A Bullet Through Plywood

Q. In a ballistics demonstration a police officer fires a bullet of mass 50.0 g with speed 200 m/s on soft plywood of thickness 2.00 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet, in m/s? expects 63.2 m/s

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Q. In a ballistics demonstration a police officer fires a bullet of mass 50.0 g $m = 50.0$ g with speed 200 m/s $v_{0} = 200$ m/s on soft plywood $F_{p} = - | F_{p} |$ of thickness 2.00 cm $x_{1} = 2.00$ cm. The bullet emerges with only 10% of its initial kinetic energy $K_{1} = 0.1 \cdot K_{0}$ . What is the emergent speed of the bullet, in m/s?

The playground

Inputs

mass

m

50 g

speed

v_{0}

200.0 m/s

thickness

x_{1}

2 cm

Outputs

emergent speed

v_{1}

The solution

Solved with KineticEnergy

Given

$m = 50.0$ g
$v_{0} = 200$ m/s
$x_{1} = 2.00$ cm
$F_{p} = - | F_{p} |$ from “on soft plywood”
$K_{1} = 0.1 \cdot K_{0}$ from “emerges with only 10% of its initial kinetic energy”
$v_{1}$ : the positive one of the values that fit from “The bullet emerges”

Assumed

$x_{0} = 0$ m

Find $v_{1}$ .

Working

$m = 50.0$ g $= 0.05$ kg
$x_{1} = 2.00$ cm $= 0.02$ m
kinetic energy of a body(bullet, through, 1) $K_{0} = \frac{m \cdot {v_{0}}^{2}}{2}$
$K_{0} = \frac{0.05 \cdot 200^{2}}{2}$
$K_{0} = 1000$ J
emerges with only 10% of its initial kinetic energy $K_{1} = 0.1 \cdot K_{0}$
$K_{1} = 0.1 \cdot 1000$
$K_{1} = 100$ J
work energy theorem(bullet, through) $K_{1} - K_{0} = W_{F,p}$
$100 - 1000 = W_{F,p}$
$W_{F,p} = −900$ J
change of position(through, x) $Δ x = x_{1} - x_{0}$
$Δ x = 0.02 - 0$
$Δ x = 0.02$ m
work of a steady force(bullet, through, x) $W_{F,p} = F_{p} \cdot Δ x$
$(−900) = F_{p} \cdot 0.02$
$F_{p} = −45000$ N
net force is the resultant(bullet, through, x) $F_{net} = F_{p} + 0$
$F_{net} = (−45000) + 0$
$F_{net} = −45000$ N
second law(through, x) $F_{net} = m \cdot a$
$(−45000) = 0.05 \cdot a$
$a = −900000$ m/s²
velocity after a distance(through, x) ${v_{1}}^{2} = {v_{0}}^{2} + 2 \cdot a \cdot (x_{1} - x_{0})$
${v_{1}}^{2} = 200^{2} + 2 \cdot (−900000) \cdot (0.02 - 0)$
$v_{1} = −63.2455532, 63.2455532$ m/sthe positive one is taken
$v_{1} = 63.2 m/s$
$v_{1} \approx 63.2$ m/sto 3 significant figures

Conclusion

Kinetic energy goes as the square of the speed, so keeping a tenth of the energy keeps the square root of a tenth of the speed: about a third, not a tenth.

The Ruby

1require_relative "../physics"23question :a_bullet_through_plywood do4  kind :numerical5  describes "In a ballistics demonstration a police officer fires a bullet of mass %{mass} with speed %{speed} on " \6            "soft plywood of thickness %{thickness}. The bullet emerges with only 10% of its initial kinetic energy. " \7            "%{emergent_speed}, in m/s?"8  given mass: "50.0 g"9  given speed: "200 m/s"10  given thickness: "2.00 cm"11  asks emergent_speed: "What is the emergent speed of the bullet"12end

1require_relative "../questions/a_bullet_through_plywood"2require_relative "../science/work_energy_and_power/kinetic_energy"34solve :a_bullet_through_plywood, using: KineticEnergy do5  objects :bullet, :plywood6  axes :x7  bullet.goes(:through)89  take mass: bullet.mass10  take speed: bullet.initial_velocity11  take thickness: bullet.final_position1213  assume bullet.initial_position => "0 m"14  deduce_from "on soft plywood", bullet.force(by: plywood).direction => -x15  deduce_from "The bullet emerges", bullet.final_velocity => :positive16  deduce_from "emerges with only 10% of its initial kinetic energy",17              bullet.kinetic_energy(at: through.end) => 0.1 * bullet.kinetic_energy(at: through.start)1819  emergent_speed = bullet.final_velocity20  answer emergent_speed:, in: "m/s", figures: 32122  conclusion "Kinetic energy goes as the square of the speed, so keeping a tenth of the energy keeps " \23             "the square root of a tenth of the speed: about a third, not a tenth."24end

1require_relative "../solutions/a_bullet_through_plywood"23play_with :a_bullet_through_plywood do4  title "A Bullet Through Plywood"5end

The laws

quantity	written	units	within	from
volume	$volume$	—	—	Vectors
mass	$m$	kg	0 … ∞	Vectors
speed	$speed$	—	—	Vectors
density	$density$	—	—	Vectors
number of moles	$n$	—	—	Vectors
angular frequency	${angular}_{frequency}$	—	—	Vectors
work	$W_{F}$	J	—	Vectors
current	$current$	—	—	Vectors
temperature	$temperature$	—	—	Vectors
pressure	$pressure$	—	—	Vectors
time	$t$	s	0 … 1e+05	Vectors
power	$power$	—	—	Vectors
total path length	$s$	—	—	Vectors
energy	$energy$	—	—	Vectors
gravitational potential	${gravitational}_{potential}$	—	—	Vectors
coefficient of friction	$μ$	—	—	Vectors
charge	$charge$	—	—	Vectors
acceleration	$a$	m/s²	—	Vectors
velocity	$v$	m/s	—	Vectors
displacement	$Δ x$	m	—	Vectors
angular velocity	${angular}_{velocity}$	—	—	Vectors
force	$F$	N	—	Vectors
angular momentum	${angular}_{momentum}$	—	—	Vectors
linear momentum	$p_{lin}$	—	—	Vectors
electric field	${electric}_{field}$	—	—	Vectors
average velocity	$v_{av}$	m/s	—	Vectors
magnetic moment	${magnetic}_{moment}$	—	—	Vectors
relative velocity	$v_{AB}$	—	—	Vectors
impulse	$impulse$	—	—	Vectors
first vector	$A$	—	0 … ∞	Vectors
second vector	$B$	—	0 … ∞	Vectors
angle between	$φ$	—	0 … 3.142	Vectors
resultant	$R$	—	0 … ∞	Vectors
resultant direction	$β$	—	0 … 3.142	Vectors
direction	$α$	—	-3.142 … 3.142	Vectors
x component	$A_{x}$	—	—	Vectors
y component	$A_{y}$	—	—	Vectors
position	$x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity
change of velocity	$Δ v$	m/s	—	Acceleration
average acceleration	$a_{av}$	m/s²	—	Acceleration
starting velocity	$v_{0}$	m/s	—	Acceleration
final velocity	$v$	m/s	—	Acceleration
gravity	$g$	m/s²	—	UniformAcceleration
net force	$F_{net}$	N	—	NewtonsSecondLaw
weight	$W$	N	—	NewtonsSecondLaw
work by gravity	$W_{g}$	J	—	Work
net work	$W_{net}$	J	—	Work
kinetic energy	$K$	J	—	KineticEnergy
change of kinetic energy	$Δ K$	J	—	KineticEnergy
displacement of bullet during through along x	$Δ x$	m	—
position of bullet at end of through along x	$x_{1}$	m	—
position of bullet at start of through along x	$x_{0}$	m	—
average velocity of bullet during through along x	$v_{av}$	m/s	—
time taken of through	$Δ t$	s	0 … 1e+05
average speed of bullet during through	$c_{av}$	m/s	—
path length of bullet during through	$d$	m	—
change of velocity of bullet during through along x	$Δ v$	m/s	—
velocity of bullet at end of through from ground along x	$v_{1}$	m/s	—
velocity of bullet at start of through from ground along x	$v_{0}$	m/s	—
average acceleration of bullet during through along x	$a_{av}$	m/s²	—
acceleration of bullet during through along x	$a$	m/s²	—
net force on bullet during through along x	$F_{net}$	N	—
mass of bullet	$m$	kg	0 … ∞
mass of plywood	$m_{p}$	kg	0 … ∞
weight on ground	$\| W_{g} \|$	N	—
weight on bullet	$\| W \|$	N	—
weight on plywood	$\| W_{p} \|$	N	—
force on bullet by plywood during through	$\| F_{p} \|$	N	—
force on bullet by plywood during through along x	$F_{p}$	N	—
work on bullet by plywood during through	$W_{F,p}$	J	—
net work on bullet during through	$W_{net}$	J	—
kinetic energy of bullet at start of through	$K_{0}$	J	—
kinetic energy of bullet at end of through	$K_{1}$	J	—
change of kinetic energy of bullet during through	$Δ K$	J	—
displacement of bullet during through	$\| Δ x \|$	m	—
position of bullet at end of through	$\| x_{1} \|$	m	—
position of bullet at start of through	$\| x_{0} \|$	m	—
average velocity of bullet during through	$\| v_{av} \|$	m/s	—
change of velocity of bullet during through	$\| Δ v \|$	m/s	—
velocity of bullet at end of through from ground	$\| v_{1} \|$	m/s	—
velocity of bullet at start of through from ground	$\| v_{0} \|$	m/s	—
average acceleration of bullet during through	$\| a_{av} \|$	m/s²	—
acceleration of bullet during through	$\| a \|$	m/s²	—
net force on bullet during through	$\| F_{net} \|$	N	—

x_{0} \equiv x_{i} (start of motion)

starting position

x \equiv x_{i} (end of motion)

final position

v_{0} \equiv v_{i} (start of motion)

starting velocity

v \equiv v_{i} (end of motion)

final velocity

t \equiv Δ t_{m}

time

A = \sqrt{{A_{x}}^{2} + {A_{y}}^{2}}

magnitude from components

d_{i} (whole) = d_{i} (part 1) + d_{i} (part 2)

path in parts

Δ t_{w} = Δ t_{p} + Δ t_{p}

time in parts

a_{i} (part 1) = a_{i} (whole)

steady through its parts

m_{b} \neq 0

has mass?

R^{2} = A^{2} + B^{2} + 2 \cdot A \cdot B \cdot \cos φ

law of cosines

R \cdot \sin β = B \cdot \sin φ

law of sines

R \cdot \cos β = A + B \cdot \cos φ

along the first

A_{x} = A \cdot \cos α

resolved along x

A_{y} = A \cdot \sin α

resolved along y

Δ x = x_{1} - x_{0}

change of position(through, x)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(through, x)

c_{av} = \frac{d}{Δ t}

path a time(through, x)

Δ v = v_{1} - v_{0}

velocity gained(through, x)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(through, x)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(through, x)

x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}

position after a time(through, x)

{v_{1}}^{2} = {v_{0}}^{2} + 2 \cdot a \cdot (x_{1} - x_{0})

velocity after a distance(through, x)

x_{1} - x_{0} = \frac{(v_{1} + v_{0}) \cdot Δ t}{2}

mean of the velocities(through, x)

F_{net} = m \cdot a

second law(through, x)

F_{net} = F_{p} + 0

net force is the resultant(bullet, through, x)

W_{F,p} = F_{p} \cdot Δ x

work of a steady force(bullet, through, x)

W_{net} = F_{net} \cdot Δ x

work of the net force(bullet, through)

K_{0} = \frac{m \cdot {v_{0}}^{2}}{2}

kinetic energy of a body(bullet, through, 1)

K_{1} = \frac{m \cdot {v_{1}}^{2}}{2}

kinetic energy of a body(bullet, through, 2)

Δ K = K_{1} - K_{0}

kinetic energy gained(bullet, through)

K_{1} - K_{0} = W_{F,p}

work energy theorem(bullet, through)

| Δ x | = \sqrt{{Δ x}^{2}}

magnitude from components(displacement of bullet during through)

| x_{1} | = \sqrt{{x_{1}}^{2}}

magnitude from components(position of bullet at end of through)

| x_{0} | = \sqrt{{x_{0}}^{2}}

magnitude from components(position of bullet at start of through)

| v_{av} | = \sqrt{{v_{av}}^{2}}

magnitude from components(average velocity of bullet during through)

| Δ v | = \sqrt{{Δ v}^{2}}

magnitude from components(change of velocity of bullet during through)

| v_{1} | = \sqrt{{v_{1}}^{2}}

magnitude from components(velocity of bullet at end of through from ground)

| v_{0} | = \sqrt{{v_{0}}^{2}}

magnitude from components(velocity of bullet at start of through from ground)

| a_{av} | = \sqrt{{a_{av}}^{2}}

magnitude from components(average acceleration of bullet during through)

| a | = \sqrt{a^{2}}

magnitude from components(acceleration of bullet during through)

| F_{net} | = \sqrt{{F_{net}}^{2}}

magnitude from components(net force on bullet during through)

| F_{p} | = \sqrt{{F_{p}}^{2}}

magnitude from components(force on bullet by plywood during through)

F_{p} = - | F_{p} |

on soft plywood

K_{1} = 0.1 \cdot K_{0}

emerges with only 10% of its initial kinetic energy

m \neq 0

has mass?(bullet)

m_{p} \neq 0

has mass?(plywood)

F_{net} = 0

free of forces?(bullet, through)