Energy of a moving object

Definition

Kinetic energy

The energy held in the kinetic store of an object because it is moving.

Every moving object has energy in its kinetic store, and the amount depends on its mass and its speed.
The kinetic energy is calculated using $v2E_k = \dfrac{1}{2}\,m\,v^2$ , which you must learn.
Here $E_k$ is the kinetic energy in joules, $J\text{J}$ , $m$ is the mass in kilograms, $kg\text{kg}$ , and $v$ is the speed in metres per second, $m/s\text{m/s}$ .
Kinetic energy is directly proportional to mass, so doubling the mass at the same speed doubles the kinetic energy.
Kinetic energy is proportional to the square of the speed, so doubling the speed makes the kinetic energy four times as large.
This is why a car travelling at twice the speed needs roughly four times the braking distance, because the brakes must remove four times as much energy.
Square the speed before multiplying, because squaring the whole of $vm\,v$ is one of the commonest slips in this calculation.
To find a speed from a known kinetic energy, rearrange the equation to $\sqrt{\dfrac{2\,E_k}{m}}$ .

Example

Kinetic energy of a trolley

A trolley of mass $kg0.80\ \text{kg}$ moves at $m/s1.5\ \text{m/s}$ .

E_k = \frac{1}{2}\,m\,v^2

E_k = 0.5 \times 0.80 \times 1.5^2

E_k = 0.5 \times 0.80 \times 2.25 = 0.90\ \text{J}

The kinetic store of the trolley holds $J0.90\ \text{J}$ .

Example

Speed from kinetic energy

A football of mass $kg0.45\ \text{kg}$ has $J45\ \text{J}$ of kinetic energy just after it is kicked.

\sqrt{\frac{2\,E_k}{m}}

\sqrt{\frac{2 \times 45}{0.45}} = \sqrt{200}

14\ \text{m/s}

The ball leaves the boot at about $m/s14\ \text{m/s}$ , given to two significant figures.

Practical

Investigating kinetic energy with light gates

Aim: to measure the kinetic energy of a trolley at the foot of a ramp and find how it changes with the release height.
Apparatus: dynamics trolley, ramp about $m1\ \text{m}$ long propped on blocks, light gate connected to a data logger, a card of measured length fixed upright on the trolley, metre rule, top-pan balance, and a box of padding at the end of the runway.
Variables: the release height is the independent variable, the speed and kinetic energy at the foot of the ramp are the dependent variables, and the trolley, the ramp surface and the card length are controlled.
Method:
- Measure the mass of the trolley with its card on a top-pan balance and convert it into kilograms.
- Measure the length of the card with a ruler, for example $cm10.0\ \text{cm}$ , and convert it to $m0.100\ \text{m}$ .
- Clamp the light gate at the foot of the ramp at a height where the card breaks the beam cleanly before any other part of the trolley does.
- Mark a release line on the ramp and measure its vertical height above the bench with the metre rule.
- Hold the trolley with its back wheels on the line and let go without pushing.
- Record the time the card takes to pass through the beam, which the data logger shows to $s0.001\ \text{s}$ .
- Release the trolley three times from the same line, then repeat from at least four other heights.
Results: the trolley passes through the gate faster when it is released from higher up, and its kinetic energy rises roughly in proportion to the release height.
Maths: the speed through the gate is $\dfrac{\text{card length}}{\text{time}}$ , and the kinetic energy is then $v2E_k = \dfrac{1}{2}\,m\,v^2$ using the mean time for each height.
Watch out: a push at release adds energy, a card that is not upright gives the wrong length, and friction means the kinetic energy at the bottom is always slightly less than the gravitational potential energy lost.
Safety: keep the padding at the end of the runway so the trolley cannot run off the bench, and keep feet out of the way in case it falls.

Energy of a stretched spring

Definition

Elastic potential energy

The energy held in the elastic store of an object that has been stretched, compressed or bent elastically.

Definition

Extension

The increase in the length of an object when it is stretched, equal to its stretched length minus its original length.

Definition

Spring constant

The force needed to produce each metre of extension in a spring, measured in newtons per metre.

A spring that has been stretched or compressed holds energy in its elastic potential energy store.
The energy stored is calculated using $x2E_e = \dfrac{1}{2}\,k\,x^2$ , which is a select and apply equation that you need to recognise rather than memorise.
Here $E_e$ is the elastic potential energy in joules, $J\text{J}$ , $k$ is the spring constant in newtons per metre, $N/m\text{N/m}$ , and $x$ is the extension in metres, $m\text{m}$ .
The equation only applies while the spring has a linear force-extension relationship, because beyond that point each extra newton stretches the spring by a larger amount.
The extension is the increase in length, so subtract the original length from the stretched length before substituting.
Extensions are usually measured in centimetres or millimetres, so convert them to metres first, for example $m5.0\ \text{cm} = 0.050\ \text{m}$ .
Because the extension is squared, doubling the extension makes the stored energy four times as large.
A stiffer spring with a larger spring constant stores more energy for the same extension.
The energy stored in a spring equals the work done in stretching it, which is why both quantities are measured in joules.

Example

Energy stored in a spring

A spring with a spring constant of $N/m40\ \text{N/m}$ is stretched from a length of $cm12.0\ \text{cm}$ to a length of $cm27.0\ \text{cm}$ .
The extension is $15.0\ \text{cm}$ , which is $m0.150\ \text{m}$ .

E_e = \frac{1}{2}\,k\,x^2

E_e = 0.5 \times 40 \times 0.150^2

E_e = 0.5 \times 40 \times 0.0225 = 0.45\ \text{J}

The elastic potential energy store of the spring holds $J0.45\ \text{J}$ .

Energy of a raised object

Definition

Gravitational potential energy

The energy held in the gravitational store of an object because of its height in a gravitational field.

Definition

Gravitational field strength

The gravitational force acting on each kilogram of mass, which is $N/kg10\ \text{N/kg}$ near the Earth's surface.

Lifting an object means doing work against its weight, so the object's gravitational potential energy store increases.
The change in gravitational potential energy is calculated using $E_p = m\,g\,h$ , which you must learn.
Here $E_p$ is the gravitational potential energy in joules, $J\text{J}$ , $m$ is the mass in kilograms, $kg\text{kg}$ , $g$ is the gravitational field strength in newtons per kilogram, $N/kg\text{N/kg}$ , and $h$ is the height in metres, $m\text{m}$ .
Near the Earth's surface $10\ \text{N/kg}$ , and a question gives you this value when you need it.
The height is the vertical height gained, so a person walking up a long gentle ramp gains the same gravitational potential energy as one climbing a steep ladder to the same level.
The equation gives the change in gravitational potential energy, so $h$ is measured from whatever starting level the question uses, such as the floor or the ground.
On the Moon, where $g$ is only about $N/kg1.6\ \text{N/kg}$ , lifting the same mass to the same height stores far less energy.
When an object falls with negligible air resistance, the gravitational potential energy it loses equals the kinetic energy it gains.
Setting the two equal, $v2m\,g\,h = \dfrac{1}{2}\,m\,v^2$ , lets you find the speed of a falling object from the height it has fallen.

Example

Energy of a lifted mass

A motor lifts a $kg2.0\ \text{kg}$ mass through a height of $m1.5\ \text{m}$ , with $10\ \text{N/kg}$ .

E_p = m\,g\,h

E_p = 2.0 \times 10 \times 1.5 = 30\ \text{J}

The gravitational potential energy store of the mass increases by $J30\ \text{J}$ .

Example

Speed after a fall

An apple of mass $kg0.20\ \text{kg}$ falls $m3.2\ \text{m}$ from a branch, with $10\ \text{N/kg}$ and negligible air resistance.

E_p = m\,g\,h = 0.20 \times 10 \times 3.2 = 6.4\ \text{J}

All of this energy goes into the kinetic store, so $JE_k = 6.4\ \text{J}$ just before the apple lands.

\sqrt{\frac{2\,E_k}{m}} = \sqrt{\frac{2 \times 6.4}{0.20}} = \sqrt{64}

8.0\ \text{m/s}

The apple hits the ground at $m/s8.0\ \text{m/s}$ .

Practical

Investigating gravitational potential energy with a motor

Aim: to compare the electrical energy supplied to a motor with the gravitational potential energy gained by the mass it lifts.
Apparatus: small electric motor clamped to the edge of a bench, string wound on the motor axle, $g100\ \text{g}$ mass hanger with extra slotted masses, joulemeter, low-voltage power supply, metre rule clamped vertically, and a cushion on the floor.
Variables: the mass lifted is the independent variable, the energy supplied and the gravitational potential energy gained are the dependent variables, and the lifting height and supply potential difference are controlled.
Method:
- Clamp the motor firmly to the bench and tie the string to the mass hanger so it rests on the floor directly below the axle with the string just taut.
- Clamp the metre rule vertically beside the string and mark a fixed lifting height, such as $m1.00\ \text{m}$ above the floor.
- Connect the motor to the power supply through the joulemeter and reset the meter to zero.
- Switch on and let the motor wind the mass upwards, switching off the moment the base of the hanger reaches the mark.
- Record the joulemeter reading, lower the mass back to the floor and repeat twice more.
- Add slotted masses and repeat the whole procedure for at least four different masses.
Results: the gravitational potential energy gained by the mass is always less than the energy recorded by the joulemeter, because the motor, the wires and the bearings warm up and the motor hums.
Maths: calculate $E_p = m\,g\,h$ with the mass in kilograms, then subtract it from the mean joulemeter reading to find the energy transferred to other stores.
Watch out: switching off late lets the mass overshoot the mark, slack string means the timing of the lift is uneven, and a mass that swings gains kinetic energy that is not counted in $E_p$ .
Safety: place a cushion beneath the masses in case the string slips or snaps, and keep hands and feet out of the area below the hanging masses.

Energy changes of a bouncing ball

A bouncing ball shows all three stores in this article changing in turn.
Held above the floor, the ball has energy in its gravitational potential energy store.
As it falls, its gravitational potential energy store decreases and its kinetic store increases.
When it hits the floor the ball squashes, so its kinetic store decreases and its elastic potential energy store increases.
As the ball regains its shape, its elastic potential energy store decreases and its kinetic store increases, so it leaves the floor moving upwards.
As it rises, its kinetic store decreases and its gravitational potential energy store increases until it stops at the top of the bounce.
The ball never returns to its starting height, because during each squash some energy goes to the thermal stores of the ball and the floor and some is carried away as sound.
Since $E_p = m\,g\,h$ and neither $m$ nor $g$ changes, the fraction of energy still in the gravitational store after one bounce equals $height\dfrac{\text{rebound height}}{\text{drop height}}$ .
This fraction shows how efficient a ball is at bouncing, so a ball that keeps a larger fraction of its energy bounces higher.

Practical

Investigating bouncing balls

Aim: to describe the energy changes of a bouncing ball and compare the fraction of energy different balls keep after one bounce.
Apparatus: a selection of balls such as a tennis ball, a rubber ball, a table tennis ball and a golf ball, two metre rules clamped vertically one above the other, a clamp stand, a hard floor tile and a phone camera that films in slow motion.
Variables: the type of ball is the independent variable, the rebound height is the dependent variable, and the drop height and floor surface are controlled.
Method:
- Clamp the metre rules vertically with the zero mark resting on the floor tile.
- Hold the first ball so that its bottom is level with the $m1.00\ \text{m}$ mark, then release it without throwing.
- Film the ball against the scale, then play the clip back frame by frame to read the height of the bottom of the ball at the top of its first bounce.
- Line the camera up level with the expected rebound height, so that parallax does not make the reading too high or too low.
- Drop each ball five times from the same height and calculate a mean rebound height, leaving out any anomalous drop.
- Repeat for every type of ball using the same tile and drop height.
Results: a lively rubber ball might rebound to about $m0.70\ \text{m}$ from a $m1.00\ \text{m}$ drop, keeping about $70%70\%$ of its energy in the gravitational store, while a soft ball rebounds far lower.
Maths: the fraction of energy kept is $height\dfrac{\text{mean rebound height}}{\text{drop height}}$ , and multiplying by $100$ gives it as a percentage.
Watch out: judging the top of the bounce by eye gives large random errors, measuring to the top of the ball instead of the bottom shifts every reading, and a warm ball bounces higher than a cold one.
Safety: clear the floor so nobody trips on a rolling ball, and do not stand on chairs or desks to reach greater drop heights.

Common Mistake

Do not forget to square the speed in $v2E_k = \frac{1}{2}\,m\,v^2$ or the extension in $x2E_e = \frac{1}{2}\,k\,x^2$ .
Do not substitute a mass in grams or an extension in centimetres, because both equations need SI units.
Do not use the original length of a spring as $x$ , because $x$ is the extension.
Do not say that a bouncing ball loses energy, because the energy is transferred to thermal stores and carried away by sound.

Exam technique

Calculating stored energy

Learn $v2E_k = \frac{1}{2}\,m\,v^2$ and $E_p = m\,g\,h$ , because you are expected to recall them, while $x2E_e = \frac{1}{2}\,k\,x^2$ is one you select and apply.
Write the equation, substitute the values, then calculate, keeping each step on a separate line.
In a two-step problem such as the speed after a fall, find the energy first and then use it to find the speed.
Round only at the end, and give the answer to the same number of significant figures as the data, usually two or three.
In a practical question, describe repeat readings, a mean and how you dealt with anomalies, as well as the variables you controlled.

Self review

State the equation for kinetic energy and the unit of each quantity.
Calculate the kinetic energy of a $kg1200\ \text{kg}$ car moving at $m/s15\ \text{m/s}$ .
Calculate the energy stored in a spring with a spring constant of $N/m200\ \text{N/m}$ and an extension of $cm4.0\ \text{cm}$ .
Calculate the gravitational potential energy gained by a $kg60\ \text{kg}$ climber who climbs $m12\ \text{m}$ , with $10\ \text{N/kg}$ .
Explain why a bouncing ball never returns to the height it was dropped from.

Energy of a moving object

Definition

Kinetic energy

The energy held in the kinetic store of an object because it is moving.

Every moving object has energy in its kinetic store, and the amount depends on its mass and its speed.
The kinetic energy is calculated using $v2E_k = \dfrac{1}{2}\,m\,v^2$ , which you must learn.
Here $E_k$ is the kinetic energy in joules, $J\text{J}$ , $m$ is the mass in kilograms, $kg\text{kg}$ , and $v$ is the speed in metres per second, $m/s\text{m/s}$ .
Kinetic energy is directly proportional to mass, so doubling the mass at the same speed doubles the kinetic energy.
Kinetic energy is proportional to the square of the speed, so doubling the speed makes the kinetic energy four times as large.
This is why a car travelling at twice the speed needs roughly four times the braking distance, because the brakes must remove four times as much energy.
Square the speed before multiplying, because squaring the whole of $vm\,v$ is one of the commonest slips in this calculation.
To find a speed from a known kinetic energy, rearrange the equation to $\sqrt{\dfrac{2\,E_k}{m}}$ .

Example

Kinetic energy of a trolley

A trolley of mass $kg0.80\ \text{kg}$ moves at $m/s1.5\ \text{m/s}$ .

E_k = \frac{1}{2}\,m\,v^2

E_k = 0.5 \times 0.80 \times 1.5^2

E_k = 0.5 \times 0.80 \times 2.25 = 0.90\ \text{J}

The kinetic store of the trolley holds $J0.90\ \text{J}$ .

Example

Speed from kinetic energy

A football of mass $kg0.45\ \text{kg}$ has $J45\ \text{J}$ of kinetic energy just after it is kicked.

\sqrt{\frac{2\,E_k}{m}}

\sqrt{\frac{2 \times 45}{0.45}} = \sqrt{200}

14\ \text{m/s}

The ball leaves the boot at about $m/s14\ \text{m/s}$ , given to two significant figures.

Practical

Investigating kinetic energy with light gates

Aim: to measure the kinetic energy of a trolley at the foot of a ramp and find how it changes with the release height.
Apparatus: dynamics trolley, ramp about $m1\ \text{m}$ long propped on blocks, light gate connected to a data logger, a card of measured length fixed upright on the trolley, metre rule, top-pan balance, and a box of padding at the end of the runway.
Variables: the release height is the independent variable, the speed and kinetic energy at the foot of the ramp are the dependent variables, and the trolley, the ramp surface and the card length are controlled.
Method:
- Measure the mass of the trolley with its card on a top-pan balance and convert it into kilograms.
- Measure the length of the card with a ruler, for example $cm10.0\ \text{cm}$ , and convert it to $m0.100\ \text{m}$ .
- Clamp the light gate at the foot of the ramp at a height where the card breaks the beam cleanly before any other part of the trolley does.
- Mark a release line on the ramp and measure its vertical height above the bench with the metre rule.
- Hold the trolley with its back wheels on the line and let go without pushing.
- Record the time the card takes to pass through the beam, which the data logger shows to $s0.001\ \text{s}$ .
- Release the trolley three times from the same line, then repeat from at least four other heights.
Results: the trolley passes through the gate faster when it is released from higher up, and its kinetic energy rises roughly in proportion to the release height.
Maths: the speed through the gate is $\dfrac{\text{card length}}{\text{time}}$ , and the kinetic energy is then $v2E_k = \dfrac{1}{2}\,m\,v^2$ using the mean time for each height.
Watch out: a push at release adds energy, a card that is not upright gives the wrong length, and friction means the kinetic energy at the bottom is always slightly less than the gravitational potential energy lost.
Safety: keep the padding at the end of the runway so the trolley cannot run off the bench, and keep feet out of the way in case it falls.

Energy of a stretched spring

Definition

Elastic potential energy

The energy held in the elastic store of an object that has been stretched, compressed or bent elastically.

Definition

Extension

The increase in the length of an object when it is stretched, equal to its stretched length minus its original length.

Definition

Spring constant

The force needed to produce each metre of extension in a spring, measured in newtons per metre.

A spring that has been stretched or compressed holds energy in its elastic potential energy store.
The energy stored is calculated using $x2E_e = \dfrac{1}{2}\,k\,x^2$ , which is a select and apply equation that you need to recognise rather than memorise.
Here $E_e$ is the elastic potential energy in joules, $J\text{J}$ , $k$ is the spring constant in newtons per metre, $N/m\text{N/m}$ , and $x$ is the extension in metres, $m\text{m}$ .
The equation only applies while the spring has a linear force-extension relationship, because beyond that point each extra newton stretches the spring by a larger amount.
The extension is the increase in length, so subtract the original length from the stretched length before substituting.
Extensions are usually measured in centimetres or millimetres, so convert them to metres first, for example $m5.0\ \text{cm} = 0.050\ \text{m}$ .
Because the extension is squared, doubling the extension makes the stored energy four times as large.
A stiffer spring with a larger spring constant stores more energy for the same extension.
The energy stored in a spring equals the work done in stretching it, which is why both quantities are measured in joules.

Example

Energy stored in a spring

A spring with a spring constant of $N/m40\ \text{N/m}$ is stretched from a length of $cm12.0\ \text{cm}$ to a length of $cm27.0\ \text{cm}$ .
The extension is $15.0\ \text{cm}$ , which is $m0.150\ \text{m}$ .

E_e = \frac{1}{2}\,k\,x^2

E_e = 0.5 \times 40 \times 0.150^2

E_e = 0.5 \times 40 \times 0.0225 = 0.45\ \text{J}

The elastic potential energy store of the spring holds $J0.45\ \text{J}$ .

Energy of a raised object

Definition

Gravitational potential energy

The energy held in the gravitational store of an object because of its height in a gravitational field.

Definition

Gravitational field strength

The gravitational force acting on each kilogram of mass, which is $N/kg10\ \text{N/kg}$ near the Earth's surface.

Lifting an object means doing work against its weight, so the object's gravitational potential energy store increases.
The change in gravitational potential energy is calculated using $E_p = m\,g\,h$ , which you must learn.
Here $E_p$ is the gravitational potential energy in joules, $J\text{J}$ , $m$ is the mass in kilograms, $kg\text{kg}$ , $g$ is the gravitational field strength in newtons per kilogram, $N/kg\text{N/kg}$ , and $h$ is the height in metres, $m\text{m}$ .
Near the Earth's surface $10\ \text{N/kg}$ , and a question gives you this value when you need it.
The height is the vertical height gained, so a person walking up a long gentle ramp gains the same gravitational potential energy as one climbing a steep ladder to the same level.
The equation gives the change in gravitational potential energy, so $h$ is measured from whatever starting level the question uses, such as the floor or the ground.
On the Moon, where $g$ is only about $N/kg1.6\ \text{N/kg}$ , lifting the same mass to the same height stores far less energy.
When an object falls with negligible air resistance, the gravitational potential energy it loses equals the kinetic energy it gains.
Setting the two equal, $v2m\,g\,h = \dfrac{1}{2}\,m\,v^2$ , lets you find the speed of a falling object from the height it has fallen.

Example

Energy of a lifted mass

A motor lifts a $kg2.0\ \text{kg}$ mass through a height of $m1.5\ \text{m}$ , with $10\ \text{N/kg}$ .

E_p = m\,g\,h

E_p = 2.0 \times 10 \times 1.5 = 30\ \text{J}

The gravitational potential energy store of the mass increases by $J30\ \text{J}$ .

Example

Speed after a fall

An apple of mass $kg0.20\ \text{kg}$ falls $m3.2\ \text{m}$ from a branch, with $10\ \text{N/kg}$ and negligible air resistance.

E_p = m\,g\,h = 0.20 \times 10 \times 3.2 = 6.4\ \text{J}

All of this energy goes into the kinetic store, so $JE_k = 6.4\ \text{J}$ just before the apple lands.

\sqrt{\frac{2\,E_k}{m}} = \sqrt{\frac{2 \times 6.4}{0.20}} = \sqrt{64}

8.0\ \text{m/s}

The apple hits the ground at $m/s8.0\ \text{m/s}$ .

Practical

Investigating gravitational potential energy with a motor

Aim: to compare the electrical energy supplied to a motor with the gravitational potential energy gained by the mass it lifts.
Apparatus: small electric motor clamped to the edge of a bench, string wound on the motor axle, $g100\ \text{g}$ mass hanger with extra slotted masses, joulemeter, low-voltage power supply, metre rule clamped vertically, and a cushion on the floor.
Variables: the mass lifted is the independent variable, the energy supplied and the gravitational potential energy gained are the dependent variables, and the lifting height and supply potential difference are controlled.
Method:
- Clamp the motor firmly to the bench and tie the string to the mass hanger so it rests on the floor directly below the axle with the string just taut.
- Clamp the metre rule vertically beside the string and mark a fixed lifting height, such as $m1.00\ \text{m}$ above the floor.
- Connect the motor to the power supply through the joulemeter and reset the meter to zero.
- Switch on and let the motor wind the mass upwards, switching off the moment the base of the hanger reaches the mark.
- Record the joulemeter reading, lower the mass back to the floor and repeat twice more.
- Add slotted masses and repeat the whole procedure for at least four different masses.
Results: the gravitational potential energy gained by the mass is always less than the energy recorded by the joulemeter, because the motor, the wires and the bearings warm up and the motor hums.
Maths: calculate $E_p = m\,g\,h$ with the mass in kilograms, then subtract it from the mean joulemeter reading to find the energy transferred to other stores.
Watch out: switching off late lets the mass overshoot the mark, slack string means the timing of the lift is uneven, and a mass that swings gains kinetic energy that is not counted in $E_p$ .
Safety: place a cushion beneath the masses in case the string slips or snaps, and keep hands and feet out of the area below the hanging masses.

Energy changes of a bouncing ball

A bouncing ball shows all three stores in this article changing in turn.
Held above the floor, the ball has energy in its gravitational potential energy store.
As it falls, its gravitational potential energy store decreases and its kinetic store increases.
When it hits the floor the ball squashes, so its kinetic store decreases and its elastic potential energy store increases.
As the ball regains its shape, its elastic potential energy store decreases and its kinetic store increases, so it leaves the floor moving upwards.
As it rises, its kinetic store decreases and its gravitational potential energy store increases until it stops at the top of the bounce.
The ball never returns to its starting height, because during each squash some energy goes to the thermal stores of the ball and the floor and some is carried away as sound.
Since $E_p = m\,g\,h$ and neither $m$ nor $g$ changes, the fraction of energy still in the gravitational store after one bounce equals $height\dfrac{\text{rebound height}}{\text{drop height}}$ .
This fraction shows how efficient a ball is at bouncing, so a ball that keeps a larger fraction of its energy bounces higher.

Practical

Investigating bouncing balls

Aim: to describe the energy changes of a bouncing ball and compare the fraction of energy different balls keep after one bounce.
Apparatus: a selection of balls such as a tennis ball, a rubber ball, a table tennis ball and a golf ball, two metre rules clamped vertically one above the other, a clamp stand, a hard floor tile and a phone camera that films in slow motion.
Variables: the type of ball is the independent variable, the rebound height is the dependent variable, and the drop height and floor surface are controlled.
Method:
- Clamp the metre rules vertically with the zero mark resting on the floor tile.
- Hold the first ball so that its bottom is level with the $m1.00\ \text{m}$ mark, then release it without throwing.
- Film the ball against the scale, then play the clip back frame by frame to read the height of the bottom of the ball at the top of its first bounce.
- Line the camera up level with the expected rebound height, so that parallax does not make the reading too high or too low.
- Drop each ball five times from the same height and calculate a mean rebound height, leaving out any anomalous drop.
- Repeat for every type of ball using the same tile and drop height.
Results: a lively rubber ball might rebound to about $m0.70\ \text{m}$ from a $m1.00\ \text{m}$ drop, keeping about $70%70\%$ of its energy in the gravitational store, while a soft ball rebounds far lower.
Maths: the fraction of energy kept is $height\dfrac{\text{mean rebound height}}{\text{drop height}}$ , and multiplying by $100$ gives it as a percentage.
Watch out: judging the top of the bounce by eye gives large random errors, measuring to the top of the ball instead of the bottom shifts every reading, and a warm ball bounces higher than a cold one.
Safety: clear the floor so nobody trips on a rolling ball, and do not stand on chairs or desks to reach greater drop heights.

Common Mistake

Do not forget to square the speed in $v2E_k = \frac{1}{2}\,m\,v^2$ or the extension in $x2E_e = \frac{1}{2}\,k\,x^2$ .
Do not substitute a mass in grams or an extension in centimetres, because both equations need SI units.
Do not use the original length of a spring as $x$ , because $x$ is the extension.
Do not say that a bouncing ball loses energy, because the energy is transferred to thermal stores and carried away by sound.

Exam technique

Calculating stored energy

Learn $v2E_k = \frac{1}{2}\,m\,v^2$ and $E_p = m\,g\,h$ , because you are expected to recall them, while $x2E_e = \frac{1}{2}\,k\,x^2$ is one you select and apply.
Write the equation, substitute the values, then calculate, keeping each step on a separate line.
In a two-step problem such as the speed after a fall, find the energy first and then use it to find the speed.
Round only at the end, and give the answer to the same number of significant figures as the data, usually two or three.
In a practical question, describe repeat readings, a mean and how you dealt with anomalies, as well as the variables you controlled.

Self review

State the equation for kinetic energy and the unit of each quantity.
Calculate the kinetic energy of a $kg1200\ \text{kg}$ car moving at $m/s15\ \text{m/s}$ .
Calculate the energy stored in a spring with a spring constant of $N/m200\ \text{N/m}$ and an extension of $cm4.0\ \text{cm}$ .
Calculate the gravitational potential energy gained by a $kg60\ \text{kg}$ climber who climbs $m12\ \text{m}$ , with $10\ \text{N/kg}$ .
Explain why a bouncing ball never returns to the height it was dropped from.

7.1.5 Kinetic, elastic and gravitational potential energy

Energy of a moving object

Kinetic energy of a trolley

Speed from kinetic energy

Investigating kinetic energy with light gates

Energy of a stretched spring

Energy stored in a spring

Energy of a raised object

Energy of a lifted mass

Speed after a fall

Investigating gravitational potential energy with a motor

Energy changes of a bouncing ball

Investigating bouncing balls

Calculating stored energy

Review 7.1.5 Kinetic, elastic and gravitational potential energy by teaching Genie

7.1.5 Kinetic, elastic and gravitational potential energy

Energy of a moving object

Kinetic energy of a trolley

Speed from kinetic energy

Investigating kinetic energy with light gates

Energy of a stretched spring

Energy stored in a spring

Energy of a raised object

Energy of a lifted mass

Speed after a fall

Investigating gravitational potential energy with a motor

Energy changes of a bouncing ball

Investigating bouncing balls

Calculating stored energy

Review 7.1.5 Kinetic, elastic and gravitational potential energy by teaching Genie

7.1.5 Kinetic, elastic and gravitational potential energy

Energy of a moving object

Kinetic energy of a trolley

Speed from kinetic energy

Investigating kinetic energy with light gates

Energy of a stretched spring

Energy stored in a spring

Energy of a raised object

Energy of a lifted mass

Speed after a fall

Investigating gravitational potential energy with a motor

Energy changes of a bouncing ball

Investigating bouncing balls

Calculating stored energy

Review 7.1.5 Kinetic, elastic and gravitational potential energy by teaching Genie

7.1.5 Kinetic, elastic and gravitational potential energy Revision Guide

7.1.5 Kinetic, elastic and gravitational potential energy

Energy of a moving object

Kinetic energy of a trolley

Speed from kinetic energy

Investigating kinetic energy with light gates

Energy of a stretched spring

Energy stored in a spring

Energy of a raised object

Energy of a lifted mass

Speed after a fall

Investigating gravitational potential energy with a motor

Energy changes of a bouncing ball

Investigating bouncing balls

Calculating stored energy

Review 7.1.5 Kinetic, elastic and gravitational potential energy by teaching Genie

7.1.5 Kinetic, elastic and gravitational potential energy Revision Guide