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{excerpt:hidden=true}Simple problem illustrating the definition of [impulse] and the utility of an [initial-state final-state diagram].{excerpt}

Consider a ball of mass {*}_m{~}b{~}{_}{*} that is moving to the right at a constant speed {*}_v{~}b{~}{_}{*} when it suddenly impacts a wall and reverses direction (still moving at the same speed).  What is the impulse delivered to the ball in the collision?

h4. Solution

{toggle-cloak:id=sys} *System:* {cloak:id=sys} The ball as a [point particle].{cloak:sys}

{toggle-cloak:id=int} *Interactions:* {cloak:id=int} During the impact, we assume that the collision [force] from the wall is vastly larger than any other [external forces|external force] on the ball, so that other forces are ignored.{cloak:int}

{toggle-cloak:id=mod} *Model:* {cloak:id=mod}[Momentum and External Force].{cloak:mod}

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

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We will solve this problem using three different approaches to illustrate alternate ways to perform vector subtraction.

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{card:label=Using Algebra}

h4. Using Algebra

{toggle-cloak:id=diag1} {color:red} *Diagrammatic Representation* {color}

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{warning}The [{_}magnitude{_}|magnitude] of the [momentum|momentum] before and after the collsion is the same ({*}({_}m{~}b{~}v{~}b{~}{_}){*}), which can easily lead to the conclusion that there has been no change.  Thinking about the situation, however, should quickly convince you that the ball has certainly been acted on by some force, which implies that a change _did_ occur.  Carefully drawing the [initial-state final-state diagram] below (taking special note of the coordinate system) shows the resolution to this difficulty.{warning}

|!ballreversei.png!|!ballreversef.png!|
||Initial State||Final State||

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{toggle-cloak:id=math1} {color:red} *Mathematical Representation* {color}

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The ball's initial {*}_x_{*} momentum is positive in our coordinates ({*}(+{_}m{~}b{~}v{~}b{~}{_}){*}), while its final {*}_x_{*} momentum is _negative_ ({*}( -- {_}m{~}b{~}v{~}b{~}{_}){*}), giving a change of:
\\
{latex}\begin{large}\[ J_{x} = -m_{b}v_{b} - m_{b}v_{b} = -2m_{b}v_{b}\]\end{large}{latex}
\\
where the negative sign indicates that the impulse is applied in the negative {*}_x_{*} direction, and so the impulse points leftward in this case. 

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{card:Using Algebra}
{card:label=Adding Vectors to get Final Momentum}
h4. Adding to get the Final Momentum

We have defined impulse as the final [momentum|momentum] minus the initial [momentum|momentum], but subtracting [vectors|vector] can be confusing.  Therefore, we will first consider a rearrangement of the definition of impulse.  We can write:
\\
{latex}\begin{large}\[ \vec{p}_{f} = \vec{p}_{i} + \vec{J} \] \end{large}{latex}
\\
Thus, we can consider the impulse as the [vector|vector] we must _add_ to the initial [momentum|momentum] to yield the final [momentum|momentum].  

We can use this formulation to draw a vector diagram representing the ball-wall collision.  Remembering the rules for adding [vectors|vector] tail-to-tip, we can draw the following diagram which includes the impulse vector:

!addimpulsevec.png!
{card:Adding Vectors to get Final Momentum}
{card:label=Subtracting Initial Momentum Vector from Final}

h4. Subtracting Initial Momentum from Final

It is also possible to draw a vector representation of the regular definition of impulse
\\
{latex}\begin{large}\[ \vec{J} = \vec{p}_{f} - \vec{p}_{i}  \] \end{large}{latex}
\\
but drawing a vector equation that includes subtraction is tricky.  We _must_ think of this equation in the following way:
\\
{latex}\begin{large}\[ \vec{J} = \vec{p}_{f} + (- \vec{p}_{i})  \] \end{large}{latex}
\\
In other words, we must think of the right hand side as the final momentum _plus_ the _negative_ of the initial momentum vector.  Since the negative of a vector is just the reversed vector, this leads to the picture:
\\
!addneginitvec.png!
\\
which gives the same impulse vector as the diagram above.

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