Principle of the conservation of the momentum problem

[mini-mynta]

Homework Statement

In an animated movie a spaceman (m=1,5 kg) stands on the slippery ice on a lake. Then he sees a soda bottle sliding towards him from west with a speed of 3,5 m/s. The spaceman catches the bottle, drinks up the content (m=0,33 kg) and throws away the empty bottle (m=0,2 kg) so that its speed relative to the ice is 3,0 m/s in southern direction. In what direction and with which speed should the spaceman slide after this, according to the laws of physics?

Homework Equations

p=mv
E_k=0,5mv²
F_\mu=\muN=\mumg
W=Fs

The Attempt at a Solution

I've tried to use the principle of the conservation of the momentum. I figured that after the spaceman has picked up the bottle he and the bottle should move on to the east and the momentum should be the same as the bottles was from the beginning: p₁=m₁v₁. Then I could calculate the spaceman's and bottle's combined velocity v₂. Then the friction would eventually stop the movement as the kinetic energy turns into heat according to W=F_\mus, W=E_k in this case. By calculating the length s of which the spaceman slides in each direction it's possible to calculate the angle since the movement forms a right triangle.
The spaceman then throws away the bottle southwards and as the bottle receives a momentum the spaceman should also start moving in the opposite direction. That way the total momentum would still be 0, and I could calculate the spaceman's velocity, and again, using W=F_\mus could calculate s, which now is the length of his northbound movement. This velocity should IMO be the final velocity answer, and using tan\alpha=s₂/s₁ I could calculate the angle of his final position compared to his initial position.

The answer should be:1,1 m/s, 18^o
I got v=0,3 m/s, \alpha=7,3^o

 

[gneill]

There was no friction mentioned in the problem statement. "Slippery ice" is a hint that friction should not need to be taken into account.

 

[mini-mynta]

Thank you, I'll try again then!