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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.)

h4. Solution {

Solution

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:

\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}

Image Added

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

Then,

we

can

substitute

into

the

equation:

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

to

obtain:

}\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:

}\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:

}\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:

}\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:

}\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.

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:

 Collision* {color}

{cloak:id=collis}

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:

}\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:

}\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:

}\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 

Note,

however,

that

we

were

asked

for

the

impulse

delivered

to

the

player's

head.

By

Newton's

3rd

Law

,

that

impulse

is

simply:

}\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.