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A typical perfectly inelastic collision in 2-D.

The running back and the defender, treated as point particles.

Any impulse resulting from external forces is neglected because the collision is assumed to be instantaneous.

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It is important to understand that the 1m and 2m legs of the triangle drawn in black are not the components of the final velocity. They are distances, not velocities. The vector triangle formed by the final velocity (shown in purple) and its x- and y- components must be similar (in the mathematical sense) to the distance triangle for the critical case in which both players pass over the point of intersection between the Goal Line and the Out of Bounds Line. Thus, in this case, the angle in the distance triangle (27° for the given distances) must be the same as the angle of the velocity vector. Of course, the Defending Player will want to be travelling faster than this, to be certain that the opposing player is driven out of bounds before he crosses the goal line!

\begin{large}\[ p_{x,i}^{\rm system}=p_{x,i}^{RB}+p_{x,i}^{DF} = p_{x,f}^{\rm system} = p_{x,f}^{RB}+p_{x,f}^{DF} \] \[p_{x,i}^{\rm system}=p_{x,i}^{RB}+p_{x,i}^{DF} = p_{x,f}^{\rm system} = p_{x,f}^{RB}+p_{x,f}^{DF} \] \end{large}

\begin{large}\[ m^{RB}v_{x,i}^{RB} = (m^{RB}+m^{DF})v_{x,f}\]
\[ m^{DF}v_{y,i}^{DF} = (m^{RB}+m^{DF}) v_{y,f}\] \end{large}

\begin{large}\[ v_{x,f} = \frac{m^{RB}v_{x,i}^{RB}}{m^{RB}+m^{DF}} \]\end{large}

\begin{large}\[ v_{y,f} = v_{x,f} \tan\theta = \frac{m^{RB}v_{x,i}^{RB}}{m^{RB}+m^{DF}} \tan\theta \]\end{large}

\begin{large}\[ v_{y,i}^{DF} = \frac{m^{RB}+m^{DF}}{m^{DF}}v_{y,f} = \frac{m^{RB}v_{x,i}^{RB}}{m^{DF}} \tan\theta = \mbox{5.6 m/s}\] \end{large}

Note that our formula indicates that a larger angle will require a larger speed for the defender.