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Two supports do not necessarily carry equal loads. Their forces depend on where the load acts. A moment equation supplies information that vertical force balance alone cannot provide.

Need help choosing a next step? Scan the contact QR code below or contact Hanlin on WeChat. Send BPHO MOMENTS 1020 and tell us your school year and question.

Define positions and assumptions

A light horizontal beam is supported at its ends, 1.0 m apart. A downward 20 N load acts 0.40 m from the left support. Neglect the beam's own weight and assume static equilibrium.

Let the upward reactions be L and R. Vertical balance gives L + R = 20 N. This is one equation for two unknowns.

Take moments about a convenient point

About the left support, L has zero lever arm. Balancing moments gives R(1.0) = 20(0.40), so R = 8 N. Therefore L = 12 N.

The nearer left support carries more. The lever arm is the perpendicular distance from the pivot to the force's line of action, not automatically any distance marked on a diagram.

Check with a different pivot

About the right support, L(1.0) = 20(0.60), again giving 12 N. Both equations agree. Include the beam's weight at its centre of mass if it is no longer negligible.

A load outside the support interval may require a different contact or attachment analysis; do not assume every calculated negative reaction is physically available. This original problem develops statics foundations rather than reproducing an official Olympiad question.

Continue reading: British Physics Olympiad Mechanics: Momentum vs Kinetic Energy.

Official information and further reading: Source 1

For an object floating at rest, buoyancy balances its weight. The displaced liquid volume need not equal the object's entire volume. Keeping these two volumes separate is the central step in this original foundation exercise.

Need help choosing a next step? Scan the contact QR code below or contact Hanlin on WeChat. Send BPHO FLOATING 1019 and tell us your school year and question.

Define the model

A uniform object has volume 0.003 m³ and density 800 kg/m³. It floats in water of density 1000 kg/m³. Ignore air buoyancy and surface-tension effects. Take g = 9.8 m/s².

The object has mass 2.4 kg and weight 23.52 N. At rest, the water must provide the same upward force.

Solve for displaced volume

Using buoyancy = water density × g × displaced volume gives a displaced volume of 0.0024 m³. Its fraction of the whole volume is 0.0024/0.003 = 0.80.

Thus 80% of the object's volume is submerged under these assumptions. For a nonuniform shape, this is a volume fraction, not automatically 80% of its height.

Check the fully submerged comparison

If held completely underwater, the object would displace 0.003 m³ and experience buoyancy of 29.4 N. This exceeds its weight, so an additional downward force would be needed to hold it still in this ideal model.

Do not use the full volume in the original floating calculation and then claim the forces balance. Draw the actual state and identify the displaced volume first. This is a teaching problem, not an official Olympiad question or an experimental instruction.

Continue reading: British Physics Olympiad Mechanics: Momentum vs Kinetic Energy.

Official information and further reading: Source 1

A source with internal resistance does not necessarily deliver its full EMF across an external load. Draw the whole model, including the internal resistance, before applying a circuit equation. That makes the voltage and energy accounting much clearer.

Need help choosing a next step? Scan the contact QR code below or contact Hanlin on WeChat. Send BPHO CIRCUITS 1018 and tell us your school year and question.

State the model and find the current

Represent a source by an ideal EMF of 12 V in series with internal resistance r = 1 ohm. Connect an external resistor R = 5 ohms, assuming constant resistances and ideal connecting wires.

The total resistance is 6 ohms, so I = 12/6 = 2 A. The terminal voltage across the load is IR = 10 V, also given by EMF - Ir = 12 - 2 = 10 V.

Follow the power

The ideal source supplies power EI = 12 × 2 = 24 W. The external resistor receives I²R = 20 W, while the internal resistance dissipates I²r = 4 W.

The powers add to 24 W. Counting 12 V across the external resistor would overstate its power and fail this conservation check. EMF is energy supplied per unit charge; despite its name, it is measured in volts, not newtons.

Change the external resistance

If R becomes 11 ohms, the total resistance is 12 ohms and the current is 1 A. The terminal voltage is now 11 V. The smaller internal voltage drop explains why terminal voltage is closer to the EMF.

The ideal source power is 12 W, divided into 11 W in the load and 1 W internally. A larger terminal voltage does not by itself imply a larger load power, because current also changes.

Keep the approximation visible

Real sources may not have constant EMF and internal resistance under all conditions. This simple model is useful within its stated assumptions, not a complete description of every battery.

This is an original calculation exercise for physics preparation, not an experimental instruction or official Olympiad question. Label the source, load and internal loss separately when reviewing your own circuit work.

Continue reading: British Physics Olympiad Mechanics: Momentum vs Kinetic Energy.

Official information and further reading: Source 1

Wave questions often contain three related quantities that answer different questions. Frequency counts cycles per second, period measures time per cycle, and wavelength measures the spatial separation of matching phases. Begin by identifying which quantity the problem gives.

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Translate the description into quantities

Consider an ideal periodic travelling wave with frequency f = 50 Hz and speed v = 120 m/s in a uniform medium. A fixed point experiences 50 complete cycles each second. The time for one cycle is T = 1/f = 0.020 s.

The wavelength is given by v = fλ, so λ = 120/50 = 2.4 m. The period is a time and the wavelength is a distance. Giving both answers in metres would reveal a conceptual error.

Use the units as a check

A hertz is one cycle per second. Dividing metres per second by cycles per second leaves metres per cycle, which corresponds to a wavelength. Multiplying frequency by period gives one: fT = 1.

The numbers also tell a consistent story. During one period, the wave pattern moves vT = 120 × 0.020 = 2.4 m, exactly one wavelength.

Reason about a change of medium

For transmission across a stationary boundary driven by the same source, the frequency remains the source frequency. If the transmitted wave speed is 80 m/s, its wavelength is 80/50 = 1.6 m. Its period remains 0.020 s.

This example assumes a suitable transmitted periodic wave and a stationary boundary. It does not determine the amplitude, reflected fraction or energy transfer from wave speed alone.

Avoid a common shortcut

It is tempting to say that a faster wave must have a higher frequency. That conclusion requires an additional condition, such as a fixed wavelength. With a fixed source frequency, a higher speed instead means a longer wavelength.

For each practice problem, write the fixed quantity before applying the equation. Then check whether the answer describes a time, distance or speed. This is an original foundation exercise, not an official Olympiad question; current activity details should be checked through the relevant organisers.

Continue reading: British Physics Olympiad Mechanics: Momentum vs Kinetic Energy.

Official information and further reading: Source 1

Spring problems are useful for British Physics Olympiad preparation because they connect a force model with an energy model. Before using conservation of mechanical energy, state what belongs to your system and which interactions could transfer energy out.

This is an original introductory exercise, not a released BPhO question.

Need help choosing a next step? Scan the contact QR code below or contact Hanlin on WeChat. Send BPHO SPRINGS and tell us your school year and question.

Specify the initial and final states

A 0.50 kg block is released from rest by an ideal spring of constant 200 N/m, initially compressed by 0.10 m. The track is horizontal and frictionless. Find the speed when the spring first reaches its natural length. Neglect the spring's mass.

Choose the block and spring as the system. Height does not change, and the stated model allows stored elastic energy to become kinetic energy.

Calculate the energy before the speed

Initial elastic energy is kx²/2 = 200 x 0.10²/2 = 1.0 J. Thus mv²/2 = 1.0 J, giving v² = 4.0 m²/s² and v = 2.0 m/s.

The compression is in metres. The energy depends on its square, so a unit error in compression has a particularly large effect.

Change one assumption

If 0.25 J is transferred into thermal energy before the same final position, the block has only 0.75 J of kinetic energy. Its speed is then sqrt(3) m/s, approximately 1.73 m/s.

Do not keep using the frictionless equation after adding a loss. Conversely, do not invent a loss simply because real springs are imperfect; use the model specified in the question.

Make a prediction before recalculating

With the original assumptions, doubling compression gives four times the energy and twice the speed. Doubling mass at fixed compression reduces speed by a factor of sqrt(2).

Write the initial energy, final energy and any transfer in separate lines during review. Then explain why gravitational energy cancels here. A method becomes transferable when you can identify the condition that would make it fail.

Continue reading: British Physics Olympiad Mechanics: Momentum vs Kinetic Energy.

Official information and further reading: Source 1 Source 2

In a collision problem, can you use both conservation of momentum and conservation of kinetic energy? Before writing equations, check the system and the collision model. Those two decisions determine which relationships you can use.

This British Physics Olympiad preparation guide uses an original two-cart exercise to practise that choice. It is suitable for students who have studied momentum and kinetic energy and want to explain their mechanics reasoning more clearly.

1. Check the conditions before choosing a conservation law

The total momentum of a chosen system is conserved when the net external impulse on it is zero. For a short collision on a horizontal track, we may model the external horizontal impulse as negligible and conserve horizontal momentum.

Kinetic energy needs a separate check. In an elastic collision, the system's total kinetic energy is conserved. In a perfectly inelastic collision, the objects stick together and generally have less kinetic energy afterwards. A collision in which objects bounce apart is not automatically elastic.

For further study of these definitions, see OpenStax's chapter on types of collisions.

2. Original exercise: two carts stick together

Cart A has mass 2 kg and moves to the right at 3 m/s. Cart B has mass 1 kg and is initially stationary ahead of A. They collide on a horizontal track and stick together.

Model the external horizontal impulse during the collision as zero. Treat the motion as one-dimensional and consider the carts' translational kinetic energy. Find their shared velocity immediately after the collision, then compare the total kinetic energy before and after.

Before reading the solution, write down your system, positive direction and reason for choosing a conservation law. This exercise is a concept check, not a prediction of a competition question or an official difficulty benchmark.

3. Define the system and keep the signs

Choose both carts together as the system and take right as positive. The large forces the carts exert on each other are internal to this system. Their horizontal impulses cancel in the total momentum balance.

Cart A alone does not conserve its momentum during the collision: Cart B exerts a force on it. Choosing the system is therefore part of the reasoning, not just a label on the diagram.

Because the carts stick together, both have the same final velocity v. Conservation of horizontal momentum gives:

2 × 3 + 1 × 0 = (2 + 1)v
6 = 3v
v = 2 m/s to the right.

The total momentum is 6 kg m/s before and after. We did not need to assume that kinetic energy was conserved to find this velocity.

4. Calculate kinetic energy independently

Before the collision, only Cart A is moving:

K_before = ½ × 2 × 3² + ½ × 1 × 0² = 9 J.

After the collision, the combined 3 kg mass moves at 2 m/s:

K_after = ½ × 3 × 2² = 6 J.

The kinetic energy decreases by 3 J, or one-third of its initial value. Energy can be transferred into deformation, thermal energy and sound. This decrease does not mean total energy has been destroyed; kinetic energy is only one form of energy.

5. What changes if the collision is perfectly elastic?

Keep the same masses and initial velocities, but now specify a perfectly elastic collision. The carts do not stick together, so give them separate final velocities v_A and v_B. Both total momentum and total kinetic energy must match their initial values:

2v_A + v_B = 6
v_A² + ½v_B² = 9

Substitute v_B = 6 - 2v_A into the energy equation. This gives v_A² - 4v_A + 3 = 0. The physical post-collision solution is v_A = 1 m/s and v_B = 4 m/s, both to the right.

The other algebraic solution, 3 m/s and 0 m/s, simply reproduces the incoming velocities. It does not describe the carts separating after impact. Checking the physical situation helps select the relevant solution.

You can verify the elastic result: the momentum is 2 × 1 + 1 × 4 = 6 kg m/s, and the kinetic energy is ½ × 2 × 1² + ½ × 1 × 4² = 9 J.

6. Three mistakes to watch for

• Using kinetic energy conservation without checking the model. In the sticking exercise, it would give a speed of √6 m/s, which fails the momentum check.
• Using speed instead of signed velocity in momentum. In one dimension, leftward and rightward motion have opposite signs. Kinetic energy uses the squared speed and is non-negative.
• Conserving the momentum of one cart. The collision force from the other cart is external to a one-cart system. Use the two-cart system and check the net external impulse.

7. Try two changed situations

Question 1: Return to the sticking model. Cart A still moves right at 3 m/s, but the 1 kg Cart B now moves left at 3 m/s. What is the shared final velocity?

Answer: v = [2 × 3 + 1 × (-3)]/3 = 1 m/s to the right. Equal initial speeds in opposite directions do not cancel the momentum here because the masses differ. The initial kinetic energy is 13.5 J and the final kinetic energy is 1.5 J.

Question 2: Cart B is stationary again, but its mass is now 2 kg. The carts stick together. Find the final velocity and decrease in kinetic energy.

Answer: The shared velocity is 6/4 = 1.5 m/s to the right. The kinetic energy changes from 9 J to 4.5 J, a decrease of 4.5 J.

For each answer, check the momentum of both carts and explain whether the collision model permits kinetic energy conservation.

8. Use the method in your BPhO preparation

When reviewing a mechanics problem, record these five items:

1. The objects included in your system.
2. The positive direction and signed initial velocities.
3. The external forces or impulses relevant to the interval.
4. The reason each conservation equation is valid.
5. A check of units, signs and physical behaviour.

The Oxford Physics resource guide lists BPhO past papers and solutions among its problem-solving resources. Choose materials that match the specific challenge or round you are preparing for, and confirm current entry requirements with the organiser in your region.

For help choosing a starting point, read our BPhO beginner's preparation guide. To practise another topic, see our series, parallel and power exercise.

Discuss your British Physics Olympiad preparation with Hanlin

Considering a BPhO challenge, or already preparing and unsure how to improve your mechanics reasoning? Scan the QR code below to contact Hanlin on WeChat. Send BPHO MOMENTUM with your year or grade, current physics course and target challenge or round.

You can also share one practice problem and your own working to help focus the discussion. If your competition choice is not yet decided, tell us what you have studied and what you would like help with.

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When preparing for the British Physics Olympiad, knowing a circuit formula is only the first step. You also need to decide which voltage and current belong to each component. A useful starting point is to compare the same resistors in two arrangements while keeping the supply conditions explicit.

State the model before using the formulas

Consider an ideal 6 V battery and two identical 6 Ω resistors. Ignore wire resistance and assume each resistor remains at 6 Ω. The battery maintains a constant terminal voltage. These assumptions matter: a real battery with internal resistance can behave differently.

For each resistor, Ohm's law is V = IR. Its electrical power can be written as P = VI, I²R or V²/R when the relevant voltage and current refer to that resistor. This is an original learning example, not a BPhO paper question.

Two resistors in series

The total resistance is 6 + 6 = 12 Ω. The circuit current is 6/12 = 0.5 A. The same current passes through both resistors, so the voltage across each is 0.5 × 6 = 3 V.

Each resistor dissipates 3 × 0.5 = 1.5 W. Together they dissipate 3 W, matching the battery's output power of 6 × 0.5 = 3 W. This agreement checks that the component calculations and total calculation are consistent.

Two resistors in parallel

Each resistor is connected across the 6 V supply. Each branch current is 6/6 = 1 A, so the battery supplies 2 A in total. The equivalent resistance is 6/2 = 3 Ω.

Each resistor dissipates 6 × 1 = 6 W, giving 12 W in total. This also matches 6 V × 2 A. Under the stated model, the parallel arrangement draws four times the total power of the series arrangement.

Why “larger resistance means more power” can mislead

The formula P = I²R shows that power increases with resistance if the current is fixed. The formula P = V²/R shows that power decreases with resistance if the voltage is fixed. There is no contradiction; the quantity held constant is different.

In our comparison, the ideal battery voltage is fixed and the circuit current changes. You cannot treat the series current of 0.5 A as if it still flows through each parallel branch. Always establish the new circuit conditions before comparing power.

A short circuit review checklist

1. Identify which components are genuinely in series or parallel.
2. Calculate the equivalent resistance before the supply current.
3. Find each component's voltage and current separately.
4. Check that branch currents add correctly and voltage drops are consistent.
5. Compare total component power with supply power under the model.

To extend the example, replace the ideal battery with a battery that has internal resistance. Predict whether the parallel circuit still receives the full 6 V before doing any calculation. This helps you notice where an assumption, rather than an algebraic step, controls the answer.

Use BPhO's official resources for the particular competition you are preparing for, and practise writing the physical reason alongside the equation. Do not infer a whole competition's difficulty from this introductory example.

New to the subject? Read our British Physics Olympiad Preparation: A Beginner's Guide to BPhO.

Official resources

• British Physics Olympiad: official Round 1 information