Answer
The magnitude of Jill's velocity is 5.46 m/s directed at an angle of $35.9^{\circ}$ south of east.
Work Step by Step
Let Jack be person A and let Jill be person B.
$m_B~v_{Bx} + m_A~v_{Ax} = m_A~v_0$
$v_{Bx} = \frac{m_A~v_0-m_A~v_{Ax}}{m_B}$
$v_{Bx} = \frac{(55.0~kg)(8.00~m/s)-(55.0~kg)(5.00~m/s)~cos(34.0^{\circ})}{48.0~kg}$
$v_{Bx} = 4.42~m/s$
Since $p_y = 0$, the north-south component of the momentum of Jack and Jill are equal in magnitude.
$m_B~v_{By} = m_A~v_{Ay}$
$v_{By} = \frac{m_A~v_{Ay}}{m_B}$
$v_{By} = \frac{(55.0~kg)(5.00~m/s)~sin(34.0^{\circ})}{48.0~kg}$
$v_{By} = 3.20~m/s$
We can find the magnitude of Jill's velocity.
$v_B = \sqrt{(4.42~m/s)^2+(3.20~m/s)^2}$
$v_B = 5.46~m/s$
We can find the angle $\theta$ south of east.
$tan(\theta) = \frac{3.20~m/s}{4.42~m/s}$
$\theta = arctan(\frac{3.20~m/s}{4.42~m/s})$
$\theta = 35.9^{\circ}$
The magnitude of Jill's velocity is 5.46 m/s directed at an angle of $35.9^{\circ}$ south of east.
