Unit 1: Mechanics and Materials

Define one positive direction, attach signs to displacement, velocity and acceleration, and convert all data to SI units. List the known values of $s,u,v,a,t$, then choose the equation that contains the required quantity but omits the unwanted unknown. Substitute signed values before solving.

A trolley starts at $u=3.0\,\mathrm{m\,s^{-1}}$ and accelerates uniformly at $2.0\,\mathrm{m\,s^{-2}}$ for $4.0\,\mathrm{s}$. Using $v=u+at$ gives $v=3.0+(2.0)(4.0)=11\,\mathrm{m\,s^{-1}}$. The positive answer means it still moves in the chosen positive direction.

Label both axes with quantity and unit, choose a usable scale, and preserve time intervals. A line above or below the time axis has a positive or negative vertical quantity; crossing the axis means that quantity changes sign. On a displacement–time graph, a turning point has zero gradient and marks an instant of zero velocity.

Translate a graph interval by interval: state the sign, whether the coordinate is constant or changing, and what that means physically. For example, a horizontal velocity–time line below the axis represents motion at constant velocity in the negative direction, not an object at rest.

For a straight segment, calculate gradient as change in vertical coordinate divided by change in time. For a curve, draw a tangent at the required instant and find the tangent's gradient using a large triangle. A changing gradient shows non-uniform velocity or acceleration.

Find area geometrically for rectangles, triangles or trapezia. Area below the time axis is negative. Therefore the area under a velocity–time graph is displacement, while total distance requires adding the magnitudes of positive and negative areas. The area under an acceleration–time graph changes the velocity: $v=u+\int a\,dt$.

A scalar is fully specified by magnitude and unit. A vector also requires direction, shown by an arrow, a bold symbol or another recognised vector notation. Two vectors are equal only if both magnitude and direction match.

A runner may travel a distance of $400\,\mathrm{m}$ yet finish with zero displacement. Similarly, speed can remain constant while velocity changes because direction changes. Choose a positive axis when using vector components in one dimension, so opposite directions receive opposite signs.

Choose two perpendicular axes that simplify the situation, such as horizontal/vertical or parallel/perpendicular to a slope. Draw the vector and its two components as a right-angled triangle. The original vector is the hypotenuse and equals the vector sum of its components.

A $50\,\mathrm{N}$ force acts $30^\circ$ above the horizontal. Its components are $F_x=50\cos30^\circ=43\,\mathrm{N}$ and $F_y=50\sin30^\circ=25\,\mathrm{N}$. Add signs after deciding which axis directions are positive.

For any angle, draw vectors to scale using head-to-tail addition: place the tail of the second at the head of the first. The resultant runs from the first tail to the final head. A parallelogram construction is equivalent. State the scale, measure both resultant length and direction, and include units.

For perpendicular vectors $A$ and $B$, use $R=\sqrt{A^2+B^2}$ and $\tan\theta=B/A$, with $\theta$ measured from the direction represented by $A$. For several vectors, first add signed components: $R_x=\sum V_x$ and $R_y=\sum V_y$, then reconstruct the resultant.

Velocities of $6.0\,\mathrm{m\,s^{-1}}$ east and $8.0\,\mathrm{m\,s^{-1}}$ north give $R=10\,\mathrm{m\,s^{-1}}$ at $\tan^{-1}(8/6)=53^\circ$ north of east.

Resolve the launch velocity first: $u_x=u\cos\theta$ and $u_y=u\sin\theta$ when $\theta$ is above horizontal. Use one shared time $t$ in the horizontal and vertical equations. Solve the component with enough data, then carry that same time into the other component.

A ball leaves horizontally at $12\,\mathrm{m\,s^{-1}}$ and falls $5.0\,\mathrm{m}$. Vertically, $5.0=\tfrac12gt^2$, so $t\approx1.01\,\mathrm{s}$. Horizontally it travels $x=u_xt\approx12.1\,\mathrm{m}$.

Choose the object or rigid body and draw only the external forces acting on it. Label each force by type and source: weight, normal contact force, tension, thrust, friction or drag. Arrow direction shows the force direction; a consistent scale may show magnitude when required.

Add forces as vectors to obtain the resultant. Balanced arrows mean zero resultant force, which permits rest or constant velocity. For an extended rigid body, forces with different lines of action may also create moments even when their vector sum is zero.

For constant mass, $\sum F=ma$. Choose an axis, resolve forces along it, and calculate the signed resultant before using the equation. If $\sum F=0$, then $a=0$: the object may be stationary or moving with constant velocity, as stated by Newton's first law.

For a falling object, weight is initially greater than upward resistance, so it accelerates downward. As speed increases, drag increases. The resultant and acceleration shrink until upward forces balance weight; the speed is then constant and is called terminal velocity.

Near Earth's surface, use the local value of $g$ supplied or an accepted value such as $9.81\,\mathrm{N\,kg^{-1}}$. A $2.4\,\mathrm{kg}$ object has weight $W=(2.4)(9.81)=24\,\mathrm{N}$ to two significant figures, directed toward Earth.

Mass normally stays the same when an object moves between gravitational fields, while weight changes with $g$. Numerically, $1\,\mathrm{N\,kg^{-1}}=1\,\mathrm{m\,s^{-2}}$, linking field strength to free-fall acceleration.

Release a dense sphere from rest and measure its fall distance $s$ and time $t$ using an electromagnet with an electronic timer, light-gate arrangement or suitably calibrated video. Repeat timings at several distances, measure from consistent reference points, and use distances large enough that timing resolution is a small fraction of $t$.

With $u=0$ and approximately constant $g$, $s=\tfrac12gt^2$. Plot $s$ on the vertical axis against $t^2$ on the horizontal axis. A straight best-fit line should have gradient $g/2$, so $g=2\times\text{gradient}$. Use a large gradient triangle and include units $\mathrm{m\,s^{-2}}$.

When body A exerts a force on body B, B simultaneously exerts a force of the same type and magnitude on A in the opposite direction. Write the pair explicitly as 'force of A on B' and 'force of B on A' to keep the bodies clear.

During a collision, a ball pushes a pin forward while the pin pushes the ball backward with equal force. The objects can have different accelerations because $a=F/m$ and their masses may differ. Each force belongs on a different free-body diagram.

Linear momentum is the vector $\mathbf p=m\mathbf v$. Its SI unit is $\mathrm{kg\,m\,s^{-1}}$ (equivalently $\mathrm{N\,s}$). Because mass is scalar, momentum points in the same direction as velocity.

Choose a positive direction before calculation. In one dimension, assign positive and negative velocities, then calculate each signed momentum. A $0.20\,\mathrm{kg}$ ball moving at $-15\,\mathrm{m\,s^{-1}}$ has $p=-3.0\,\mathrm{kg\,m\,s^{-1}}$; the minus sign specifies direction.

The same momentum can arise from a small mass at high speed or a large mass at low speed. Momentum is not kinetic energy: momentum depends linearly on velocity and is a vector, whereas kinetic energy depends on speed squared and is a scalar.

If the resultant external force on a system is negligible during an interaction, total linear momentum is constant. In one dimension, choose a positive direction and write $\sum mv\text{ before}=\sum mv\text{ after}$ using signed velocities for every object.

Define the system and the short interaction interval, list masses and velocities before and after, then solve the single momentum equation. For a $2.0\,\mathrm{kg}$ cart at $3.0\,\mathrm{m\,s^{-1}}$ sticking to a stationary $1.0\,\mathrm{kg}$ cart, $6.0=(3.0)v$, so $v=2.0\,\mathrm{m\,s^{-1}}$ in the original direction.

During the interaction, Newton's third-law internal forces are equal and opposite and act for the same time, producing equal and opposite changes of momentum. Internal transfers therefore leave the system total unchanged; an external impulse would change it.

The moment of a force about an axis is $M=Fx$, where $x$ is the perpendicular distance from the axis to the force's line of action. Its unit is $\mathrm{N\,m}$. Label a moment clockwise or anticlockwise.

Mark the pivot, extend the force arrow into its line of action, and draw the shortest perpendicular from the pivot to that line. Multiply the force by this distance. Equivalently, resolve the force perpendicular to a known position vector and multiply that component by the distance from the pivot.

A $40\,\mathrm{N}$ force acts perpendicular to a handle $0.30\,\mathrm{m}$ from its pivot, giving $M=(40)(0.30)=12\,\mathrm{N\,m}$. If the same force acts obliquely, its perpendicular component—and hence its moment—is smaller.

The centre of gravity is the point through which the resultant weight of an extended body may be treated as acting. Include that weight at the correct position when constructing the free-body diagram.

For static equilibrium, both the resultant force and resultant moment are zero. Choose any convenient pivot and apply the principle of moments: total clockwise moment equals total anticlockwise moment. Selecting a pivot through unknown reaction forces often removes their moments from the equation.

Draw and label all forces and perpendicular distances, select a pivot, assign clockwise/anticlockwise senses, form the moment equation, and then use horizontal or vertical force balance if another unknown remains. Check that the answer could physically keep the body supported.

For a constant force, the work transferred is $\Delta W=F\Delta s$ when force and displacement are parallel. If the angle between them is $\theta$, use $\Delta W=F\Delta s\cos\theta$. Work is energy transferred and is measured in joules.

Work is positive when the force component points along the displacement, negative when it opposes motion, and zero when it is perpendicular. For example, a $60\,\mathrm{N}$ force pulling $5.0\,\mathrm{m}$ at $30^\circ$ to the motion does $(60)(5.0)\cos30^\circ=260\,\mathrm{J}$ of work.

A resultant force doing work changes the object's kinetic energy. Work against friction transfers mechanical energy to internal energy; this energy has not vanished.

The kinetic energy of a body of mass $m$ moving at speed $v$ is $E_k=\tfrac12mv^2$. It is a scalar measured in joules, so use the magnitude of velocity and SI units.

For a $0.80\,\mathrm{kg}$ object moving at $6.0\,\mathrm{m\,s^{-1}}$, $E_k=\tfrac12(0.80)(6.0)^2=14.4\,\mathrm{J}$. Doubling speed at fixed mass multiplies kinetic energy by four; doubling mass at fixed speed doubles it.

To find speed, rearrange before substituting: $v=\sqrt{2E_k/m}$. The positive square root gives speed; attach direction only if a separate velocity statement is required.

Near Earth's surface, the change in gravitational potential energy is $\Delta E_{grav}=mg\Delta h$, where $\Delta h$ is the signed change in vertical height. A rise gives positive change; a fall gives negative change.

Lifting a $3.0\,\mathrm{kg}$ load vertically by $1.5\,\mathrm{m}$ changes its GPE by $(3.0)(9.81)(1.5)=44\,\mathrm{J}$ to two significant figures. The result is independent of the path taken between the same starting and finishing heights.

Potential energy itself depends on the chosen zero level, but the change between two heights is physically meaningful. State or infer which final height is above the other before assigning the sign.

Energy cannot be created or destroyed. Choose a system and compare its initial and final energy stores, including energy transferred by work. A useful accounting statement is $E_{initial}+W_{in}=E_{final}+E_{transferred\ out}$.

Identify the start and finish states, write only the relevant terms such as $\tfrac12mv^2$, $mg\Delta h$ and $F\Delta s$, and keep dissipated energy in the balance. Solve symbolically where possible, then check units and whether the magnitude is physically possible.

If a descending object loses $120\,\mathrm{J}$ of GPE and gains $90\,\mathrm{J}$ of KE, the remaining $30\,\mathrm{J}$ has been transferred to other stores, for example by work against resistance. Total energy is still conserved.

Power measures how quickly energy is transferred: $P=E/t$ or $P=W/t$. One watt is one joule per second. These equations give average power over the stated interval.

A motor transfers $18\,\mathrm{kJ}$ in $30\,\mathrm{s}$. Convert first: $P=18000/30=600\,\mathrm{W}$. At the same transferred energy, shorter time means greater average power; greater power does not by itself mean greater total energy.

Use $E/t$ when energy transferred is known and $W/t$ when the transfer is expressed as work done. Rearrange to $E=Pt$ or $t=E/P$ as needed, preserving consistent units.

Efficiency is a ratio with no unit. Multiply by $100\%$ only when a percentage is requested. A device receiving $2.5\,\mathrm{kW}$ and delivering $1.8\,\mathrm{kW}$ usefully has $\eta=1.8/2.5=0.72=72\%$.

Identify input and useful output before substitution. Useful output equals efficiency times total input; total input equals useful output divided by efficiency. The non-useful part is transferred to other stores, often internal energy of the device and surroundings.

Density is mass per unit volume: $\rho=m/V$. In SI units, mass is in kilograms, volume in cubic metres and density in $\mathrm{kg\,m^{-3}}$. The relationship can be rearranged to $m=\rho V$ or $V=m/\rho$.

Identify the material volume represented by the data, convert each measurement before substitution, and check powers of ten carefully. For a regular solid, calculate volume from its dimensions; for an irregular solid, a measured displacement volume may be used.

A sample has mass $0.54\,\mathrm{kg}$ and volume $2.0\times10^{-4}\,\mathrm{m^3}$. Its density is $0.54/(2.0\times10^{-4})=2.7\times10^3\,\mathrm{kg\,m^{-3}}$.

A body in a fluid experiences an upward force because fluid pressure is greater at greater depth. The resulting upthrust equals the weight of fluid displaced: $U=m_{fluid}g=\rho_{fluid}V_{displaced}g$.

A fully submerged object displaces $3.0\times10^{-4}\,\mathrm{m^3}$ of water of density $1000\,\mathrm{kg\,m^{-3}}$. Its upthrust is $(1000)(3.0\times10^{-4})(9.81)=2.9\,\mathrm{N}$.

For a small sphere moving slowly through a fluid with laminar flow, Stokes' law gives viscous drag $F=6\pi\eta rv$. Here $\eta$ is dynamic viscosity in $\mathrm{Pa\,s}$, $r$ is sphere radius and $v$ is speed relative to the fluid. Drag acts opposite to relative motion.

At terminal speed for a falling sphere, $W=U+F_{drag}$. Therefore $F_{drag}=W-U$ can be inserted into Stokes' law to find viscosity or speed. For the liquids considered here, increasing temperature reduces viscosity, so a given sphere reaches a greater terminal speed.

Measure the sphere's diameter in several orientations with a micrometer and average it to obtain $r$. Obtain the sphere's weight and the liquid density, or measure the quantities needed to calculate them. Keep the transparent liquid column vertical and record its temperature.

Place at least three horizontal markers on the tube. Release the sphere centrally without pushing it. Put the first timing marker far enough below the surface for acceleration to have ended. Confirm terminal motion by showing that equal distances take equal times, or that speeds in consecutive marked regions agree.

Measure a marker separation $s$ with a metre rule and time the same point on the sphere crossing the two markers. Repeat and average, then calculate $v=s/t$. Reduce reaction-time error with a longer timed distance or calibrated video, while keeping the sphere in the terminal-speed region.

At terminal speed, use $W=U+6\pi\eta rv$, with $U=\rho_{liquid}V_{sphere}g$, and rearrange for $\eta$. Keep all quantities in SI units and compare repeats before quoting an appropriately precise result.

Hooke's law is $\Delta F=k\Delta x$: change in force is proportional to change in extension or compression while the response remains proportional. The stiffness $k=\Delta F/\Delta x$ is measured in $\mathrm{N\,m^{-1}}$.

Measure extension from the unloaded length, not the total length. On a force-against-extension graph, $k$ is the gradient of the straight proportional section. Rearrange to $\Delta x=\Delta F/k$ when stiffness and force change are known.

A spring length changes from $0.180\,\mathrm{m}$ to $0.230\,\mathrm{m}$ when force increases by $4.0\,\mathrm{N}$. Since $\Delta x=0.050\,\mathrm{m}$, $k=4.0/0.050=80\,\mathrm{N\,m^{-1}}$.

Stress measures force intensity; strain measures fractional length change. Young modulus compares them in the linear elastic region and measures material stiffness: a larger $E$ means more stress is required for the same strain.

Convert diameter to radius and calculate $A=\pi r^2$ for a circular wire. Use extension rather than final length in strain. Equivalently, substituting the definitions gives $E=FL/(A\Delta L)$.

In the initial straight region, gradient $\Delta F/\Delta x$ is stiffness. A curved graph beyond the proportional limit may still include some elastic behaviour, so the limit of proportionality and elastic limit are different ideas.

To decide whether deformation is elastic or plastic, consider what happens after the force is removed. Returning to zero deformation indicates elastic recovery; a non-zero intercept on the deformation axis indicates permanent change.

Plot tensile or compressive stress on the vertical axis against strain on the horizontal axis. Because both quantities account for sample dimensions, the graph describes material response more directly than a force-extension graph.

When comparing samples, use the same feature: gradient for stiffness, breaking stress for resistance to fracture by stress, and strain coordinate for deformation. Read values with axis multipliers and units before calculating.

Clamp a long, thin wire securely and measure its original test length $L$. Measure diameter with a micrometer at several positions and in different orientations, then average and calculate $A=\pi d^2/4$. A long, thin wire gives a larger measurable extension for a given load, reducing percentage uncertainty in extension.

Add known masses gradually, allow the wire to settle, and record extension $\Delta L$ from a fiducial marker or pointer. Calculate tensile force $F=mg$. Repeat readings where practical and keep loads within the linear elastic region so the wire returns to its original length.

Plot stress $F/A$ against strain $\Delta L/L$; the best-fit gradient of the straight region is $E$. Alternatively, plot $F$ against $\Delta L$: if its gradient is $k$, then $E=kL/A$. Use a large gradient triangle and propagate the measured units consistently.

Check micrometer zero error, avoid parallax at the extension scale, and use a tray or shield beneath suspended masses. Do not exceed a safe load or stand below the load. Diameter uncertainty matters strongly because area depends on $d^2$.

A small increase in extension requires work $F\,d x$. Therefore the elastic strain energy change is the area under the force-extension graph between the chosen extensions. Force in newtons multiplied by extension in metres gives joules.

For a linear graph starting at the origin, the area is a triangle: $\Delta E_{el}=\tfrac12F\Delta x$. For example, reaching $F=30\,\mathrm{N}$ at $\Delta x=0.080\,\mathrm{m}$ stores $0.5(30)(0.080)=1.2\,\mathrm{J}$.

For a non-linear graph, divide the region into narrow strips and add trapezium areas, or count grid squares with the axis scale applied. Narrower strips follow the curve more closely. State that the result is an estimate and retain sensible precision.