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{td:alignwidth=center275|bgcolor=#F2F2F2}*[Examples from Momentum]*
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{table}{excerpt:hidden=true}A typical perfectly inelastic collision in 2-D.{excerpt}
!football1.png!
A 220 lb running back is running along the sideline of the football field toward the goal line.  The running back is 1.0 m from the sideline running at 8.0 m/s toward the goal.  A 180 lb defender runs across the field toward the running back 2.0 m from the goal line.  Just before the collision, both players leap into the air.  The running back's horizontal speed remains 8.0 m/s.  When the players collide, the defender clings to the running back resulting in a [totally inelastic collision].  What is the minimum possible horizontal velocity for the defender in order to ensure the players cross out of bounds before crossing the goal line?
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h4. Solution

{toggle-cloak:id=sys} *System:*
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The running back and the defender, treated as [point particles|point particle].

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{toggle-cloak:id=int} *Interactions:*
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Any impulse resulting from external forces is neglected because the collision is assumed to be instantaneous.

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{toggle-cloak:id=mod} *Model:*

{cloak:id=mod}[Momentum and External Force].
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{toggle-cloak:id=app} *Approach:*
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{toggle-cloak:id=diag} {color:red}{*}Diagrammatic Representation{*}{color}
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We are interested only in the horizontal components of the velocity.  If we treat the players as point masses, then the defender's goal is for the horizontal velocity after the collision to be directed at a minimum of 30 degrees from the running back's original direction of motion, as shown in the figure.

!football2.jpg!
{warning}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 (30° 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\!
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{toggle-cloak:id=math} {color:red}{*}Mathematical Representation{*}{color}
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Since we are assuming external forces are negligible during the collision, we will have constant horizontal momentum:
{latex}\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}{latex}
where RB denotes the running back and DF the defender.  For the case at hand, these equations can be simplified to read:
{latex}\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}{latex}
We can immediately solve the _x_ direction equation:
{latex}\begin{large}\[ v_{x,f} = \frac{m^{RB}v_{x,i}^{RB}}{m^{RB}+m^{DF}} \]\end{large}{latex}
but the _y_ equation contains two unknowns.  We can resolve this problem by using the constraints (discussed above) that in order to end up out of bounds:
{latex}\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}{latex}
With this information, we can solve to find:
{latex}\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}{latex}
so that the horizontal component of the defender's velocity must be at least 5.6 m/s.
{tip}Note that our formula indicates that a larger angle will require a larger speed for the defender.
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