Option C: The Physics of Sports

What you'll learn

How centre of gravity and moments explain stability, toppling and muscle forces.
How impulse and coefficient of restitution describe impacts and bounces.
How moment of inertia, torque and angular momentum explain spin in sport.
How projectile motion, Bernoulli’s equation and drag affect balls, athletes and equipment.

Starting point: forces, distances and energy

This option applies familiar mechanics to sporting situations. You will keep using forces in newtons (N), distances in metres (m), time in seconds (s), mass in kilograms (kg), energy in joules (J), and speed in metres per second (m s $^{-1}$ ).

The key habit is to draw a clear diagram: forces, distances, velocities and axes. Most mistakes in this topic come from using the right equation with the wrong direction or the wrong perpendicular distance.

Stability and toppling

Definition

Centre of gravity

The centre of gravity is the single point through which the weight of a body may be considered to act. The line of action of the weight is the vertical line through the centre of gravity.

The base of support is the region between the contact points with the ground. An athlete is more stable if:

the line of action of their weight falls inside the base of support;
their base of support is wide;
their centre of gravity is low.

Toppling begins when the line of action of the weight falls outside the base. The body then has a turning effect about the outer contact point, called the pivot.

Stability, toppling and forearm moments in sports

Example

Checking whether an athlete topples

A 65 kg climber leans so that the vertical line through their centre of gravity is 0.08 m outside the outer edge of their foot. Estimate the moment of their weight about the foot.

The pivot is the outer edge of the foot, so the perpendicular distance from the pivot to the line of action of weight is $0.08\,\text{m}$ .
Calculate the weight using $W = m g$ :

65\,\text{kg} \times 9.81\,\text{N kg}^{-1} = 638\,\text{N}

Calculate the moment using $M = F d$ :

638\,\text{N} \times 0.08\,\text{m} = 51\,\text{N m}

The line of action is outside the base, so this moment tends to rotate the climber and cause toppling.

Moments in muscles and sporting equipment

A moment is the turning effect of a force about a pivot. It is calculated using:

M = F d

where $d$ is the perpendicular distance from the pivot to the line of action of the force.

Key Idea

Principle of moments

For an object in rotational equilibrium, the total clockwise moment about a pivot equals the total anticlockwise moment about the same pivot.

In the body, muscles often act very close to joints. This gives a small moment arm, so the muscle force may be much larger than the load force. In sailing, a similar idea applies: wind on the sail can produce a heeling moment, while the keel and crew position provide a restoring moment.

Example

Finding a biceps force

A hand holds a 120 N weight. The weight acts 0.32 m from the elbow pivot. The biceps force acts 0.040 m from the elbow. Find the biceps force needed for equilibrium.

Set anticlockwise moment equal to clockwise moment about the elbow:

F_{\text{biceps}}d_{\text{biceps}} = F_{\text{load}}d_{\text{load}}

Substitute the values:

F_{\text{biceps}} \times 0.040\,\text{m} = 120\,\text{N} \times 0.32\,\text{m}

Rearrange and calculate:

F_{\text{biceps}} = \frac{120\,\text{N} \times 0.32\,\text{m}}{0.040\,\text{m}} = 960\,\text{N}

The muscle force is much larger than the load because its perpendicular distance from the pivot is much smaller.

Common Mistake

Using the length instead of the perpendicular distance

The distance in a moment calculation is not always the length of the limb, mast or lever. It must be the shortest perpendicular distance from the pivot to the force’s line of action.

Impulse in impacts

Momentum is mass multiplied by velocity: $p = m v$ . Since velocity has direction, momentum has direction too.

Newton’s second law can be written for a collision as:

F t = m v - m u

Here, $F$ is the average resultant force, $t$ is contact time, $m$ is mass, $u$ is initial velocity and $v$ is final velocity. The quantity $F t$ is called impulse, measured in newton seconds (N s).

In sport, increasing contact time reduces average force for the same change in momentum. That is why a goalkeeper “gives” with the ball and why mats reduce impact forces.

Example

Calculating the average force on a ball

A 0.16 kg ball approaches a bat at 25 m s $^{-1}$ and leaves in the opposite direction at 35 m s $^{-1}$ . Contact time is 0.0050 s. Take the leaving direction as positive.

Assign signs to the velocities: $-25\,\text{m s}^{-1}$ and $+35\,\text{m s}^{-1}$ .
Use $F t = m v - m u$ :

\times 0.0050\,\text{s} = 0.16\,\text{kg}(35 - (-25))\,\text{m s}^{-1}

Calculate $F$ :

\frac{0.16\,\text{kg} \times 60\,\text{m s}^{-1}}{0.0050\,\text{s}} = 1.9 \times 10^3\,\text{N}

Bounces and coefficient of restitution

Definition

Coefficient of restitution

The coefficient of restitution, $e$ , measures how elastic a collision is:

\frac{\text{relative speed after collision}}{\text{relative speed before collision}}

For a ball dropped vertically onto a fixed floor:

\sqrt{\frac{h}{H}}

where $H$ is the drop height and $h$ is the bounce height. A value close to 1 means a very bouncy collision; a value close to 0 means much kinetic energy has been transferred to thermal energy, sound and deformation.

Example

Finding the coefficient of restitution from bounce height

A ball is dropped from 1.80 m and rebounds to 1.15 m. Find $e$ .

Identify the heights: $1.80\,\text{m}$ and $1.15\,\text{m}$ .
Substitute into the bounce-height equation:

\sqrt{\frac{1.15\,\text{m}}{1.80\,\text{m}}}

Calculate:

\approx 0.80

Rotation: moment of inertia, torque and angular momentum

Definition

Moment of inertia

The moment of inertia, $I$ , is a measure of how difficult it is to change an object’s angular velocity about a particular axis. It depends on both mass and how far that mass is from the axis. Its unit is kg m $^{2}$ .

For spheres:

shell=23mr2\begin{aligned} I_{\text{solid sphere}} &= \frac{2}{5}mr^2 \\ I_{\text{thin spherical shell}} &= \frac{2}{3}mr^2 \end{aligned}

A shell has more mass further from the centre, so it has a larger moment of inertia than a solid sphere with the same mass and radius.

Angular velocity, $ω\omega$ , is the rate of change of angle, in radians per second (rad s $^{-1}$ ). Angular acceleration, $α\alpha$ , is the rate of change of angular velocity:

\alpha = \frac{\omega_2 - \omega_1}{t}

A torque, $τ\tau$ , is the rotational equivalent of a resultant force:

\tau = I\alpha

Angular momentum, $L$ , is:

I\omega

and rotational kinetic energy is:

\text{rotational } KE = \frac{1}{2}I\omega^2

Example

Spin-up torque and rotational kinetic energy

A thin spherical shell ball has mass 0.058 kg and radius 0.033 m. It is spun from 20 rad s $^{-1}$ to 110 rad s $^{-1}$ in 0.15 s. Find its moment of inertia, angular acceleration, torque and increase in rotational kinetic energy.

Use the thin shell equation:

\frac{2}{3}mr^2 = \frac{2}{3}(0.058\,\text{kg})(0.033\,\text{m})^2 = 4.2 \times 10^{-5}\,\text{kg m}^2

Calculate angular acceleration:

\alpha = \frac{110 - 20}{0.15}\,\text{rad s}^{-2} = 600\,\text{rad s}^{-2}

Use $τ=Iα\tau = I\alpha$ :

\tau = 4.2 \times 10^{-5}\,\text{kg m}^2 \times 600\,\text{rad s}^{-2} = 2.5 \times 10^{-2}\,\text{N m}

Calculate the increase in rotational kinetic energy:

\Delta KE = \frac{1}{2}I(\omega_2^2 - \omega_1^2) = \frac{1}{2}(4.2 \times 10^{-5})(110^2 - 20^2) = 0.25\,\text{J}

Conservation of angular momentum

If the external torque on a rotating system is zero, angular momentum is conserved:

I_1\omega_1 = I_2\omega_2

This explains why a diver, skater or gymnast spins faster when they tuck in: their moment of inertia decreases, so their angular velocity increases.

Example

A diver tucking during rotation

A diver has moment of inertia 18 kg m $^{2}$ and angular velocity 1.5 rad s $^{-1}$ . In a tuck, their moment of inertia becomes 6.0 kg m $^{2}$ . Find the new angular velocity.

Apply conservation of angular momentum:

I_1\omega_1 = I_2\omega_2

Substitute the values:

18\,\text{kg m}^2 \times 1.5\,\text{rad s}^{-1} = 6.0\,\text{kg m}^2 \times \omega_2

Rearrange:

\omega_2 = \frac{18 \times 1.5}{6.0}\,\text{rad s}^{-1} = 4.5\,\text{rad s}^{-1}

Common Mistake

Conserving rotational kinetic energy in a tuck

Angular momentum is conserved only when external torque is negligible. Rotational kinetic energy does not have to stay constant because the athlete can do work using muscles while changing shape.

Energy in sporting motion

Use conservation of energy by choosing the relevant stores:

linear kinetic energy: $12mv2\frac{1}{2}mv^2$ ;
rotational kinetic energy: $12Iω2\frac{1}{2}I\omega^2$ ;
gravitational potential energy: $m g h$ ;
elastic potential energy, for a Hooke’s law spring: $12kx2\frac{1}{2}kx^2$ .

Example

Energy of a rolling ball launched by a spring

A 0.40 kg solid spherical training ball of radius 0.060 m is launched up a ramp by a spring of stiffness 300 N m $^{-1}$ , compressed by 0.10 m. At a height of 0.20 m, find the speed of the rolling ball. Assume no energy losses and rolling without slipping, so $r\omega$ .

Write the energy balance:

\frac{1}{2}kx^2 = mgh + \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

For a solid sphere, substitute $\frac{2}{5}mr^2$ and $ω=vr\omega = \frac{v}{r}$ :

\frac{1}{2}I\omega^2 = \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{5}mv^2

Calculate available kinetic energy:

\frac{1}{2}kx^2 - mgh = \frac{1}{2}(300)(0.10)^2 - (0.40)(9.81)(0.20) = 0.715\,\text{J}

Combine linear and rotational kinetic energy:

0.715\,\text{J} = \left(\frac{1}{2} + \frac{1}{5}\right)(0.40\,\text{kg})v^2

1.6\,\text{m s}^{-1}

Projectile motion in sport

A projectile is an object moving through the air under gravity alone, if air resistance is ignored. Split the motion into horizontal and vertical components.

Horizontal acceleration is zero, so horizontal velocity is constant.
Vertical acceleration is $- g$ , where $9.81\,\text{m s}^{-2}$ .
Use the usual constant-acceleration equations vertically, such as $v = u + a t$ and $\frac{1}{2}at^2$ .

Projectile motion and airflow effects in sport

Example

Range of a kicked ball

A ball is kicked at 20 m s $^{-1}$ at 35° to the horizontal and lands at the same height. Ignore air resistance. Find its time of flight and range.

Resolve the initial velocity:

u_x = 20\cos 35^\circ = 16.4\,\text{m s}^{-1}

u_y = 20\sin 35^\circ = 11.5\,\text{m s}^{-1}

For landing at the same height, the vertical displacement is zero:

u_y t - \frac{1}{2}gt^2

so the non-zero time is:

\frac{2u_y}{g} = \frac{2(11.5)}{9.81}\,\text{s} = 2.34\,\text{s}

Use horizontal motion for the range:

u_x t = 16.4\,\text{m s}^{-1} \times 2.34\,\text{s} = 38.4\,\text{m}

Bernoulli lift and drag

Bernoulli’s equation in this option is:

p_0 - \frac{1}{2}\rho v^2

where $p$ is pressure, $p_0$ is a constant for the flow, $ρ\rho$ is fluid density and $v$ is flow speed. Faster air corresponds to lower pressure. In sport, pressure differences can help explain lift or sideways forces on balls, sails and aerofoils.

Example

Pressure difference from Bernoulli’s equation

Air of density 1.2 kg m $^{-3}$ moves at 18 m s $^{-1}$ over one side of a ball and 12 m s $^{-1}$ over the other. Estimate the pressure difference.

The faster side has lower pressure, so use:

\Delta p = \frac{1}{2}\rho(v_{\text{fast}}^2 - v_{\text{slow}}^2)

Substitute values:

\Delta p = \frac{1}{2}(1.2)(18^2 - 12^2)\,\text{Pa}

Calculate:

\Delta p = 108\,\text{Pa}

The force acts from the higher-pressure side towards the lower-pressure side.

Drag is the resistive force caused by motion through a fluid. Its magnitude is:

F_D = \frac{1}{2}\rho v^2AC_D

where $A$ is frontal area and $C_D$ is the dimensionless drag coefficient.

Example

Calculating drag on a cyclist

A cyclist has frontal area 0.45 m $^{2}$ , drag coefficient 0.90, and speed 12 m s $^{-1}$ . Air density is 1.2 kg m $^{-3}$ . Find the drag force.

Substitute into $FD=12ρv2ACDF_D = \frac{1}{2}\rho v^2AC_D$ :

F_D = \frac{1}{2}(1.2)(12^2)(0.45)(0.90)\,\text{N}

Calculate:

F_D = 35\,\text{N}

Notice the speed dependence: if the cyclist doubled their speed, the drag force would become four times larger, assuming $C_D$ stays constant.

Common Mistake

Limits of the air-flow equations

Bernoulli’s equation assumes steady, idealised flow, and the drag equation uses an empirical drag coefficient. Real sports flows can be turbulent, so these models are useful approximations rather than perfect descriptions.

Practical and data skills

You may be asked to process real sporting data. Common approaches include:

using video frames to measure projectile position against time, then plotting horizontal and vertical motion separately;
plotting bounce height $h$ against drop height $H$ , where the gradient is $e^2$ ;
plotting drag force $F_D$ against $v^2$ , where the gradient is $12ρACD\frac{1}{2}\rho AC_D$ ;
estimating centre of gravity using balance methods or plumb lines, then discussing uncertainty in position measurements.

Tip

Graph strategy

If an equation predicts direct proportionality, plot the dependent variable against the whole changing quantity. For drag, plot $F_D$ against $v^2$ , not just against $v$ .

Exam technique

In the exam

Draw a diagram first: mark the pivot, perpendicular distances, velocity directions, centre of gravity or force directions before substituting numbers.
Use signs carefully in impulse and projectile questions; opposite directions must have opposite signs.
State assumptions when needed, especially “neglect air resistance”, “no external torque”, “rolling without slipping”, or “steady flow”.

Self review

Check yourself

Why does widening an athlete’s stance make toppling less likely?
A skater pulls their arms in and spins faster. Which quantity is conserved, and what changes?
How could you use a graph of bounce height against drop height to find the coefficient of restitution?

What you'll learn

How centre of gravity and moments explain stability, toppling and muscle forces.
How impulse and coefficient of restitution describe impacts and bounces.
How moment of inertia, torque and angular momentum explain spin in sport.
How projectile motion, Bernoulli’s equation and drag affect balls, athletes and equipment.

Starting point: forces, distances and energy

Stability and toppling

Definition

Centre of gravity

The centre of gravity is the single point through which the weight of a body may be considered to act. The line of action of the weight is the vertical line through the centre of gravity.

The base of support is the region between the contact points with the ground. An athlete is more stable if:

the line of action of their weight falls inside the base of support;
their base of support is wide;
their centre of gravity is low.

Toppling begins when the line of action of the weight falls outside the base. The body then has a turning effect about the outer contact point, called the pivot.

Stability, toppling and forearm moments in sports

Example

Checking whether an athlete topples

A 65 kg climber leans so that the vertical line through their centre of gravity is 0.08 m outside the outer edge of their foot. Estimate the moment of their weight about the foot.

The pivot is the outer edge of the foot, so the perpendicular distance from the pivot to the line of action of weight is $0.08\,\text{m}$ .
Calculate the weight using $W = m g$ :

65\,\text{kg} \times 9.81\,\text{N kg}^{-1} = 638\,\text{N}

Calculate the moment using $M = F d$ :

638\,\text{N} \times 0.08\,\text{m} = 51\,\text{N m}

The line of action is outside the base, so this moment tends to rotate the climber and cause toppling.

Moments in muscles and sporting equipment

A moment is the turning effect of a force about a pivot. It is calculated using:

M = F d

where $d$ is the perpendicular distance from the pivot to the line of action of the force.

Key Idea

Principle of moments

For an object in rotational equilibrium, the total clockwise moment about a pivot equals the total anticlockwise moment about the same pivot.

Example

Finding a biceps force

A hand holds a 120 N weight. The weight acts 0.32 m from the elbow pivot. The biceps force acts 0.040 m from the elbow. Find the biceps force needed for equilibrium.

Set anticlockwise moment equal to clockwise moment about the elbow:

F_{\text{biceps}}d_{\text{biceps}} = F_{\text{load}}d_{\text{load}}

Substitute the values:

F_{\text{biceps}} \times 0.040\,\text{m} = 120\,\text{N} \times 0.32\,\text{m}

Rearrange and calculate:

F_{\text{biceps}} = \frac{120\,\text{N} \times 0.32\,\text{m}}{0.040\,\text{m}} = 960\,\text{N}

The muscle force is much larger than the load because its perpendicular distance from the pivot is much smaller.

Common Mistake

Using the length instead of the perpendicular distance

The distance in a moment calculation is not always the length of the limb, mast or lever. It must be the shortest perpendicular distance from the pivot to the force’s line of action.

Impulse in impacts

Momentum is mass multiplied by velocity: $p = m v$ . Since velocity has direction, momentum has direction too.

Newton’s second law can be written for a collision as:

F t = m v - m u

In sport, increasing contact time reduces average force for the same change in momentum. That is why a goalkeeper “gives” with the ball and why mats reduce impact forces.

Example

Calculating the average force on a ball

A 0.16 kg ball approaches a bat at 25 m s $^{-1}$ and leaves in the opposite direction at 35 m s $^{-1}$ . Contact time is 0.0050 s. Take the leaving direction as positive.

Assign signs to the velocities: $-25\,\text{m s}^{-1}$ and $+35\,\text{m s}^{-1}$ .
Use $F t = m v - m u$ :

\times 0.0050\,\text{s} = 0.16\,\text{kg}(35 - (-25))\,\text{m s}^{-1}

Calculate $F$ :

\frac{0.16\,\text{kg} \times 60\,\text{m s}^{-1}}{0.0050\,\text{s}} = 1.9 \times 10^3\,\text{N}

Bounces and coefficient of restitution

Definition

Coefficient of restitution

The coefficient of restitution, $e$ , measures how elastic a collision is:

\frac{\text{relative speed after collision}}{\text{relative speed before collision}}

For a ball dropped vertically onto a fixed floor:

\sqrt{\frac{h}{H}}

Example

Finding the coefficient of restitution from bounce height

A ball is dropped from 1.80 m and rebounds to 1.15 m. Find $e$ .

Identify the heights: $1.80\,\text{m}$ and $1.15\,\text{m}$ .
Substitute into the bounce-height equation:

\sqrt{\frac{1.15\,\text{m}}{1.80\,\text{m}}}

Calculate:

\approx 0.80

Rotation: moment of inertia, torque and angular momentum

Definition

Moment of inertia

For spheres:

shell=23mr2\begin{aligned} I_{\text{solid sphere}} &= \frac{2}{5}mr^2 \\ I_{\text{thin spherical shell}} &= \frac{2}{3}mr^2 \end{aligned}

A shell has more mass further from the centre, so it has a larger moment of inertia than a solid sphere with the same mass and radius.

Angular velocity, $ω\omega$ , is the rate of change of angle, in radians per second (rad s $^{-1}$ ). Angular acceleration, $α\alpha$ , is the rate of change of angular velocity:

\alpha = \frac{\omega_2 - \omega_1}{t}

A torque, $τ\tau$ , is the rotational equivalent of a resultant force:

\tau = I\alpha

Angular momentum, $L$ , is:

I\omega

and rotational kinetic energy is:

\text{rotational } KE = \frac{1}{2}I\omega^2

Example

Spin-up torque and rotational kinetic energy

Use the thin shell equation:

\frac{2}{3}mr^2 = \frac{2}{3}(0.058\,\text{kg})(0.033\,\text{m})^2 = 4.2 \times 10^{-5}\,\text{kg m}^2

Calculate angular acceleration:

\alpha = \frac{110 - 20}{0.15}\,\text{rad s}^{-2} = 600\,\text{rad s}^{-2}

Use $τ=Iα\tau = I\alpha$ :

\tau = 4.2 \times 10^{-5}\,\text{kg m}^2 \times 600\,\text{rad s}^{-2} = 2.5 \times 10^{-2}\,\text{N m}

Calculate the increase in rotational kinetic energy:

\Delta KE = \frac{1}{2}I(\omega_2^2 - \omega_1^2) = \frac{1}{2}(4.2 \times 10^{-5})(110^2 - 20^2) = 0.25\,\text{J}

Conservation of angular momentum

If the external torque on a rotating system is zero, angular momentum is conserved:

I_1\omega_1 = I_2\omega_2

This explains why a diver, skater or gymnast spins faster when they tuck in: their moment of inertia decreases, so their angular velocity increases.

Example

A diver tucking during rotation

A diver has moment of inertia 18 kg m $^{2}$ and angular velocity 1.5 rad s $^{-1}$ . In a tuck, their moment of inertia becomes 6.0 kg m $^{2}$ . Find the new angular velocity.

Apply conservation of angular momentum:

I_1\omega_1 = I_2\omega_2

Substitute the values:

18\,\text{kg m}^2 \times 1.5\,\text{rad s}^{-1} = 6.0\,\text{kg m}^2 \times \omega_2

Rearrange:

\omega_2 = \frac{18 \times 1.5}{6.0}\,\text{rad s}^{-1} = 4.5\,\text{rad s}^{-1}

Common Mistake

Conserving rotational kinetic energy in a tuck

Angular momentum is conserved only when external torque is negligible. Rotational kinetic energy does not have to stay constant because the athlete can do work using muscles while changing shape.

Energy in sporting motion

Use conservation of energy by choosing the relevant stores:

linear kinetic energy: $12mv2\frac{1}{2}mv^2$ ;
rotational kinetic energy: $12Iω2\frac{1}{2}I\omega^2$ ;
gravitational potential energy: $m g h$ ;
elastic potential energy, for a Hooke’s law spring: $12kx2\frac{1}{2}kx^2$ .

Example

Energy of a rolling ball launched by a spring

Write the energy balance:

\frac{1}{2}kx^2 = mgh + \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

For a solid sphere, substitute $\frac{2}{5}mr^2$ and $ω=vr\omega = \frac{v}{r}$ :

\frac{1}{2}I\omega^2 = \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{5}mv^2

Calculate available kinetic energy:

\frac{1}{2}kx^2 - mgh = \frac{1}{2}(300)(0.10)^2 - (0.40)(9.81)(0.20) = 0.715\,\text{J}

Combine linear and rotational kinetic energy:

0.715\,\text{J} = \left(\frac{1}{2} + \frac{1}{5}\right)(0.40\,\text{kg})v^2

1.6\,\text{m s}^{-1}

Projectile motion in sport

A projectile is an object moving through the air under gravity alone, if air resistance is ignored. Split the motion into horizontal and vertical components.

Horizontal acceleration is zero, so horizontal velocity is constant.
Vertical acceleration is $- g$ , where $9.81\,\text{m s}^{-2}$ .
Use the usual constant-acceleration equations vertically, such as $v = u + a t$ and $\frac{1}{2}at^2$ .

Projectile motion and airflow effects in sport

Example

Range of a kicked ball

A ball is kicked at 20 m s $^{-1}$ at 35° to the horizontal and lands at the same height. Ignore air resistance. Find its time of flight and range.

Resolve the initial velocity:

u_x = 20\cos 35^\circ = 16.4\,\text{m s}^{-1}

u_y = 20\sin 35^\circ = 11.5\,\text{m s}^{-1}

For landing at the same height, the vertical displacement is zero:

u_y t - \frac{1}{2}gt^2

so the non-zero time is:

\frac{2u_y}{g} = \frac{2(11.5)}{9.81}\,\text{s} = 2.34\,\text{s}

Use horizontal motion for the range:

u_x t = 16.4\,\text{m s}^{-1} \times 2.34\,\text{s} = 38.4\,\text{m}

Bernoulli lift and drag

Bernoulli’s equation in this option is:

p_0 - \frac{1}{2}\rho v^2

Example

Pressure difference from Bernoulli’s equation

Air of density 1.2 kg m $^{-3}$ moves at 18 m s $^{-1}$ over one side of a ball and 12 m s $^{-1}$ over the other. Estimate the pressure difference.

The faster side has lower pressure, so use:

\Delta p = \frac{1}{2}\rho(v_{\text{fast}}^2 - v_{\text{slow}}^2)

Substitute values:

\Delta p = \frac{1}{2}(1.2)(18^2 - 12^2)\,\text{Pa}

Calculate:

\Delta p = 108\,\text{Pa}

The force acts from the higher-pressure side towards the lower-pressure side.

Drag is the resistive force caused by motion through a fluid. Its magnitude is:

F_D = \frac{1}{2}\rho v^2AC_D

where $A$ is frontal area and $C_D$ is the dimensionless drag coefficient.

Example

Calculating drag on a cyclist

A cyclist has frontal area 0.45 m $^{2}$ , drag coefficient 0.90, and speed 12 m s $^{-1}$ . Air density is 1.2 kg m $^{-3}$ . Find the drag force.

Substitute into $FD=12ρv2ACDF_D = \frac{1}{2}\rho v^2AC_D$ :

F_D = \frac{1}{2}(1.2)(12^2)(0.45)(0.90)\,\text{N}

Calculate:

F_D = 35\,\text{N}

Notice the speed dependence: if the cyclist doubled their speed, the drag force would become four times larger, assuming $C_D$ stays constant.

Common Mistake

Limits of the air-flow equations

Practical and data skills

You may be asked to process real sporting data. Common approaches include:

using video frames to measure projectile position against time, then plotting horizontal and vertical motion separately;
plotting bounce height $h$ against drop height $H$ , where the gradient is $e^2$ ;
plotting drag force $F_D$ against $v^2$ , where the gradient is $12ρACD\frac{1}{2}\rho AC_D$ ;
estimating centre of gravity using balance methods or plumb lines, then discussing uncertainty in position measurements.

Tip

Graph strategy

If an equation predicts direct proportionality, plot the dependent variable against the whole changing quantity. For drag, plot $F_D$ against $v^2$ , not just against $v$ .

Exam technique

In the exam

Draw a diagram first: mark the pivot, perpendicular distances, velocity directions, centre of gravity or force directions before substituting numbers.
Use signs carefully in impulse and projectile questions; opposite directions must have opposite signs.
State assumptions when needed, especially “neglect air resistance”, “no external torque”, “rolling without slipping”, or “steady flow”.

Self review

Check yourself

Why does widening an athlete’s stance make toppling less likely?
A skater pulls their arms in and spins faster. Which quantity is conserved, and what changes?
How could you use a graph of bounce height against drop height to find the coefficient of restitution?

What you'll learn

Starting point: forces, distances and energy

Stability and toppling

Moments in muscles and sporting equipment

Impulse in impacts

Bounces and coefficient of restitution

Rotation: moment of inertia, torque and angular momentum

Conservation of angular momentum

Energy in sporting motion

Projectile motion in sport

Bernoulli lift and drag

Practical and data skills

Review Option C: The Physics of Sports by teaching Genie

Remember key concepts with flashcards

Option C: The Physics of Sports

What you'll learn

Starting point: forces, distances and energy

Stability and toppling

Moments in muscles and sporting equipment

Impulse in impacts

Bounces and coefficient of restitution

Rotation: moment of inertia, torque and angular momentum

Conservation of angular momentum

Energy in sporting motion

Projectile motion in sport

Bernoulli lift and drag

Practical and data skills

Review Option C: The Physics of Sports by teaching Genie

Remember key concepts with flashcards

Option C: The Physics of Sports

What you'll learn

Starting point: forces, distances and energy

Stability and toppling

Moments in muscles and sporting equipment

Impulse in impacts

Bounces and coefficient of restitution

Rotation: moment of inertia, torque and angular momentum

Conservation of angular momentum

Energy in sporting motion

Projectile motion in sport

Bernoulli lift and drag

Practical and data skills

Review Option C: The Physics of Sports by teaching Genie

Remember key concepts with flashcards

Option C: The Physics of Sports Revision Guide

Option C: The Physics of Sports

What you'll learn

Starting point: forces, distances and energy

Stability and toppling

Moments in muscles and sporting equipment

Impulse in impacts

Bounces and coefficient of restitution

Rotation: moment of inertia, torque and angular momentum

Conservation of angular momentum

Energy in sporting motion

Projectile motion in sport

Bernoulli lift and drag

Practical and data skills

Review Option C: The Physics of Sports by teaching Genie

Remember key concepts with flashcards

Option C: The Physics of Sports Revision Guide