Relative Velocity of Astronauts After a Throw and Catch

[greyradio]

[SOLVED] Momentum problem

Two astronauts float in space, at rest relative to each other. The first astronaut, mass 63.4 kg, holds a ball of mass 7.28 kg, which she throws at speed 6.17 m/s. The second astronaut, mass 83.1 kg, then catches the ball. Find the speed one astronaut is moving relative to the other after one throw and one catch.

P = mv
M1V1 + M2V2 = M1Vf + M2Vf

I found the momentum of the first astronaut throwing the ball which is:

7.28(6.12) = 63.4 vf

7.28(6.12)/63.4 = vf
-.7027 m/s = vf

Also the momentum of the second astronaut catching the ball:

83.1(0) + 7.28(6.17)/ 83.1 + 7.28 = vf
.497 m/s = vf

but I am rather confused about what its asking for. It asking for the speed one astronaut is moving relative to the other. Does it matter which astronaut's speed I use or do I need to use a relative velocity equation?

If I need to use a relative velocity equation would it be something like:

V astro b relative to a = V astro a relative to b + V of astro A and B

 

[Dick]

You probably don't want to write 6.12 when the problem statement says 6.17. I get v=-0.7085m/s for the first astronaut. For the second astronaut your numbers don't add up at all. But to answer your question the relative velocity of the two is the difference between the two velocities if you keep the sign straight.

 

[Snazzy]

Well, for your question, you can use $$(m_1+m_2)v_i = m_1v_1_f+m_2v_2_f$$, where I don't quite get the same answer as you for the first astronomer's velocity.

And then if you calculate the second astronomer's velocity, you get an answer which is consistent with what you got.

Then the relative velocity of astronaut B to astronaut A
$$=v_B-v_A$$, where $$v_A$$ is a negative number, therefore
$$v_B_t_o_A=v_B+|v_A|$$

 

[greyradio]

thanks it worked out.