A Bullet Through Plywood

solved KineticEnergy

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?

Inputs

mass

m

50 g

speed

v_{0}

200.0 m/s

thickness

x_{1}

2 cm

Outputs

emergent speed

v_{1}

How it's solved

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 web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

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

solution.rb

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

playground.rb

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

The laws it drew on

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)