Describing movement

Linear motion means moving from one position to another. Rotation about an axis is Angular motion. An athlete can do both at once. During a rotating jump, for example, the body travels through the air while turning.

Speed is distance travelled per unit time. Displacement per unit time is Velocity, which includes direction. Acceleration is change in velocity per unit time. Speeding up counts as acceleration, but so does slowing down or changing direction.

Section 15(a) of the SEHS data booklet provides:

Watch the notation: the booklet uses $s$ for speed in the first equation but displacement in the second.

Using section 15(a), an athlete who travels 24.0 m in 4.00 s has an average speed of $24.0/4.00 = 6.00\ \text{m s}^{-1}$. If the displacement is also 24.0 m forwards, the average velocity is $6.00\ \text{m s}^{-1}$ forwards. Returning to the starting point changes the result: average velocity for the whole journey is zero, although average speed is not.

If an athlete accelerates from rest to $6.00\ \text{m s}^{-1}$ forwards in 3.00 s, the calculation simplifies to $a=v/t=6.00/3.00=2.00\ \text{m s}^{-2}$ forwards.

Newton’s laws and resultant force

Newton’s first law states that an object remains at rest or continues at constant velocity unless a resultant external force acts on it. Constant velocity requires no resultant force. It doesn't require all forces to be absent.

Newton’s second law links resultant force to acceleration. For a given mass, a larger resultant force produces a larger acceleration in the direction of that force.

When objects interact, they exert forces of equal size and opposite direction on one another. This is Newton’s third law. The forces act on different objects, so they do not cancel when we analyse the forces on just one object.

A runner pushes backwards on the ground; the ground pushes forwards on the runner. The diagram separates this interaction pair from the opposing forces acting on the runner: the forward ground force and backward air resistance. Adding these forces gives the resultant force.

When forces act along one line, choose a positive direction, then add the forces with their signs. A forward ground force of 240 N and backward air resistance of 40 N give a resultant force of 200 N forwards. Equal opposing forces would give zero acceleration—not necessarily zero velocity.

Section 15(a) gives:

The momentum form gives average resultant force over a time interval.

Using section 15(a), a 60.0 kg athlete experiencing a resultant force of 200 N accelerates at $a=F/m=200/60.0=3.33\ \text{m s}^{-2}$.

Weight is the gravitational force acting on an object. Don't confuse it with mass: mass describes the amount of matter and is measured in kg, whereas weight is a force measured in N.

Using section 15(a) and $g=9.81\ \text{m s}^{-2}$ from section 14, a 60.0 kg athlete has a weight of $60.0\times9.81=589\ \text{N}$, acting downwards.

Stability

Stability is resistance to losing balance or being displaced from a balanced position. The point representing the distribution of an object’s mass is its centre of mass. A vertical line through this point is the line of gravity. The support base is the area bounded by the points contacting the supporting surface.

Four factors affect stability:

• Height of the centre of mass: lowering it generally increases resistance to toppling. A defender bends the knees before contact.
• Size of the support base: a wider base gives the line of gravity more distance to move before it reaches an edge. Widen the base in the direction of the expected disturbance.
• Position of the line of gravity: balance is easier to maintain while it falls within the support base. Moving it towards an edge leaves less margin before toppling. Beyond an edge, it tends to cause toppling unless a new support is established.
• Mass: a larger mass needs more force to produce the same acceleration. Under otherwise similar conditions, this makes an athlete harder to displace.

The figure compares an upright, narrow stance with a lowered, wider stance. Leaning can also move the line of gravity beyond the support base, as shown. Greater stability helps an athlete resist contact. Someone preparing to move quickly, though, may deliberately adopt a less stable position.

Summing joint forces

The principle of summing joint forces involves coordinating contributions from several joints to increase the overall effect of a movement. Use only one joint and you limit what the rest of the body could contribute.

In a powerful two-handed overhead throw, the athlete can start with leg drive, continue through the trunk, then use the shoulders, elbows and wrists. With correct sequencing, later segments build on the movement already generated. A forceful upward push may instead involve several joints extending together. The aim is to coordinate their contributions, not simply to move every joint as fast as possible at the same time.

Linear momentum, impulse and direction

Linear momentum is the product of mass and velocity. Multiplying force by the time interval over which it acts gives Impulse.

Section 15(a) gives:

The resultant impulse equals the change in momentum.

Using section 15(a), a 60.0 kg athlete moving at $5.00\ \text{m s}^{-1}$ forwards has momentum of $60.0\times5.00=300\ \text{kg m s}^{-1}$ forwards. An average resultant force of 200 N acting forwards for 0.300 s supplies an impulse of $200\times0.300=60.0\ \text{N s}$ forwards.

A greater impulse produces a greater change in linear momentum. To increase impulse, an athlete can increase average resultant force or extend the useful force application time. Combining both also works. A longer push helps only if the force remains effective.

The principle of impulse direction means that the change in momentum follows the direction of the resultant impulse. An upward resultant impulse changes vertical momentum; a forward resultant impulse changes forward momentum. When an athlete pushes down against the ground in a jump, the ground reaction force acts upwards. To find the resultant impulse, consider this upward ground force alongside the athlete’s downward weight.

Producing and controlling rotation

An eccentric force has a line of action that does not pass through the centre of mass. It produces a turning effect and may also change linear motion.

A force directed through the centre of mass does not initiate rotation about that point. An off-centre force can. At take-off, an athlete can arrange for the ground force’s line of action to pass away from the centre of mass. The figure contrasts a force through the centre of mass with an eccentric force and its turning effect.

Angular momentum describes the quantity of rotational motion. It is conserved when no additional external eccentric forces act. Once airborne, an athlete can bring body segments closer to the rotation axis to turn faster. Moving them farther away slows the turn. Neither change creates additional angular momentum, so rotation must be established appropriately before leaving the supporting surface.

Section 15(a) gives the relationship:

The booklet places this relationship under angular velocity, but $v$ represents linear speed; $\omega$ represents angular velocity.

Using section 15(a), a point on rotating equipment at a radius of 0.800 m, completing one revolution in 0.500 s, has $v=2\pi(0.800)/0.500=10.1\ \text{m s}^{-1}$. Its angular velocity is $\omega=v/r=12.6\ \text{rad s}^{-1}$. At the same angular velocity, points farther from the axis travel faster.

Trigonometry is not assessed. Calculations are restricted to three terms at standard level and four at higher level.

Key terms: speed, linear velocity, angular velocity, acceleration, resultant force, weight, stability, centre of mass, support base, line of gravity, summing joint forces, impulse, linear momentum, impulse direction, eccentric force, angular momentum.

Momentum changes during contact

During a collision, bodies exert forces on one another over a contact interval. Each receives an impulse, changing its momentum. Because the forces on the two bodies are equal and opposite, their impulses are equal and opposite too.

Section 15(a) provides $p=mv$, $J=F\Delta t$ and $F=\Delta p/\Delta t$. Combine these relationships to get:

Collision calculations are limited to one dimension. Choose a positive direction and keep the signs of all velocities. A rebound reverses the direction of motion; it doesn't just reduce the speed.

Worked collision example

A 0.200 kg training ball strikes a fixed wall. Take motion towards the wall as positive: the incoming velocity is $+6.00\ \text{m s}^{-1}$, and the rebound velocity is $-4.00\ \text{m s}^{-1}$. Contact lasts 0.0200 s. In the before-and-after diagram, the ball changes direction and receives a negative impulse.

Using section 15(a):

$\Delta p = 0.200[-4.00-(+6.00)] = -2.00\ \text{kg m s}^{-1}$

So the impulse is $-2.00\ \text{N s}$, acting away from the wall. The average resultant force during contact is:

$F = \frac{-2.00}{0.0200} = -100\ \text{N}$

Increasing the contact time reduces the average force for the same momentum change. When receiving a ball, a player can move the hands with it during contact. This spreads the momentum change over a longer interval. Shorten the contact time, and the same momentum change produces a larger average force.

Coefficient of restitution

The coefficient of restitution compares the relative speed of separation with the relative speed of approach in a collision. It has no units and indicates how strongly the bodies rebound.

Section 15(a) gives:

For the training ball and fixed wall, take the ball as body $a$ and the wall as body $b$. The wall’s velocity is zero both before and after contact. Using section 15(a):

$C_R = \frac{0-(-4.00)}{6.00-0} = 0.667$

For a stationary surface, this is also the rebound speed divided by the incoming speed. With the same incoming speed, a larger coefficient gives a greater rebound speed. Zero means there is no relative separation speed immediately after impact. At a value of one, the relative separation and approach speeds are equal.

Both the ball and the contacting surface affect the rebound. A ball–surface combination that rebounds strongly may suit a fast passing drill. One with a smaller coefficient may make the ball easier to receive and control.

Key terms: collision, change in momentum, impulse, coefficient of restitution, relative speed of approach, relative speed of separation.

Static and dynamic friction

Friction is a contact force that opposes relative sliding between surfaces, or their tendency to slide. It acts along the contacting surfaces. That direction isn't necessarily opposite the athlete’s overall movement.

When surfaces do not slide relative to one another, Static friction provides grip. During a running push-off, for example, the shoe tends to slide backwards. The ground’s static friction force therefore acts forwards on the shoe.

Dynamic friction acts when surfaces slide relative to one another and opposes that sliding. A training sled moving forwards, for example, experiences a backward frictional force from the ground.

The coefficient of friction is a dimensionless measure of the frictional properties of two contacting materials. Static and dynamic coefficients describe different contact conditions. For many material pairs, the static coefficient is greater than the dynamic coefficient: starting the slide requires more force than keeping it going.

Applying the booklet equations

Section 15(a) gives:

$F_f = \mu_d F_N$

The normal force acts perpendicular to the contacting surface.

Static friction adjusts to provide the force needed to prevent sliding, up to a maximum of $\mu_sF_N$. It doesn't automatically equal that maximum. In the booklet model, calculate dynamic friction using $\mu_dF_N$. The coefficient alone does not determine the frictional force; the normal force matters too.

In the sled diagram, normal force acts upwards and weight acts downwards. The applied force points forwards, while friction points backwards. With a horizontal pull on a horizontal surface and no vertical acceleration, normal force equals weight.

Worked friction example

A sled has a normal force of 500 N. For this sled–surface combination, $\mu_s=0.600$ and $\mu_d=0.400$.

Using section 15(a), the maximum static friction is:

$F_{f,\text{max}} = 0.600\times500 = 300\ \text{N}$

With a horizontal pull of 200 N, static friction acts backwards with a force of 200 N. The sled stays at rest. At the limiting condition, static friction reaches 300 N; a pull that exceeds this resistance starts the sled sliding.

Once the sled is sliding, dynamic friction is:

$F_f = 0.400\times500 = 200\ \text{N}$

A 200 N horizontal pull would then balance the frictional force. If there are no other horizontal forces, the sled can move at constant velocity.

Modifying friction for performance

The materials in contact determine the coefficients, but surface condition matters too. Moisture, wear or contamination can change how the materials interact. Friction belongs to the contact pair, not to a shoe, ball or surface on its own.

How much friction is desirable depends on the task:

• Increase grip when force must be transmitted without slipping. Suitable shoe–surface materials help an athlete accelerate and stop, as well as change direction. A textured equipment grip can help prevent the hand from sliding.
• Reduce friction when sliding is useful. A smooth, appropriately prepared sliding surface can reduce resistance, so more of the athlete’s effort contributes to movement.

More friction isn't always better. Coaches and equipment designers must match the contact conditions to the movement: a rapid cut needs secure traction, while equipment intended to glide needs low sliding resistance.

Key terms: frictional force, coefficient of friction, static friction, dynamic friction, materials in contact, normal force.

Mechanical work

Work is energy transferred when a force moves an object through a distance in the force’s direction. Without displacement, the force does no mechanical work on the object. The athlete may still expend energy maintaining that force.

Section 15(b) of the SEHS data booklet gives:

Use a constant force or an appropriate average force over the distance. In these examples, force and movement lie along the same line; trigonometry is not assessed.

Using section 15(b), an athlete who applies an average forward force of 300 N to move a training sled 2.40 m does:

$W = 300\times2.40 = 720\ \text{J}$

The result gives the work done by the applied force. That work does not necessarily all become increased speed. If the sled moves at constant velocity, work against resistance balances the applied force’s work.

Energy transformations

Doing work transfers energy and changes its form. Kinetic energy is energy associated with motion; gravitational potential energy is energy associated with position in a gravitational field.

In a sled push, chemical energy available to the athlete becomes kinetic energy if the sled accelerates. It also becomes thermal energy as resistance is overcome. During a lift, chemical energy is transferred into gravitational potential energy as the load rises, while some becomes thermal energy. The flow diagram shows these alternative outcomes of the athlete doing work.

Technique and equipment do not create energy. Better technique and design can transfer energy more effectively to the intended movement and reduce unwanted transformations.

Power and work intensity

Power is the rate at which work is done. Completing a large amount of work slowly differs from completing the same work quickly; power captures that difference.

Section 15(b) gives:

Dividing average work by elapsed time gives average power. The force–velocity form expresses power when force and velocity are aligned.

Using section 15(b), a sled push that transfers 720 J in 1.20 s gives:

$P = \frac{720}{1.20} = 600\ \text{W}$

The sled’s average speed is $2.40/1.20=2.00\ \text{m s}^{-1}$. With a constant 300 N applied force, the alternative calculation gives the same result:

$P = 300\times2.00 = 600\ \text{W}$

Transfer the same 720 J in 0.900 s and average power is $720/0.900=800\ \text{W}$. The work hasn't changed. The work rate is higher, though, so mechanical power output can indicate work intensity during exercise. It is not the same as the body's total energy expenditure.

Optimizing power output

Power depends on both force and movement velocity. A large force with barely any movement does not necessarily produce high power. Moving quickly against very little force may give a low power output too.

Correct technique helps an athlete apply force effectively while maintaining useful movement velocity. During a sled drive, coordinated joint action and a push directed along the sled’s movement can increase useful work per unit time. Poor timing wastes effort, as does directing force where it cannot assist movement.

Effective equipment design helps match the athlete’s force and movement speed to the task. A suitable cycle gear, for example, allows an athlete to combine pedal force with an effective pedalling speed. A gear that's too difficult to turn may reduce movement velocity; one offering too little resistance may limit useful force. Reducing unwanted friction in moving equipment can also help transfer the athlete’s power to the intended movement.

Key terms: work, energy transformation, power, power output, work intensity, correct technique, sports equipment.