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Notes · Physical EducationUK · A-Levels

Linear, Angular and Projectile Motion

This chapter analyses the three kinds of motion seen in sport. It defines the linear quantities and momentum and reads the graphs of motion, treats angular motion and the conservation of angular momentum that governs spins and somersaults, and analyses projectile motion - the factors that shape a flight path and the resolution of the forces acting in the air - with the calculations the examination requires.

4 sections·~16 min reading time·3 competencies·Level Standard 2 · Advanced 2

T·0666 / 17
Exam profile
AO1 · Define the linear, angular and projectile quantities and state the conservation of angular momentumAO2 · Apply the quantities, graphs and factors to named sporting movementsAO3 · Analyse and evaluate motion, calculating velocity, momentum and angular momentum and interpreting graphs
Operators:defineexplainapplycalculateanalyseevaluate

basic level

AS-Level requires the linear quantities, momentum and the interpretation of motion graphs, and the basic factors affecting a projectile.

higher level

The full A-Level requires angular motion and the conservation of angular momentum, and the resolution of the forces on a projectile.

Depth

Reading depth: In depth

Text

Text size: Standard

Contents · 4 sections▾
  1. Linear, Angular and Projectile Motion
    • 01Linear motion, momentum and motion graphs◐
    • 02Angular motion and the conservation of angular momentum●
    • 03Projectile motion and factors affecting flight◐
    • 04Resolving the forces on a projectile●
§ 01

Linear motion, momentum and motion graphs#

●●○StandardLPAQA 7582 3.2.3.2LPDfE GCE PE - linear motion

Velocity-time graph of a 100 m sprint

Sprint velocity against timeGraph of velocity, roots at x = 0, y-intercept at y = 0, increasing, on the interval x from 0 to 1024681024681012maximum velocityvelocityVelocity / m s−1Time / s
Fig. 1Velocity rises steeply from the blocks as the sprinter accelerates, then levels off at maximum velocity; the gradient is the acceleration and the area beneath the line is the distance run.

Key points

Linear motion is movement in a straight (or curved) line where all parts of the body move the same distance in the same time, and it is described by a set of related quantities. Distance is the total path length (a scalar), while displacement is the straight-line distance and direction from start to finish (a vector). Speed is the rate of change of distance (speed=distance/time\text{speed} = \text{distance} / \text{time}speed=distance/time, a scalar), while velocity is the rate of change of displacement (a vector). Acceleration is the rate of change of velocity, a=(v−u)/ta = (v - u) / ta=(v−u)/t; a negative value (deceleration) means slowing down.
Momentum is the quantity of motion a body has, defined as the product of its mass and its velocity: p=mvp = mvp=mv, measured in kilogram metres per second, and it is a vector in the direction of the velocity. Momentum matters in sport because a body with large momentum is hard to stop or deflect - a heavy, fast rugby player carries great momentum - and because, in a collision or a strike, it is the change in momentum (the impulse) that determines the outcome. A large mass or a high velocity, or both, gives large momentum.
The motion of a performer is often shown on graphs, and reading them is a key skill. On a distance-time (or displacement-time) graph the gradient is the speed (or velocity): a horizontal line means the body is stationary, a straight slope means constant speed, and a curve means changing speed. On a velocity-time graph the gradient is the acceleration and the area under the line is the distance travelled: a horizontal line means constant velocity, a positive slope means acceleration and a negative slope means deceleration.
These graphs tell the story of a race. A 100 m sprinter's velocity-time graph rises steeply from the blocks as they accelerate hard, then flattens as they approach and hold their maximum velocity, and may dip slightly at the end as they fatigue. The steep early gradient shows the large initial acceleration, the flat section shows top speed, and the area under the whole graph is the 100 m distance. Interpreting the phase of a race from the gradient and shape of a motion graph, and calculating speed, velocity, acceleration and momentum, is exactly what this section tests.
velocity=displacementtime\text{velocity} = \dfrac{\text{displacement}}{\text{time}}velocity=timedisplacement​

Velocity

The rate of change of displacement (a vector).

a=v−uta = \dfrac{v - u}{t}a=tv−u​

Acceleration

The change in velocity divided by the time taken.

p=mvp = mvp=mv

Momentum

Momentum equals mass multiplied by velocity (a vector in the direction of the velocity).

Worked example

Comparing momentum

A 95 kg forward runs at 7.0 m s-1 and a 78 kg winger runs at 9.0 m s-1. Calculate each player's momentum and state who is harder to stop.

  1. 01Forward's momentum

    p = mv = 95 x 7.0.

    pforward=95×7.0=665 kg m s−1p_{\text{forward}} = 95 \times 7.0 = 665\ \text{kg m s}^{-1}pforward​=95×7.0=665 kg m s−1
  2. 02Winger's momentum

    p = mv = 78 x 9.0.

    pwinger=78×9.0=702 kg m s−1p_{\text{winger}} = 78 \times 9.0 = 702\ \text{kg m s}^{-1}pwinger​=78×9.0=702 kg m s−1
  3. 03Compare

    The winger has the greater momentum (702 against 665 kg m s-1), so despite the lower mass, the higher velocity makes the winger marginally harder to stop.

Result: The forward has 665 kg m s-1 and the winger 702 kg m s-1; the winger's higher velocity gives them the greater momentum.

Exam focus

  • Distinguish distance/displacement and speed/velocity, and calculate acceleration and momentum.
  • Interpret distance-time and velocity-time graphs (gradient = speed/acceleration; area under a velocity-time graph = distance).

Typical mistakes

  • Confusing distance/speed (scalars) with displacement/velocity (vectors).
  • Reading a velocity-time graph as a distance-time graph - on a velocity-time graph the gradient is acceleration and the area is distance.

Active revision

A 90 kg rugby player runs at 8 m s-1. Calculate their momentum, and explain, using a velocity-time graph, how a sprinter's acceleration changes during a 100 m race.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for physical education (Department for Education) · AQA A-level Physical Education 7582 specification (AQA)

§ 02

Angular motion and the conservation of angular momentum#

●●●AdvancedLPAQA 7582 3.2.3.2LPDfE GCE PE - angular motion

Conservation of angular momentum in a spin

Angular momentum conserved (L = I x omega constant)Graph, Tuck in (mass near axis) → Moment of inertia decreases, Moment of inertia decreases → Angular velocity increases (fast spin), Extend / open out → Moment of inertia increases, Moment of inertia increases → Angular velocity decreases (slow, control landing)Tuck in (massnear axis)Moment ofinertiadecreasesAngular velocityincreases (fastspin)Extend /open outMoment ofinertiaincreasesAngular velocitydecreases (slow,control landing)
Fig. 2With angular momentum conserved in flight, reducing the moment of inertia (tucking in) increases the angular velocity, and extending the body decreases it.

Key points

Angular motion is movement in a circle or rotation about an axis, and it has its own set of quantities that mirror the linear ones. Angular displacement is the angle turned through, angular velocity is the rate of change of angular displacement (how fast the body spins, in radians per second), and angular acceleration is the rate of change of angular velocity. Angular motion is produced by a torque (a turning moment - an eccentric force applied off-centre), just as linear motion is produced by a force.
The angular equivalent of mass is the moment of inertia, the resistance of a body to a change in its state of rotation. Unlike mass, the moment of inertia depends not only on how much mass a body has but on how that mass is distributed relative to the axis of rotation: mass held further from the axis gives a much larger moment of inertia, and mass tucked close to the axis a much smaller one. A performer changes their moment of inertia dramatically by changing their body shape - extending or tucking the limbs.
Angular momentum is the quantity of angular motion, the product of the moment of inertia and the angular velocity (L=IωL = I \omegaL=Iω). The crucial principle is that angular momentum is conserved: once a performer leaves the ground, with no external torque acting (air resistance being negligible), their angular momentum stays constant throughout the flight. Because the product IωI\omegaIω is fixed, changing the moment of inertia must change the angular velocity in the opposite way - the basis of every somersault and spin.
This explains the classic examples. A figure skater spinning with arms outstretched has a large moment of inertia and a low angular velocity; pulling the arms and legs in towards the axis reduces the moment of inertia, so - because angular momentum is conserved - the angular velocity rises sharply and the skater spins faster. A diver or gymnast generates angular momentum at take-off, then tucks to reduce the moment of inertia and spin fast through the somersaults, and opens out to increase the moment of inertia and slow the rotation for a controlled landing. Calculating the change in angular velocity from the conservation of L=IωL = I\omegaL=Iω is the demanding quantitative skill here.
L=IωL = I \omegaL=Iω

Angular momentum

Angular momentum L equals the moment of inertia I multiplied by the angular velocity omega; it is conserved in flight.

Worked example

Change in angular velocity of a spinning skater

A skater spins with a moment of inertia of 4.8 kg m2 at an angular velocity of 2.5 rad s-1. They pull their arms in, reducing the moment of inertia to 1.6 kg m2. Calculate the new angular velocity.

  1. 01State the conserved quantity

    With no external torque, angular momentum L = I x angular velocity is conserved, so I1 x omega1 = I2 x omega2.

  2. 02Find the constant angular momentum

    L = 4.8 x 2.5 = 12 kg m2 s-1.

  3. 03Solve for the new angular velocity

    omega2 = L / I2 = 12 / 1.6.

    ω2=121.6=7.5 rad s−1\omega_2 = \dfrac{12}{1.6} = 7.5\ \text{rad s}^{-1}ω2​=1.612​=7.5 rad s−1

Result: The angular velocity rises from 2.5 to 7.5 rad s-1 - three times faster, because the moment of inertia was reduced to a third while angular momentum stayed constant.

Exam focus

  • Define moment of inertia and angular momentum and explain the conservation of angular momentum in flight.
  • Apply the conservation of angular momentum (and calculate with L = I x angular velocity) to spins and somersaults.

Typical mistakes

  • Saying angular momentum changes when a skater pulls their arms in - angular momentum is conserved; it is the moment of inertia and angular velocity that change (in opposite directions).
  • Treating the moment of inertia as fixed - it depends on the distribution of mass, so tucking or extending the body changes it greatly.

Active revision

Explain, using the conservation of angular momentum, how a trampolinist speeds up and then slows their rotation during a somersault.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for physical education (Department for Education) · AQA A-level Physical Education 7582 specification (AQA)

§ 03

Projectile motion and factors affecting flight#

●●○StandardLPAQA 7582 3.2.3.2LPDfE GCE PE - projectile motion

The parabolic flight path of a projectile

Projectile parabola (45 degree release)Graph of flight path, roots at x = 0, 20, maximum at (10, 5), y-intercept at y = 0, on the interval x from 0 to 205101520123456peak (max height)landingflight pathHeight / mHorizontal distance / m
Fig. 3Ignoring air resistance, a projectile follows a symmetrical parabola: the horizontal velocity is constant while gravity slows, stops and reverses the vertical velocity. Here the release angle is 45 degrees.

Key points

A projectile is any body (a shot, a javelin, a jumper, a ball) released into the air and then acted on only by gravity and air resistance. Ignoring air resistance, a projectile follows a symmetrical parabolic flight path, because its horizontal and vertical motions are independent: horizontally there is no force, so the horizontal velocity stays constant, while vertically gravity produces a constant downward acceleration that slows the upward motion, brings the projectile momentarily to rest at the peak, and then speeds its descent.
Three factors at the moment of release determine the horizontal distance a projectile travels. The speed of release is the most important: greater release speed gives a longer flight and greater range (for a given angle). The angle of release determines the shape of the path; for a projectile landing at the same height it is released, the optimum angle for maximum range is 45 degrees, which balances the time in the air against the horizontal speed. The height of release also matters: releasing from above the landing level (as a shot-putter does from shoulder height) increases the range and lowers the optimum angle to a little below 45 degrees.
These factors explain the technique of throwing and jumping events. A shot-putter maximises range by generating the greatest possible release speed, releasing from as high as they can, and using an angle a little under 45 degrees because the shot lands below the release height. A long-jumper drives for a fast, powerful take-off (speed of release) at an angle that keeps them in the air long enough to travel forward. The performer manipulates the three release factors within the limits of their event to maximise (or, in some sports, to shape) the flight.
The parabola is only the ideal, air-resistance-free path, and this is an important evaluation point. For light objects with a large surface area (a shuttlecock, a table-tennis ball) or for very fast projectiles, air resistance is significant and distorts the flight into an asymmetrical path - rising more steeply than it falls - so the true trajectory is not a perfect parabola. For heavy, dense projectiles (a shot) air resistance is small and the parabola is a good model. Recognising when the parabolic model applies, and analysing how the release factors shape a flight, is the applied skill of this section.
Worked example

Explaining the optimum angle for a shot-put

A shot-putter releases the shot from a height of about 2 m and it lands on the ground. Explain why the optimum angle of release is a little below 45 degrees rather than exactly 45 degrees.

  1. 01Recall the equal-height case

    When release and landing heights are equal, 45 degrees maximises range because it optimally balances the flight time against the horizontal velocity.

  2. 02Account for the release height

    The shot is released about 2 m above the ground, so it has extra height to fall; this gives it more flight time than the equal-height case for the same angle.

  3. 03Adjust the angle

    Because the extra fall already provides flight time, the putter can afford a slightly lower angle, which gives a greater horizontal velocity component and so a longer range.

Result: Because the shot is released above the landing level, the optimum angle is a little under 45 degrees - typically around 40-42 degrees - trading a little height for more horizontal velocity.

Exam focus

  • Explain how the speed, angle and height of release affect the horizontal distance of a projectile.
  • Explain why the ideal flight path is a parabola and when air resistance makes it asymmetrical.

Typical mistakes

  • Saying the optimum release angle is always 45 degrees - it is 45 degrees only when release and landing heights are equal; a higher release (e.g. a shot-put) lowers it.
  • Treating the horizontal velocity as changing in flight - with no horizontal force it stays constant; only the vertical velocity changes.

Active revision

Explain how a shot-putter should manipulate the speed, angle and height of release to maximise the distance of the throw.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for physical education (Department for Education) · AQA A-level Physical Education 7582 specification (AQA)

§ 04

Resolving the forces on a projectile#

●●●AdvancedLPAQA 7582 3.2.3.2LPDfE GCE PE - forces on a projectile

Resolving a release velocity (parallelogram of forces)

Resolution of a release velocityGeometric figure, O, v (release velocity), v, v cos(theta), v sin(theta)Ov (releasevelocity)x1x2vvcos(theta)vsin(theta)theta
Fig. 4The release velocity v is the resultant of its horizontal component v cos(theta) and vertical component v sin(theta); the components form a right-angled triangle with v.

Key points

Two forces act on a projectile in flight: weight, the constant downward gravitational force (W=mgW = mgW=mg), and air resistance, a drag force that opposes the direction of motion. For a heavy, dense projectile such as a shot, weight dominates and air resistance is negligible, giving the near-perfect parabola of the previous section; for a light projectile with a large surface area, air resistance is significant and reshapes the path. Analysing a flight means identifying these forces and resolving them into components.
The velocity of release, and any force, can be resolved into a horizontal and a vertical component using the parallelogram of forces (or the equivalent right-angled triangle). A release velocity vvv at an angle θ\thetaθ to the horizontal has a horizontal component vcos⁡θv\cos\thetavcosθ and a vertical component vsin⁡θv\sin\thetavsinθ. These two components then behave independently: the horizontal component stays constant (no horizontal force, ignoring air resistance), while the vertical component is steadily reduced by gravity. Splitting the motion this way is what makes projectile problems solvable.
The parallelogram (or triangle) of forces is also used to combine forces or velocities into a single resultant. Drawing the two component vectors as adjacent sides of a parallelogram, the diagonal from their common origin is the resultant, whose magnitude and direction can be found by construction or by using R=vx2+vy2R = \sqrt{v_x^2 + v_y^2}R=vx2​+vy2​​ and θ=tan⁡−1(vy/vx)\theta = \tan^{-1}(v_y / v_x)θ=tan−1(vy​/vx​). When air resistance is added to weight, this same method finds the net force on the projectile and hence how the flight departs from the parabola.
The applied significance is that resolving the release velocity predicts the whole flight. The horizontal component determines how far the projectile travels each second (and, over the flight time, the range); the vertical component determines how high it rises and, through gravity, how long it stays in the air. A performer's choice of release angle is really a choice about how to divide their release speed between these two components - more vertical for height and flight time, more horizontal for immediate distance - which is why the optimum angle balances the two. Resolving a release velocity into components and finding a resultant is a standard quantitative task.
vx=vcos⁡θ,vy=vsin⁡θv_x = v\cos\theta, \qquad v_y = v\sin\thetavx​=vcosθ,vy​=vsinθ

Resolving a release velocity

Horizontal and vertical components of a velocity at angle theta to the horizontal.

R=vx2+vy2,θ=tan⁡−1 ⁣vyvxR = \sqrt{v_x^2 + v_y^2}, \qquad \theta = \tan^{-1}\!\dfrac{v_y}{v_x}R=vx2​+vy2​​,θ=tan−1vx​vy​​

Resultant from components

The magnitude and direction of the resultant of two perpendicular components.

Worked example

Resolving a javelin's release velocity

A javelin is released at 28 m s-1 at an angle of 40 degrees to the horizontal. Calculate the horizontal and vertical components of the release velocity. (Use cos 40 = 0.766, sin 40 = 0.643.)

  1. 01Horizontal component

    v_x = v cos(theta) = 28 x 0.766.

    vx=28×0.766=21.4 m s−1v_x = 28 \times 0.766 = 21.4\ \text{m s}^{-1}vx​=28×0.766=21.4 m s−1
  2. 02Vertical component

    v_y = v sin(theta) = 28 x 0.643.

    vy=28×0.643=18.0 m s−1v_y = 28 \times 0.643 = 18.0\ \text{m s}^{-1}vy​=28×0.643=18.0 m s−1
  3. 03Interpret

    The horizontal component (21.4 m s-1) stays constant and carries the javelin forward, while the vertical component (18.0 m s-1) is reduced by gravity, determining the height and flight time.

Result: The release velocity resolves into 21.4 m s-1 horizontally and 18.0 m s-1 vertically - the horizontal component carries range, the vertical component gives height and time in the air.

Exam focus

  • Resolve a release velocity into horizontal (v cos theta) and vertical (v sin theta) components.
  • Use the parallelogram/triangle of forces to combine weight and air resistance into a resultant, or to find a resultant velocity.

Typical mistakes

  • Swapping sine and cosine - the horizontal component (along the angle from the horizontal) uses cosine, the vertical uses sine.
  • Adding the horizontal and vertical components arithmetically instead of combining them as perpendicular vectors.

Active revision

A javelin is released at 25 m s-1 at 38 degrees to the horizontal. Resolve the release velocity into its horizontal and vertical components.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for physical education (Department for Education) · AQA A-level Physical Education 7582 specification (AQA)

Contents

Section -- / 04

    • 01Linear motion, momentum and motion graphs◐
    • 02Angular motion and the conservation of angular momentum●
    • 03Projectile motion and factors affecting flight◐
    • 04Resolving the forces on a projectile●

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Linear, Angular and Projectile Motion

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References & sources

Sources

Department for Education

  • GCE AS and A level subject content for physical education

AQA

  • AQA A-level Physical Education 7582 specification

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