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Jack (mass 52.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. he collides with jill (mass 49.0 kg ), who is initially at rest. after the collision, jack is traveling at 5.00 m/s in a direction 34.0∘ north of east. ignore friction. part a what is the direction of the jill's velocity after the collision?

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Demiurgos
We must write down laws of conservation of momenta and energy. 
For the law of conservation of momenta will we will use two axes. One will be x-axis that will correspond to the east, and the other one will be y-axis corresponding to the north. Jack will be marked as 1 and Jill will be marked as 2.
Law of conservation of energy:
[tex]\frac{m_1v_1^2}{2}=\frac{m_1v'_1^2}{2}+ \frac{m_2v_2^2}{2}\\ m_2v_2^2=m_1v_1^2-m_1v_1'^2\\ v_2=\sqrt{\frac{m_1v_1^2-m_1v_1'^2}{m_2}[/tex]
This will give us Jill's velocity after the colision.
[tex]v_2=6.43\frac{m}{s}[/tex]
Law of conservation of momenta:
[tex]x: m_1v_1=m_1v_1'cos(34)+m_2v_2cos(\theta)\\ y: 0=m_1v_1'sin(34)-m_2v_2sin(\theta)\\[/tex]
We will use the second equation to get the angle at which the Jill is traveling:
[tex]0=m_1v_1'sin(34)-m_2v_2sin(\theta)\\ m_2v_2sin(\theta)=m_1v_1'sin(34)\\ sin(\theta)=\frac{m_1v_1'sin(34)}{m_2v_2}\\ \theta=sin^{-1}(\frac{m_1v_1'sin(34)}{m_2v_2})[/tex]
When we plug all the number we get:
[tex]\theta=27.45^\circ[/tex]
Please note that this is the angle below the x-axis.
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