A Ball Hit Straight Back

solved Momentum

Q. A batsman hits back a ball straight in the direction of the bowler without changing its initial speed $v_{1} = - v_{0}$ of 12 m/s $v_{0} = 12$ m/s. The mass of the ball is 0.15 kg $m = 0.15$ kg; assume linear motion of the ball. What is the impulse imparted to the ball, in N s?

Inputs

speed

v_{0}

12.0 m/s

mass

m

0.15 kg

Outputs

impulse

| J |

How it's solved

Solved with Momentum

Given

$v_{0} = 12$ m/s
$m = 0.15$ kg
$v_{1} = - v_{0}$ from “without changing its initial speed”

Find $| J |$ .

Working

without changing its initial speed $v_{1} = - v_{0}$
$v_{1} = - 12$
$v_{1} = −12$ m/s
momentum at the end(hit, x) $p_{1} = m \cdot v_{1}$
$p_{1} = 0.15 \cdot (−12)$
$p_{1} = −1.8$ kg m/s
momentum at the start(hit, x) $p_{0} = m \cdot v_{0}$
$p_{0} = 0.15 \cdot 12$
$p_{0} = 1.8$ kg m/s
impulse is the change of momentum(hit, x) $J = p_{1} - p_{0}$
$J = (−1.8) - 1.8$
$J = −3.6$ kg m/s
magnitude from components(impulse on ball during hit) $| J | = \sqrt{J^{2}}$
$| J | = \sqrt{{(−3.6)}^{2}}$
$| J | = 3.6$ N s
$| J | = 3.6$ kg m/s $= 3.6$ N s

Conclusion

The bat turns the ball's momentum right round, so it has to undo the momentum it came in with and give it as much again the other way: twice the momentum.

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_ball_hit_straight_back do4  kind :numerical5  describes "A batsman hits back a ball straight in the direction of the bowler without changing its initial speed " \6            "of %{speed}. The mass of the ball is %{mass}; assume linear motion of the ball. %{impulse}, in N s?"7  given speed: "12 m/s"8  given mass: "0.15 kg"9  asks impulse: "What is the impulse imparted to the ball"10end

solution.rb

1require_relative "../questions/a_ball_hit_straight_back"2require_relative "../science/laws_of_motion/momentum"34solve :a_ball_hit_straight_back, using: Momentum do5  objects :ball6  axes :x7  ball.goes(:hit)89  take speed: ball.initial_velocity10  take mass: ball.mass1112  deduce_from "without changing its initial speed", ball.final_velocity => -ball.initial_velocity1314  impulse = ball.impulse.magnitude15  answer impulse:, in: "N s"1617  conclusion "The bat turns the ball's momentum right round, so it has to undo the momentum it came in " \18             "with and give it as much again the other way: twice the momentum."19end

playground.rb

1require_relative "../solutions/a_ball_hit_straight_back"23play_with :a_ball_hit_straight_back do4  title "A Ball Hit Straight Back"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 hit along x	$Δ x$	m	—
position of ball at end of hit along x	$x_{1}$	m	—
position of ball at start of hit along x	$x_{0}$	m	—
average velocity of ball during hit along x	$v_{av}$	m/s	—
time taken of hit	$Δ t$	s	0 … 1e+05
average speed of ball during hit	$c_{av}$	m/s	—
path length of ball during hit	$d$	m	—
change of velocity of ball during hit along x	$Δ v$	m/s	—
velocity of ball at end of hit from ground along x	$v_{1}$	m/s	—
velocity of ball at start of hit from ground along x	$v_{0}$	m/s	—
average acceleration of ball during hit along x	$a_{av}$	m/s²	—
acceleration of ball during hit along x	$a$	m/s²	—
net force on ball during hit along x	$F_{net}$	N	—
mass of ball	$m$	kg	0 … ∞
momentum of ball at start of hit along x	$p_{0}$	kg m/s	—
momentum of ball at end of hit along x	$p_{1}$	kg m/s	—
impulse on ball during hit along x	$J$	kg m/s	—
weight on ground	$\| W_{g} \|$	N	—
weight on ball	$\| W \|$	N	—
displacement of ball during hit	$\| Δ x \|$	m	—
position of ball at end of hit	$\| x_{1} \|$	m	—
position of ball at start of hit	$\| x_{0} \|$	m	—
average velocity of ball during hit	$\| v_{av} \|$	m/s	—
change of velocity of ball during hit	$\| Δ v \|$	m/s	—
velocity of ball at end of hit from ground	$\| v_{1} \|$	m/s	—
velocity of ball at start of hit from ground	$\| v_{0} \|$	m/s	—
average acceleration of ball during hit	$\| a_{av} \|$	m/s²	—
acceleration of ball during hit	$\| a \|$	m/s²	—
net force on ball during hit	$\| F_{net} \|$	N	—
momentum of ball at start of hit	$\| p_{0} \|$	kg m/s	—
momentum of ball at end of hit	$\| p_{1} \|$	kg m/s	—
impulse on ball during hit	$\| J \|$	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(hit, x)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(hit, x)

c_{av} = \frac{d}{Δ t}

path a time(hit, x)

Δ v = v_{1} - v_{0}

velocity gained(hit, x)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(hit, x)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(hit, x)

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

position after a time(hit, x)

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

velocity after a distance(hit, x)

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

mean of the velocities(hit, x)

F_{net} = m \cdot a

second law(hit, x)

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

momentum at the start(hit, x)

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

momentum at the end(hit, x)

J = p_{1} - p_{0}

impulse is the change of momentum(hit, x)

J = F_{net} \cdot Δ t

impulse is force over time(hit, x)

| Δ x | = \sqrt{{Δ x}^{2}}

magnitude from components(displacement of ball during hit)

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

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

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

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

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

magnitude from components(average velocity of ball during hit)

| Δ v | = \sqrt{{Δ v}^{2}}

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

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

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

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

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

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

magnitude from components(average acceleration of ball during hit)

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

magnitude from components(acceleration of ball during hit)

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

magnitude from components(net force on ball during hit)

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

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

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

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

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

magnitude from components(impulse on ball during hit)

v_{1} = - v_{0}

without changing its initial speed

m \neq 0

has mass?(ball)

F_{net} = 0

free of forces?(ball, hit)