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Grade 11 Physics

Grade 11 Physics on Temari has 8 revision cards, arranged by the chapters of the Ethiopian national curriculum. Every card says when the rule applies, what each symbol in it stands for, and the mistake students most often make with it. They are free to read and need no account.

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6 September 2026
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Across the whole subject

EquationThe quantity it leaves outReach for it when
v=u+atv = u + atssno distance is given or wanted
s=ut+12at2s = ut + \tfrac{1}{2}at^{2}vvthe final velocity is unknown
v2=u2+2asv^{2} = u^{2} + 2asttno time is given or wanted
s=(u+v2)ts = \left(\dfrac{u+v}{2}\right)taathe acceleration is unknown

When you use it

Use for any body moving in a straight line with constant acceleration. List what the question gives you, spot the quantity it never mentions, and take the equation that leaves that quantity out.

Watch out

All four hold only while the acceleration stays constant. A body on a curve, or one whose acceleration changes, needs calculus instead. Take displacement, velocity and acceleration as positive in one direction and negative in the other, or a body thrown upwards gives an answer with the wrong sign.

What you are findingFormulaConditions it holds under
Net force on a massFnet=maF_{net} = mamass constant, FF the resultant
Force from momentumFnet=ΔpΔtF_{net} = \dfrac{\Delta p}{\Delta t}holds even when mass changes
Action and reactionF12=F21F_{12} = -F_{21}always, on two different bodies
Linear momentump=mvp = mva vector, along the velocity
ImpulseJ=FΔt=ΔpJ = F\,\Delta t = \Delta pconstant force over Δt\Delta t
WeightW=mgW = mggg is local, not a constant of nature

When you use it

Use when a question asks what a force does to a body: how fast it speeds up, how hard two bodies push on each other, or what a collision leaves behind. Momentum is conserved in every collision; kinetic energy is conserved only in an elastic one.

Watch out

The two forces of the third law act on two different bodies, so they never cancel each other out. Students subtract them from one another and get zero acceleration for a book resting on a table, when the pair that actually balances is the table's push and the Earth's pull, both acting on the book.

What you are findingFormulaConditions it holds under
Work by a constant forceW=FscosθW = Fs\cos\thetaθ\theta between force and displacement
Kinetic energyKE=12mv2KE = \tfrac{1}{2}mv^{2}speeds well below light speed
Gravitational potential energyPE=mghPE = mghnear the ground, gg uniform
Work and energy theoremWnet=ΔKEW_{net} = \Delta KEalways, for any net force
Average powerP=WtP = \dfrac{W}{t}work done over a whole interval
Instantaneous powerP=FvcosθP = Fv\cos\thetaat one moment
Efficiencyη=useful outputtotal input\eta = \dfrac{\text{useful output}}{\text{total input}}never greater than 1

When you use it

Use when a question gives you distances and speeds but no time, or asks how much fuel, food or electricity a job costs. The work and energy theorem is often a shortcut past the equations of motion, because it never asks how long anything took.

Watch out

A force at right angles to the motion does no work at all, so the tension in a string swinging a stone in a circle adds no energy however fast it goes. The cosine is what carries this, and dropping it is the commonest lost mark in the topic.

Straight lineTurningWhat changes
ssθ\thetametres become radians
v=stv = \dfrac{s}{t}ω=θt\omega = \dfrac{\theta}{t}v=rωv = r\omega
aaα\alphaa=rαa = r\alpha
mmI=mr2I = \sum mr^{2}mass becomes moment of inertia
F=maF = maτ=Iα\tau = I\alphaforce becomes torque
p=mvp = mvL=IωL = I\omegamomentum becomes angular momentum
KE=12mv2KE = \tfrac{1}{2}mv^{2}KE=12Iω2KE = \tfrac{1}{2}I\omega^{2}the same energy, counted round an axis

When you use it

Use when a wheel, a disc or a rod turns about a fixed axis. Every equation you already know for straight line motion has a turning twin, so solve the rotating problem by writing the straight line one and swapping each quantity for its partner in this table.

Watch out

The angle must be in radians before any of the links between the two columns holds. A question given in degrees or in revolutions per minute has to be converted first, and forgetting that is what makes an answer come out a factor of 57 too large.

BodyAxisMoment of inertia
Thin rod, length LLthrough the centre, across the rod112mL2\tfrac{1}{12}mL^{2}
Thin rod, length LLthrough one end, across the rod13mL2\tfrac{1}{3}mL^{2}
Hoop or thin ring, radius rrthrough the centre, along the axismr2mr^{2}
Solid disc or cylinder, radius rrthrough the centre, along the axis12mr2\tfrac{1}{2}mr^{2}
Solid sphere, radius rrthrough a diameter25mr2\tfrac{2}{5}mr^{2}
Hollow sphere, radius rrthrough a diameter23mr2\tfrac{2}{3}mr^{2}

When you use it

Use whenever a torque question needs a number for I, or when two bodies race down a slope and you have to say which reaches the bottom first. The smaller the fraction, the closer the mass sits to the axis and the easier the body is to spin.

Watch out

Each of these belongs to one axis only. The same rod is three times harder to swing about its end than about its middle, and reading the wrong row is the usual error. For any other axis parallel to one of these, add md2md^{2}, where d is the distance between the two axes.

F=0andτ=0\sum F = 0 \quad \text{and} \quad \sum \tau = 0
F\sum F
the sum of every force on the body, taken direction by directionN\mathrm{N}
τ\sum \tau
the sum of every turning effect about any one chosen pointNm\mathrm{N\,m}

When you use it

Use for a ladder against a wall, a beam on two supports, a signboard on a bracket: anything at rest or moving steadily. Both conditions must hold at once, which is what lets you find two unknown forces from one diagram.

Watch out

The first condition alone is not enough. Two equal and opposite forces applied at different points sum to zero and still spin the body, which is why a steering wheel turns. Take moments about the point where an unknown force acts and that force drops out of the equation, leaving one unknown to solve for.

What you are findingFormulaConditions it holds under
PressureP=FAP = \dfrac{F}{A}force at right angles to the area
Pressure at a depthP=ρghP = \rho g hgauge pressure, fluid at rest
UpthrustFb=ρfgVdispF_{b} = \rho_{f}\,g\,V_{disp}ρf\rho_{f} is the FLUID's density
Flow rate stays equalA1v1=A2v2A_{1}v_{1} = A_{2}v_{2}one pipe, nothing leaking out
BernoulliP+12ρv2+ρgh=constantP + \tfrac{1}{2}\rho v^{2} + \rho g h = \text{constant}steady flow, no viscosity

When you use it

Use for anything floating, sinking, or flowing through a pipe: a dam wall, a hydraulic jack, a boat, water speeding up through a narrow section.

Watch out

The density in the upthrust equation is the fluid's, not the object's, and the volume is the volume pushed aside rather than the whole object. A block half under water displaces half its own volume. Depth pressure also depends on depth alone, so a narrow tube and a wide lake at the same depth read the same.

What you are findingFormulaConditions it holds under
Stressσ=FA\sigma = \dfrac{F}{A}force spread over the cross section
Strainϵ=ΔLL0\epsilon = \dfrac{\Delta L}{L_{0}}a ratio, so it has no unit
Young's modulusE=σϵE = \dfrac{\sigma}{\epsilon}below the elastic limit only
Extension of a wireΔL=FL0AE\Delta L = \dfrac{FL_{0}}{AE}same conditions as above
Energy storedU=12FΔLU = \tfrac{1}{2}F\,\Delta Lthe work the stretch cost

When you use it

Use when a wire, cable or beam is stretched and the question asks how far it gives, how much load it will take, or which of two materials is stiffer.

Watch out

Young's modulus belongs to the material and not to the piece: a thick steel cable and a thin steel wire share one value of E, and it is the area in the equation that makes the thick one stretch less. Past the elastic limit the ratio stops being constant and none of these rows holds any more.

The same subject in other years

An exam paper keeps asking for what the year below taught. Those cards are here too.

Questions students ask

Is there a national exam in Grade 11?
No. Ethiopia sets national exams in Grade 6, Grade 8 and Grade 12 only. These cards are for your school's own exams, and for the national exam that comes a few years later.
Where do these cards come from?
They are written against the Ministry of Education textbook for Grade 11 Physics, and every formula, constant and table row is checked again before a card goes up.
Is this free?
Yes. Every card here is free to read and the printable sheet is free to download. Neither needs an account.
When was this last checked?
6 September 2026. Cards arrive chapter by chapter, and the line under each one says when that card was last read through.

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