Collision with orientation, finding velocity

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SuperHero
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Homework Statement


During the warm-up for the 1990 Olympics free skating competition, Debi Thomas, mass 60kg, moving at 5 m/s and looking intently at her coach, collides with Katerina Witt, mass 65 kg,who is smiling at a photographer. The two skaters were moving at right angles to each other.After the collision, they hang on to each other and jointly move at an angle of 37 relative to Debi's original direction. How fast was Katerina traveling before the collision?


The Attempt at a Solution


http://s1302.beta.photobucket.com/user/Rameel17/media/Untitled_zps4d1ae6dc.png.html

This is the diagram that i came up with and here is my calculation:

m1v1 + m2v2 = mtv0
(60kg)(5m/s[E]) + (65kg)(v2) = (125kg)(vo)[E 37 N]

So what i did was i divided the m1v1 and m2v2 separately with the total something like this:

m2v1 = mtv0
(60kg)(5m/s[E]) = (125kg)(vo)[E 37 N]
300 kgm/s [E]/125kg = (vo)(cos 37[E])
2.4 m/s[E]/ (cos 37[E]) = vo
vo = 3.00512558m/s

so i found the final total velocity, now to find the intial katerine's velocity

m2v2[N] = mtv0[E 37 N]
(65kg)(v2) = (125kg)(3.00512558m/s)(sin 37)
v2 = 3.4779177 m/s

Thats the answer.
is my process correct?
 
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haruspex said:
The equations and the thinking behind them are all correct. The verbal explanation could be better. It's more usual to phrase it in terms of components of momentum in the two directions.

Hello, Thank you so much for the reply. What do you mean by conservation of momentum phrases?
 
In the OP you wrote
SuperHero said:
So what i did was i divided the m1v1 and m2v2 separately with the total something like this:
You would not in general be able to separate them so easily, so the reader might suspect you don't understand what you are doing. It would be more convincing to write "by conservation of momentum in the direction of v1..." etc.