# Astronaut momentum problem

A 60kg astronaut floating in space simultaneously tosses away a 14-kg oxygen tank and a 5.8kg camera. The tank moves in the x direction at 1.6 m/s, and the astronaut recoils at .85m/s in a direction 200degrees counterclockwise from the x axis. Find the velocity of the camera.

I can't find a way to start off this problem. I drew a diagram and everthing.

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hage567
Homework Helper
Well, you know momentum must be conserved. So what is the total momentum before the toss and after? Find the expressions for each direction.

Well, you know momentum must be conserved. So what is the total momentum before the toss and after? Find the expressions for each direction.

The total momentum before the toss = 0 because there is no velocity.
After the toss,
Ptotal=Pastro+Ptank+Pcam

Resolve the momenta into components.

Resolve the momenta into components.
mtotvtot=mastrovastro+mtankvtank+mcamvcam

like that?

mtotvtot=mastrovastro+mtankvtank+mcamvcam

like that?
Nope, along the x and y axes.

Since the total momentum is conserved, they must be conserved along the axes, too. So
initial momentum along x axis = final momentum along x axis. Similarly for y.

x: mtvtx=mavax+mcvcx+mtankvtankx
y: mtvty=mavay+mcvcy

i think this is what you mean right?

Yes, that's right.

Make sure you get the signs right when putting in the values of the velocities.

Yes, that's right.

Make sure you get the signs right when putting in the values of the velocities.

ok, so i'm solving for the velocity of the camera but i don't have the total momentum, so how can i solve for the camera's velocity with two unknowns?

What two unknowns? Read the question again. There's only one unknown per equation, and you can solve for them with the given info.

What two unknowns? Read the question again. There's only one unknown per equation, and you can solve for them with the given info.
Total momentum, and then the final velocity for the camera.

Can i say

-mavax=mcvcamx+mtankvtankx ???

Sure you can. The initial momenta along both directions is zero, remember?

Sure you can. The initial momenta along both directions is zero, remember?

right, but i thought i was solving for the final velocity

right, but i thought i was solving for the final velocity
Yes you are. It is because of the zero initial momentum (mtvtx = mtvty = 0) that you are able to write the equation as your previous post.

can u say ( also have this problem, and this is how i was working it out)

m_ast*v_ast*cos(200) + m_tank*v_tank + m_cam*v_cam*cos(20) = 0

these two equation actualy equal each other (produce same answer)

-mavax=mcvcamx+mtankvtankx

all withrespect to the x axis

can someone explain...
why do you have to use the y axis or even worry about the x axis in this problem... you can relate teh whole thing to the x component w/ trig?

x: mtvtx=mavax+mcvcx+mtankvtankx
y: mtvty=mavay+mcvcy

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