Motion, Forces and Energy

1.1 Physical quantities and measurement techniques

Units and prefixes

Every physical quantity is a number and a unit. The SI units you need first are the metre (m) for length, the kilogram (kg) for mass, the second (s) for time and the kelvin (K) for temperature.

Measuring length

• A ruler or tape measures lengths from about a millimetre up. Put your eye directly above the mark, so the reading is not shifted by looking from the side (this error is called parallax).
• A micrometer measures very small lengths, such as the thickness of a wire, to $0.01\ \text{mm}$.

Reading a micrometer. Add the reading on the main (sleeve) scale to the reading on the rotating thimble.

• The sleeve scale shows whole millimetres and half millimetres.
• The thimble has 50 divisions, and one full turn moves the spindle $0.5\ \text{mm}$, so each thimble division is $0.01\ \text{mm}$.

$5.5 + 28 \times 0.01 = 5.78\ \text{mm}$

Measuring volume

• A measuring cylinder measures the volume of a liquid. Stand it on a level surface and read the scale at the bottom of the curved surface of the liquid (the meniscus), with your eye level with it.
• To find the volume of an irregular solid by displacement, record the water level, lower the solid in gently, and record the new level. The volume of the solid is the difference.

$65 - 40 = 25\ \text{cm}^3$

Measuring time

Use a clock or a stopwatch (digital timers read to $0.01\ \text{s}$). A person's reaction time when starting and stopping a stopwatch is about $0.2\ \text{s}$, so for a short interval the error is large compared with the time being measured.

Averaging to measure small quantities

To measure something small, measure many of them together and divide.

• A short time. To find the period of a pendulum (the time for one complete swing to and fro), time 20 oscillations and divide by 20. This spreads the reaction-time error over all 20.
• A small distance. To find the thickness of one sheet of paper, measure a stack of 100 sheets and divide by 100.

$\frac{32.0}{20} = 1.60\ \text{s}$

Scalars and vectors

• A scalar has magnitude (size) only.
• A vector has magnitude and direction.

The resultant of two vectors at right angles. The resultant is the single vector that has the same effect as the two together. For vectors at right angles, draw them as the two short sides of a right-angled triangle. The resultant is the hypotenuse, and you can find its size with Pythagoras' theorem. You can also draw it to scale and measure it.

$\sqrt{3^2 + 4^2} = \sqrt{25} = 5\ \text{N}$

Its direction makes an angle $\tan^{-1}\left(\frac{4}{3}\right) = 53^\circ$ with the $3\ \text{N}$ force.

1.2 Motion

Speed, velocity and acceleration

• Speed is the distance travelled per unit time. It is a scalar.
• Velocity is the change in displacement per unit time. It is a vector, so it has a direction.

Acceleration is measured in $\text{m/s}^2$. A deceleration is a negative acceleration.

• Uniform acceleration means the velocity changes by the same amount every second (for example, an object in free fall).
• Non-uniform acceleration means the acceleration itself is changing (for example, a car pulling away from the lights while the driver keeps changing gear).

$a = \frac{25 - 5}{8} = 2.5\ \text{m/s}^2$

$a = \frac{0 - 20}{5} = -4\ \text{m/s}^2$

The negative sign shows deceleration.

Distance–time graphs

The gradient of a distance–time graph is the speed.

Speed–time graphs

The gradient of a speed–time graph is the acceleration. The area under the graph is the distance travelled.

• First $10\ \text{s}$: $a = \dfrac{20 - 0}{10} = 2\ \text{m/s}^2$.
• Next $20\ \text{s}$: constant speed, so $a = 0$.
• Last $10\ \text{s}$: $a = \dfrac{0 - 20}{10} = -2\ \text{m/s}^2$.

Distance is the area under the graph, split into a triangle, a rectangle and a triangle:

$\tfrac{1}{2} \times 10 \times 20 + 20 \times 20 + \tfrac{1}{2} \times 10 \times 20 = 100 + 400 + 100 = 600\ \text{m}$

$\text{average speed} = \frac{600}{40} = 15\ \text{m/s}$

Free fall

Near the Earth's surface, an object falling with no air resistance has a constant acceleration of free fall, $g \approx 9.8\ \text{m/s}^2$. In air, drag (air resistance) reduces the acceleration. In a vacuum, a feather and a coin fall together.

$\Delta v = a\,\Delta t = 10 \times 3 = 30\ \text{m/s}$

1.3 Mass and weight

• Mass is a measure of the quantity of matter in an object. It is measured in kilograms. It is a scalar and does not change from place to place.
• Inertia is the resistance of an object to a change in its state of rest or motion. The bigger the mass, the bigger the inertia.
• Weight is the gravitational force on an object. It is a vector (it points towards the centre of the Earth) and is measured in newtons.

A gravitational field is a region in which a mass experiences a force because of gravitational attraction.

Measuring. A beam balance or equal-arm balance compares weights, so it also compares masses (because $g$ is the same for both pans). An electronic balance reads mass directly. A force meter (newton meter) measures weight.

• On the Earth: $W = 60 \times 10 = 600\ \text{N}$.
• On the Moon: $W = 60 \times 1.6 = 96\ \text{N}$.
• Her mass is still $60\ \text{kg}$. Mass does not change.

1.4 Density

Density is measured in $\text{kg/m}^3$ or $\text{g/cm}^3$. Water has a density of $1.0\ \text{g/cm}^3$, which is $1000\ \text{kg/m}^3$. To change $\text{g/cm}^3$ into $\text{kg/m}^3$, multiply by $1000$.

An object whose density is less than the liquid's density floats in it.

Finding the density of

• a liquid. Find the mass of an empty measuring cylinder. Pour in some liquid and read its volume. Find the mass of the cylinder with the liquid. The mass of the liquid is the difference. Divide by the volume.
• a regularly shaped solid. Measure its mass with a balance. Measure its dimensions with a ruler and calculate the volume (for a cuboid, length $\times$ width $\times$ height). Divide.
• an irregular solid that sinks. Measure its mass. Find its volume by displacement in a measuring cylinder. Divide.

$V = 8 \times 5 \times 2 = 80\ \text{cm}^3 \qquad \rho = \frac{240}{80} = 3.0\ \text{g/cm}^3$

$V = 60 - 40 = 20\ \text{cm}^3 \qquad \rho = \frac{54}{20} = 2.7\ \text{g/cm}^3$

1.5.1 Balanced and unbalanced forces

Types of force

A free-body diagram shows one object on its own, with an arrow for each force acting on it. The length of an arrow shows the size of the force, and its direction shows the direction of the force. For a book resting on a table there are two arrows: weight down and the contact force from the table up, of equal length.

Newton's first law

A force can change the velocity of an object by changing its speed or its direction of motion, or both.

Resultant force

For forces along the same straight line, add forces in the same direction and subtract forces in opposite directions. If the resultant force is zero, the forces are balanced: the object is at rest or moving at constant velocity.

Newton's second law

The force is in newtons (N), the mass in kilograms and the acceleration in $\text{m/s}^2$.

$F = 3000 - 600 = 2400\ \text{N} \qquad a = \frac{2400}{1200} = 2.0\ \text{m/s}^2$

Newton's third law

The two forces of a pair are always the same type of force and act on different objects.

1.5.2 Friction

• Friction is a force that opposes motion between two surfaces in contact. It can also produce heating (rub your hands together).
• Drag is friction in a liquid or a gas. It increases with speed.

Terminal velocity

When a skydiver jumps, the only force at first is weight, so the acceleration is $g$. As the speed rises, so does the air resistance.

1. Weight is constant and drag grows as speed grows.
2. The resultant force (weight minus drag) gets smaller, so the acceleration gets smaller.
3. When drag equals weight, the resultant force is zero and the acceleration is zero.
4. The skydiver now falls at a constant speed. This is the terminal velocity.

When the parachute opens, the drag suddenly becomes bigger than the weight. The skydiver decelerates until drag again equals weight, at a much smaller terminal velocity.

Stopping distance of a vehicle

• Thinking distance is the distance travelled between the driver seeing a hazard and pressing the brake.
• Braking distance is the distance travelled while the brakes slow the vehicle down.

1.5.3 Elastic deformation

Forces can change the size and the shape of an object. When a spring is stretched by a load, its extension is its new length minus its original length.

$x = 0.050\ \text{m} \qquad k = \frac{4.0}{0.050} = 80\ \text{N/m}$

Load–extension graphs

Experiment. Hang a spring from a clamp next to a ruler. Add loads of $1\ \text{N}$ at a time and measure the length each time. Subtract the original length to get the extension, and plot extension against load.

• At first the graph is a straight line through the origin. The extension is directly proportional to the load (doubling the load doubles the extension).
• The limit of proportionality is the point where the line starts to curve. Beyond it the spring stretches by more than it should for each extra newton.

1.5.4 Circular motion

An object moves in a circle only when a force acts at right angles to its motion, pointing towards the centre. This force changes the direction of the velocity all the time, but not the speed. Examples are the tension in a string whirling a ball, friction on the tyres of a car going round a bend, and gravity on a satellite.

For an object moving in a circle:

• if the force increases (with the mass and the radius the same), the speed increases;
• if the force increases (with the mass and the speed the same), the radius decreases;
• a bigger mass needs a bigger force to keep the same speed and radius.

If the force stops (for example, the string breaks), the object carries on in a straight line in the direction it was moving at that moment.

1.5.5 Turning effect of forces

The moment of a force is its turning effect about a point called the pivot.

Everyday examples: opening a door (push far from the hinges), using a spanner, a see-saw.

$300 \times 2.0 = 400 \times d \quad\Rightarrow\quad d = \frac{600}{400} = 1.5\ \text{m}$

Verifying the principle. Balance a metre rule on a pivot at its centre. Hang a weight on each side and move them until the rule is horizontal. Measure each distance from the pivot, calculate the two moments, and show that they are equal.

1.5.6 Centre of gravity

The centre of gravity of an object is the point where the whole weight of the object appears to act.

Finding the centre of gravity of a flat shape (a lamina) with a plumb line.

1. Make a small hole near the edge of the shape and hang it from a pin so it can swing freely.
2. Hang a plumb line (a string with a weight) from the same pin. When it stops, draw a line on the shape along the string.
3. Repeat from two other holes.
4. The lines cross at the centre of gravity.

Stability. An object is more stable if it has a low centre of gravity and a wide base. If the line of the weight, drawn down from the centre of gravity, falls outside the base when the object is tilted, it topples over. This is why a double-decker bus is tested by tilting it, and why racing cars are low and wide.

1.6 Momentum

Conservation of momentum. When objects interact and no external force acts, the total momentum before equals the total momentum after.

$1000 \times 6.0 + 500 \times 0 = (1000 + 500)\,v \quad\Rightarrow\quad v = \frac{6000}{1500} = 4.0\ \text{m/s}$

$F = \frac{\Delta p}{\Delta t} = \frac{0.40 \times 20 - 0}{0.020} = 400\ \text{N}$

1.7.1 Energy

Energy is stored in different ways:

• kinetic (moving objects)
• gravitational potential (objects raised in a gravitational field)
• chemical (fuels, food, batteries)
• elastic (strain) (stretched or squashed springs)
• nuclear (the nucleus of an atom)
• electrostatic (between charges)
• internal (thermal) (the random motion of particles)

Energy is transferred between stores by forces (mechanical work done), electrical currents (electrical work done), heating, and by waves (light, sound and other waves).

The unit of energy is the joule (J).

$E_k = \tfrac{1}{2} \times 2.0 \times 3.0^2 = 9.0\ \text{J}$

$\Delta E_p = 5.0 \times 10 \times 2.0 = 100\ \text{J}$

The loss of potential energy equals the gain of kinetic energy:

$mgh = \tfrac{1}{2}mv^2 \quad\Rightarrow\quad v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20\ \text{m/s}$

The mass cancels, so the answer does not depend on it.

1.7.2 Work

Work done is the energy transferred by a force.

$W = 50 \times 4.0 = 200\ \text{J}$

1.7.3 Energy resources

• Non-renewable resources will run out: fossil fuels (coal, oil, natural gas) and nuclear fuel.
• Renewable resources are replaced as fast as they are used: biofuels, hydroelectric, solar, geothermal, wind, tides and waves.

In most power stations a fuel heats water in a boiler to make steam. The steam turns a turbine, which turns a generator that produces electricity.

1.7.4 Efficiency

No machine is $100\%$ efficient. The rest of the energy is wasted, usually as thermal energy in the surroundings.

$\frac{350}{500} \times 100\% = 70\%$

$\frac{6}{60} \times 100\% = 10\%$

1.7.5 Power

$W = Fd = 600 \times 10 = 6000\ \text{J} \qquad P = \frac{6000}{20} = 300\ \text{W}$

1.8 Pressure

The same force gives a bigger pressure on a smaller area. This is why a knife is sharp (small area, large pressure), and why skis and snowshoes are wide (large area, small pressure, so you do not sink).

$p = \frac{800}{0.040} = 20\,000\ \text{Pa}$

Pressure on a surface produces a force at right angles to that surface. You can show this with a container with holes in its side: water spurts out at right angles to the surface at each hole.

Atmospheric pressure and the barometer

The air above us has weight, so the atmosphere produces a pressure of about $1.0 \times 10^{5}\ \text{Pa}$ at sea level. A mercury barometer is a glass tube sealed at the top, filled with mercury and turned upside down in a dish of mercury. The mercury falls until its column is about $760\ \text{mm}$ tall. The pressure of the mercury column at the bottom balances the atmospheric pressure, so the height of the column measures the atmospheric pressure.

Pressure in a liquid

Pressure in a liquid increases with depth and with the density of the liquid.

$\Delta p = 1000 \times 10 \times 10 = 1.0 \times 10^{5}\ \text{Pa}$