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{excerpt:hidden=true}How far will two children slide after a perfectly inelastic collision? {excerpt}
While a family is taking a walk on a frozen pond, the two small children (a boy and a girl) manage to run into each other. They become entangled, resulting in a [totally inelastic collision]. The boy has a mass of 15 kg and was initially running at 1.0 m/s and the girl has a mass of 20 kg and was initially running at 2.0 m/s. Before their collision, the relative angle between their velocities was 45°. Assuming that the coefficient of friction between the resulting boy+girl combination and the ice is 0.15, how far do they slide after the collision before coming to rest?
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h4. Solution

{toggle-cloak:id=sys} *System:* {cloak:id=sys}Boy and girl as [point particles|point particle].{cloak}

{toggle-cloak:id=int} *Interactions:* {cloak:id=int}Impulse from external influences is neglected during the collision under the assumption that the collision is instantaneous. After the collision, the system experiences external influences from the earth (gravity, [conservative|work#nonconservative]) and the ice (normal force and friction, each [non-conservative|non-conservative force]).{cloak}

{toggle-cloak:id=mod} *Models:* {cloak:id=mod}[Momentum and External Force] followed by [Mechanical Energy, External Work, and Internal Non-Conservative Work].{cloak}

{toggle-cloak:id=app} *Approach:* 

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{toggle-cloak:id=diag1} {color:red}{*}Diagrammatic Representation, Part 1{*}{color}
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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.

!pond1.jpg!
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{toggle-cloak:id=math1} {color:red}{*}Mathematical Representation, Part 1{*}{color}
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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|external force] create a negligible [impulse].

{latex}\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}{latex}
Rewriting in terms of the masses and velocities, and substituting the appropriate zeros gives:
{latex}\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}{latex}
We have all the givens we need to solve these equations directly for the final velocity of the system:
{latex}\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}{latex}
so that the magnitude of the final velocity is:
{latex}\begin{large}\[ v_{f} = \sqrt{(\mbox{1.24 m/s})^2+(\mbox{0.808 m/s})^{2}} = \mbox{1.48 m/s} \]\end{large}{latex}
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:
 
{latex}\begin{large}\[tan(\phi) = \frac{v_{y,f}}{v_{x,f}} = \frac{0.808}{1.24} = 0.653 \]\end{large}{latex}
 
{latex}\begin{large}\[\phi = 33.16\]\end{large}{latex}
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{toggle-cloak:id=diag2} {color:red}{*}Diagrammatic Representation, Part 2{*}{color}
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!pond2.jpg!

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|gravitation (universal)#negpe] 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.
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{toggle-cloak:id=math2} {color:red}{*}Mathematical Representation, Part 2{*}{color}
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The relevant equation is:
{latex}\begin{large} \[ K_{f} = K_{i} + \int F_{f} \:ds = K_{i}-\int \mu_{k}N\:ds\]\end{large}{latex}
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:
{latex}\begin{large}\[ 0 = \frac{1}{2}(m^{B}+m^{G})v_{i}^{2} - \mu (m^{B}+m^{G})gs \]\end{large}{latex}
Again, recognizing that the initial velocity for the slide is the final velocity of the collision, we can solve to find:
{latex}\begin{large}\[ s = \frac{v_{i}^{2}}{2 \mu g} = \mbox{0.745 m} \]\end{large}{latex}{note}Note that because the frictional force is directed opposite the direction of motion, the final answer is positive, as we expect for a distance. {note}
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