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Solution

We begin by analyzing the collision using momentum. When a problem gives a relative angle, it is important to develop a coordinate system to orient ourselves as we solve. We therefore begin with a picture. We have arbitrarily assigned the boy to move along the x-axis, and the girl to have positive x- and y-velocity components.

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With our picture developed, we can write the equations of constant momentum, since we are assuming that the collision occurs over such a short time that external forces create a negligible impulse.

\begin{large}\[ p^{B}_{x,i} + p^{G}_{x,i} = p^{B+G}_{x,f}\]
\[p^{B}_{y,i} + p^{G}_{y,i} = p^{B+G}_{y,f} \]\end{large}

Rewriting in terms of the masses and velocities, and substituting the appropriate zeros gives:

\begin{large}\[ m^{B}v^{B}_{x,i} + m^{G}v^{G}_{x,i} = (m^{B}+m^{G})v_{x,f} \]
\[ m^{G}v^{G}_{y,i} = (m^{B}+m^{G})v_{y,f} \]\end{large}

We have all the givens we need to solve these equations directly for the final velocity of the system:

\begin{large}\[ v_{x,f} = \frac{m^{B}v^{B}_{x,i}+m^{G}v^{G}_{x,i}}{m^{B}+m^{G}} = \mbox{1.24 m/s}\]
\[ v_{y,f} = \frac{m^{G}v^{G}_{y,i}}{m^{B}+m^{G}} = \mbox{0.808 m/s} \]\end{large}

so that the magnitude of the final velocity is:

\begin{large}\[ v_{f} = \sqrt{(\mbox{1.24 m/s})^2+(\mbox{0.808 m/s})^{2}} = \mbox{1.48 m/s} \]\end{large}

Notice that the final velocity of the entangled boy and girl lies neither along the original velocity of the boy or that of the girl, but between these two initial velocities. From the components of the final velocity vector we can determine the angle between the final velocity and the boy's initial velocity as:
 

\begin{large}\[tan(\phi) = \frac{v_{y,f}}{v_{x,f}} = \frac{0.808}{1.24} = 0.653 \]\end{large}

\begin{large}\[\phi = 33.16\]\end{large}

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This final velocity for the collision of the boy and girl is the initial velocity for their post-collision slide. We can now consider a free body diagram for the boy+girl system as they slide across the ice. Assuming the pond is flat, the gravitational potential energy will be constant and the normal force will do no work. Friction will do negative work, as it is directed at 180° from the direction of motion of the system. Since only one force can do work, the question of how far the system slides after the collision is easily addressed using work and energy.

The relevant equation is:

\begin{large} \[ K_{f} = K_{i} + \int F_{f} \:ds = K_{i}-\int \mu_{k}N\:ds\]\end{large}

Here we have used a minus sign in the second expression because the frictional force is directed against the direction of motion, and reduces the overall kinetic energy. Considering the free body diagram plus the fact that the boy and girl are not accelerating in the z-direction, it is clear that the normal force and the weight of the system must balance. Further, we want to know how far the boy and girl slide before they stop, indicating that v_f should be zero for this part of the problem. Thus, we can write:

\begin{large}\[ 0 = \frac{1}{2}(m^{B}+m^{G})v_{i}^{2} - \mu (m^{B}+m^{G})gs \]\end{large}

Again, recognizing that the initial velocity for the slide is the final velocity of the collision, we can solve to find:

\begin{large}\[ s = \frac{v_{i}^{2}}{2 \mu g} = \mbox{0.745 m} \]\end{large}