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?

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.