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Excerpt
hiddentrue

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

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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).

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||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}
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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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Ball as a .

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idinta
Interactions:
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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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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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idproj
Phase 1: Projectile Motion

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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}


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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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idcollis
Phase 2: Collision

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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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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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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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idmodb
Model:
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.

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Approach:

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