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Excerpt
hiddentrue

Consider the impulse and average force delivered to the head of a player performing a "header" in soccer.

Top View

Side View

Image Added

Image Added

Composition Setup
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).
 Deck of Cards
idbigdeck
 Wiki Markup
||Top View||Side View||
|!soccer1.png!|!soccer2.png!|
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}
 Card
labelPart A
Part A
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.)
Solution
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idsysa
 System: 
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idsysa
Ball as a .
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idinta
 Interactions: 
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idinta
During its projectile motion toward the player, the ball is subject to an external influence from the earth(gravity). During the collision, we assume that the force from the player's head (contact force) is much larger than gravity.
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idmoda
 Models: 
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idmoda
Projectile Motion (One-Dimensional Motion with Constant Velocity in the y direction and One-Dimensional Motion with Constant Acceleration in the z direction) plus .
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idappa
 Approach:
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idclarify
 An Important Clarification: Why consider the ball to be the system?
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 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. 
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clarify
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idproj
 Phase 1: Projectile Motion
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idproj
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
titleGivens
 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}

 Image Added 
 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} \[ t = \frac{y_{f} - 
x
y_{i}}{v_{y}} = \frac{y_{f}}{v_{y}} \] \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°
45°, 
 
 which 
 
 tells 
 
 us 
 _
v
_~z
z,
i~ 
i = 
 _
v
_~y~
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
z,
f~
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
z,
f~ 
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}
above? 

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proj
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idcollis
 Phase 2: Collision
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idcollis
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 
 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}
 positive. 
 Info
Technically, we have not found Ibh, but rather the total impulse on the ball during the collision. If the collision is long enough, gravity's contribution to this impulse might be non-negligible. How much of a difference would be made in the result for Ibh by including the effects of gravity assuming a collision time of 0.050 s? 
Note, however, that we were asked for the impulse delivered to the player's head. By Newton's 3rd 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 
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.  We will use the definition of [impulse].

Approach:  To find the average force, we write:

{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}


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appa
 Card
labelPart B
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?
Solution
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idsysb
 System: 
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Player's head as a .
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 Interactions: 
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We are only interested in the influence from the soccer ball (collision force).
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 Model: 
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.
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 Approach:
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