Conservation Of Momentum In Space

linds43

22nd November 2011 - 03:33 AM

I need help with this question! I don't really understand the concept of conserving angular and linear momentum while in outer space..

You (101.6kg) and your partner (110kg) are in outer space. As a result of an accident you are drifting away from the space station at 0.1m/s to the left. Your partner is stationary, in between you and the space station. Your partner is 100m away from the space station. The space station cannot move and no one is on board to come and get you and your partner. Your partner (stationary) is badly injured and unable to move at all. You (drifting away) are facing your partner and the space station and holding a 20kg tool box.

Using the concepts of conservation of linear and angular momentum, explain how you could get yourself and your partner back to your space station.

I'm trying to do this question, but I don't even know where to begin.. How can I use the toolbox to help me move in the other direction, back towards my partner and the space station? Will just pushing the toolbox in one direction cause me to move in the other direction? Or will that not work because I am 101.6kg and the toolbox is only 20kg?

AlexG

22nd November 2011 - 06:28 AM

QUOTE

Will just pushing the toolbox in one direction cause me to move in the other direction? Or will that not work because I am 101.6kg and the toolbox is only 20kg?

Of course it will work. Since you are about 5 times the mass of the toolbox, you will move with about 1/5th of the toolbox's velocity.

Robittybob1

22nd November 2011 - 07:41 AM

QUOTE (AlexG+Nov 22 2011, 06:28 AM)

Of course it will work. Since you are about 5 times the mass of the toolbox, you will move with about 1/5th of the toolbox's velocity.

So you were thinking the momentum is transferred? Is it momentum or energy that is transferred.
Certainly for every force there is an equal and opposite reaction So forces balanced.
F = M * A M1 * A1 = M2 * A2
A1/A2 = M2/M1 where M2 = 100 (rounded) and M1 = 20 kg
A1/A2 = 5

Velocity = A * T so if we compare their velocities

Velocity1/velocity2= A1 * T / A2 * T

Velocity1/velocity2= A1 / A2 (Since the time of acceleration is the same for both.)

So their Velocities are protional to their Accelerations
So the velocity of M1 will be 5 times M2. V1 = 5 * V2

Let V2 = 1 then V1 = 5

Their energies E = 1/2 m * V^2

E1/E2 = 1/2 m1 * V1^2 / 1/2 m2 * V2^2

E1/E2 = 20 * 25/ 100

E1/E2 = 5/1

Does that mean 5 times as much kinetic energy goes into the lighter mass??

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