Two Billiard Balls Rebound

solved Momentum

Q. Two billiard balls each of mass 0.05 kg $m = 0.05$ kg moving in opposite directions with speed 6 m/s $v_{0} = 6$ m/s collide and rebound with the same speed $v_{1} = - v_{0}$ . What is the impulse imparted to each ball due to the other, in N s?

Inputs

mass

m

0.05 kg

speed

v_{0}

6.0 m/s

Outputs

impulse

| J |

How it's solved

Solved with Momentum

Given

$m = 0.05$ kg
$v_{0} = 6$ m/s
$v_{1} = - v_{0}$ from “rebound with the same speed”

Find $| J |$ .

Working

rebound with the same speed $v_{1} = - v_{0}$
$v_{1} = - 6$
$v_{1} = −6$ m/s
momentum at the end(colliding, x) $p_{1} = m \cdot v_{1}$
$p_{1} = 0.05 \cdot (−6)$
$p_{1} = −0.3$ kg m/s
momentum at the start(colliding, x) $p_{0} = m \cdot v_{0}$
$p_{0} = 0.05 \cdot 6$
$p_{0} = 0.3$ kg m/s
impulse is the change of momentum(colliding, x) $J = p_{1} - p_{0}$
$J = (−0.3) - 0.3$
$J = −0.6$ kg m/s
magnitude from components(impulse on ball during colliding) $| J | = \sqrt{J^{2}}$
$| J | = \sqrt{{(−0.6)}^{2}}$
$| J | = 0.6$ N s
$| J | = 0.6$ kg m/s $= 0.6$ N s

Conclusion

Each ball is turned right round, so the other must take away all its momentum and give it as much again the other way; the two impulses are equal and opposite.

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 :two_billiard_balls_rebound do4  kind :numerical5  describes "Two billiard balls each of mass %{mass} moving in opposite directions with speed %{speed} collide and " \6            "rebound with the same speed. %{impulse}, in N s?"7  given mass: "0.05 kg"8  given speed: "6 m/s"9  asks impulse: "What is the impulse imparted to each ball due to the other"10end

solution.rb

1require_relative "../questions/two_billiard_balls_rebound"2require_relative "../science/laws_of_motion/momentum"34solve :two_billiard_balls_rebound, using: Momentum do5  objects :ball6  axes :x7  ball.goes(:colliding)89  take mass: ball.mass10  take speed: ball.initial_velocity1112  deduce_from "rebound with the same speed", ball.final_velocity => -ball.initial_velocity1314  impulse = ball.impulse.magnitude15  answer impulse:, in: "N s"1617  conclusion "Each ball is turned right round, so the other must take away all its momentum and give it as " \18             "much again the other way; the two impulses are equal and opposite."19end

playground.rb

1require_relative "../solutions/two_billiard_balls_rebound"23play_with :two_billiard_balls_rebound do4  title "Two Billiard Balls Rebound"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	$work$	—	—	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	$J$	kg m/s	—	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
momentum	$p$	kg m/s	—	Momentum
starting momentum	$p_{0}$	kg m/s	—	Momentum
final momentum	$p$	kg m/s	—	Momentum
displacement of ball during colliding along x	$Δ x$	m	—
position of ball at end of colliding along x	$x_{1}$	m	—
position of ball at start of colliding along x	$x_{0}$	m	—
average velocity of ball during colliding along x	$v_{av}$	m/s	—
time taken of colliding	$Δ t$	s	0 … 1e+05
average speed of ball during colliding	$c_{av}$	m/s	—
path length of ball during colliding	$d$	m	—
change of velocity of ball during colliding along x	$Δ v$	m/s	—
velocity of ball at end of colliding from ground along x	$v_{1}$	m/s	—
velocity of ball at start of colliding from ground along x	$v_{0}$	m/s	—
average acceleration of ball during colliding along x	$a_{av}$	m/s²	—
acceleration of ball during colliding along x	$a$	m/s²	—
net force on ball during colliding along x	$F_{net}$	N	—
mass of ball	$m$	kg	0 … ∞
momentum of ball at start of colliding along x	$p_{0}$	kg m/s	—
momentum of ball at end of colliding along x	$p_{1}$	kg m/s	—
impulse on ball during colliding along x	$J$	kg m/s	—
weight on ground	$\| W_{g} \|$	N	—
weight on ball	$\| W \|$	N	—
displacement of ball during colliding	$\| Δ x \|$	m	—
position of ball at end of colliding	$\| x_{1} \|$	m	—
position of ball at start of colliding	$\| x_{0} \|$	m	—
average velocity of ball during colliding	$\| v_{av} \|$	m/s	—
change of velocity of ball during colliding	$\| Δ v \|$	m/s	—
velocity of ball at end of colliding from ground	$\| v_{1} \|$	m/s	—
velocity of ball at start of colliding from ground	$\| v_{0} \|$	m/s	—
average acceleration of ball during colliding	$\| a_{av} \|$	m/s²	—
acceleration of ball during colliding	$\| a \|$	m/s²	—
net force on ball during colliding	$\| F_{net} \|$	N	—
momentum of ball at start of colliding	$\| p_{0} \|$	kg m/s	—
momentum of ball at end of colliding	$\| p_{1} \|$	kg m/s	—
impulse on ball during colliding	$\| J \|$	kg m/s	—

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

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

starting momentum

p \equiv p_{i} (end of motion)

final momentum

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(colliding, x)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(colliding, x)

c_{av} = \frac{d}{Δ t}

path a time(colliding, x)

Δ v = v_{1} - v_{0}

velocity gained(colliding, x)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(colliding, x)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(colliding, x)

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

position after a time(colliding, x)

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

velocity after a distance(colliding, x)

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

mean of the velocities(colliding, x)

F_{net} = m \cdot a

second law(colliding, x)

p_{0} = m \cdot v_{0}

momentum at the start(colliding, x)

p_{1} = m \cdot v_{1}

momentum at the end(colliding, x)

J = p_{1} - p_{0}

impulse is the change of momentum(colliding, x)

J = F_{net} \cdot Δ t

impulse is force over time(colliding, x)

| Δ x | = \sqrt{{Δ x}^{2}}

magnitude from components(displacement of ball during colliding)

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

magnitude from components(position of ball at end of colliding)

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

magnitude from components(position of ball at start of colliding)

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

magnitude from components(average velocity of ball during colliding)

| Δ v | = \sqrt{{Δ v}^{2}}

magnitude from components(change of velocity of ball during colliding)

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

magnitude from components(velocity of ball at end of colliding from ground)

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

magnitude from components(velocity of ball at start of colliding from ground)

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

magnitude from components(average acceleration of ball during colliding)

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

magnitude from components(acceleration of ball during colliding)

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

magnitude from components(net force on ball during colliding)

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

magnitude from components(momentum of ball at start of colliding)

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

magnitude from components(momentum of ball at end of colliding)

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

magnitude from components(impulse on ball during colliding)

v_{1} = - v_{0}

rebound with the same speed

m \neq 0

has mass?(ball)

F_{net} = 0

free of forces?(ball, colliding)