Linear Momentum Conservation

[pinkerpikachu]

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

 

[tiny-tim]

Hi pinkerpikachu!

First off, I want to know if that picture is correct …

Yes!

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.

But I can't really follow what you've done …

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? )

 

[baba89]

.

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

Have something to add?

Similar discussions for: Linear Momentum Conservation. Homework Statement

Linear Momentum Conservation

Linear Momentum, Conservation

Linear Momentum and Collision, Conservation

Conservation of Linear Momentum

Linear Momentum and Conservation