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h2. Part A
Suppose a soccer player is taking a corner kick.  The 0.450 kg ball is launched at 45° from ground level and travels straight across the field (in the _y_ direction in the diagram) until it is contacted by an attacking player's head at a height of 2.0 m above the ground 20.0 m horizontally from the point of the kick.  After the header, the ball is traveling at the same speed as just before the header, but it is moving purely horizontally downfield (the _x_ direction in the diagram).  What is the magniude of the impulse delivered to the player's head by the ball during the header?  (Ignore the effects of air resistance for this estimate.)

System:  Ball as a [point particle] subject to external influences from the earth (gravity) and the player's head (collision force).

Models:  Projectile Motion ([One-Dimensional Motion with Constant Velocity|1-D Motion (Constant Velocity)] in the _y_ direction and [One-Dimensional Motion with Constant Acceleration|1-D Motion (Constant Acceleration)] in the _z_ direction) plus [Momentum and Impulse].

Approach:  {note}Although we are asked for the impulse acting on the player's head, it is simpler to calculate the impulse delivered to the ball by the player's head and then find the desired quantity using [Newton's 3rd Law|Newton's Third Law].{note}

The problem has two parts.  First, we use the methods of projectile motion to determine the velocity of the soccer ball immediately prior to the collision.  Note that because this problem uses two horizontal coordinates, the projectile motion occurs in the _yz_ plane, with gravity in the -- _z_ direction.  Choosing the kick to originate from the point (0,0,0) at time _t_ = 0, our givens are:

{panel:title=Givens}{latex}\begin{large}\[ y_{i} = 0 \]\[z_{i} = 0\]\[y_{f} = \mbox{20 m}\]\[z_{f} = \mbox{2 m} \]
\[ a_{z} = -\mbox{9.8 m/s}^{2}\] \end{large}{latex}{panel}

The most direct way to proceed is to express the time (which we do not need to solve for) in terms of the y-velocity:

{latex}\begin{large} \[ t = \frac{y_{f} - x_{i}}{v_{y}} = \frac{y_{f}}{v_{y}} \] \end{large}{latex}

Then, we can substitute into the equation:

{latex}\begin{large}\[ z_{f} = z_{i} + v_{z,i}t + \frac{1}{2}a_{z}t^{2} \]\end{large}{latex}

to obtain:

{latex}\begin{large}\[ z_{f} = \frac{v_{z,i}}{v_{y}} y_{f} + \frac{1}{2}a_{z}\left(\frac{y_{f}}{v_{y}}\right)^{2}\]\end{large}{latex}

In this equation, we can use the fact that the launch angle is 45°, which tells us _v_~z,i~ = _v_~y~, so:

{latex}\begin{large}\[ z_{f} = y_{f}+\frac{1}{2}a_{z}\frac{y_{f}^{2}}{v_{y}^{2}} \] \end{large}{latex}

This equation is solved to obtain:

{latex}\begin{large}\[ v_{y}= v_{z,i} =\pm \sqrt{\frac{a_{z}y_{f}^{2}}{2(z_{f}-y_{f})}} = \mbox{10.4 m/s}\]\end{large}{latex}

We choose the plus sign, since we have set up our coordinates such that the ball will move in the + _y_ direction.

We are not finished, since we also need _v_~z,f~, the z-velocity at the end of the projectile motion and at the beginning of the ball's collision with the player's head.  To find this velocity, we can use:

{latex}\begin{large}\[ v_{z,f}^{2} = v_{z,i}^{2} + 2a_{z}(z_{f}-z_{i}) = a_{z}\frac{y_{f}^{2} - 4 z_{f}y_{f} + 4z_{f}^{2}}{2(z_{f}-y_{f})}\]\end{large}{latex}

giving:

{latex}\begin{large}\[ v_{z,f} = \pm (y_{f}-2z_{f})\sqrt{\frac{a_{z}}{2(z_{f}-y_{f})}} = - \mbox{8.35 m/s}\]\end{large}{latex}

We choose the sign that makes _v_~z,f~ negative, presuming that the ball is on the way down.

{tip}Can you prove to yourself that the minus sign is the only consistent choice when taking the square root above?{tip}

We have now completed the analysis of the projectile motion.  Using the fact that the final velocity of the projectile motion will equal the initial velocity of the collision with the player's head, we summarize the initial velocity of the ball for the collision:

{latex}\begin{large}\[ v_{x,i} = \mbox{0 m/s} \]
\[ v_{y,i} = \mbox{10.4 m/s}\]
\[ v_{z,i} = \mbox{- 8.34 m/s}\]\end{large}{latex}

The magnitude of the initial velocity is then:

{latex}\begin{large}\[ v_{i} = \sqrt{v_{x,i}^{2}+v_{y,i}^{2}+v_{z,i}^{2}} = \mbox{13.3 m/s}\]\end{large}{latex}

Thus, from the information given in the problem, we will take the final velocity of the ball immediately following the collision with the player's head to be:

{latex}\begin{large}\[ v_{x,f} = \mbox{13.3 m/s} \]
\[ v_{y,f} = \mbox{0 m/s}\]
\[ v_{z,f} = \mbox{0 m/s}\]\end{large}{latex}

The impulse delivered to the ball during its contact with the player's head is therefore:

{latex}\begin{large}\[ I_{bh} = \Delta \vec{p} = m_{ball}((v_{x,f}-v_{x,i})\hat{x}+(v_{y,f}-v_{y,i})\hat{y} + (v_{z,f}-v_{z,i})\hat{z}) = m_{ball}(v_{x,f}\hat{x} - v_{y,i}\hat{y} - v_{z,i}\hat{z}) = \mbox{6.0 kg m/s}\:\hat{x} -\mbox{4.7 kg m/s}\:\hat{y} + \mbox{3.8 kg m/s}\:\hat{z} \]\end{large}{latex}

{tip}It is important to think carefully about the expected signs when calculating a change.  The ball ends up with a positive x-momentum, so the x-impulse is positive.  The ball _loses_ a positive y-momentum, so the y-impulse is negative, the ball _loses_ a _negative_ z-momentum, so the z-impulse is _positive_.{tip}

Note, however, that we were asked for the impulse delivered to the player's head.  By [Newton's 3rd Law|Newton's Third Law], that impulse is simply:

{latex}\begin{large}\[ I_{hb} = -I_{bh} = -\mbox{6.0 kg m/s}\:\hat{x} +\mbox{4.7 kg m/s}\:\hat{y} - \mbox{3.8 kg m/s}\:\hat{z} \]\end{large}{latex}


The magnitude of this impulse is 8.48 kg m/s.  

h2. Part B

Assuming a (generous) collision time of 50 ms, what is the _average_ magnitude of the force imparted to the player's head by the ball during the collision described in Part A?

System: Player's head as a [point particle].  We will ignore all influences other than the collision force the soccer ball applies to the head, since we are only interested in the size of that force.

Model:  None.[Momentum  We will use the definition of [impulseand Impulse].

Approach:  To find the average force, we write use the Law of Interaction:

{latex}\begin{large}\[ I_{hb} = \int F_{hb}\:dt \equiv \bar{F}_{hb} \Delta t \]\end{large}{latex}

Thus,

{latex}\begin{large}\[ \bar{F}_{hb} = \frac{I_{hb}}{\Delta t} = \mbox{170 N} = \mbox{38 lbs}\]\end{large}{latex}