration is determined to be 140 kg. Understanding these principles is crucial for solving similar physics problems effectively.
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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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:)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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