A-Level · 25 September 2026 · 6 min read
A-Level mechanics questions: choose the model before the equation
A long mechanics solution often goes wrong before the first substitution. The student chooses a familiar equation without deciding whether it fits the motion. Correct algebra then carries an unsuitable model all the way to the answer.
These original practice examples focus on that choice. The aim is to explain why a method applies, how to keep signs consistent, and what would invalidate the calculation. This is especially useful when moving from isolated exercises to International AS and A-Level questions that combine several ideas.
Choose the system and account for external forces
Imagine two trolleys connected by a light string on a horizontal track. If you treat both trolleys as one system, the string tension is internal and cancels when their equations are added. If you examine one trolley alone, that same tension is an external force on the chosen object and belongs on its free-body diagram. The physics has not changed; the system boundary has.
For every force, identify what exerts it and on which object. Weight and the normal reaction are not a Newton's-third-law pair because they act on the same object. Their magnitudes may happen to be equal in a particular situation, but equality is a conclusion from vertical equilibrium, not a universal rule about contact forces.
Use a lift to separate velocity from acceleration
A 60 kg person stands on a scale in a lift accelerating upwards at 1.2 m s⁻². Taking upwards as positive, N − mg = ma. With g = 9.81 m s⁻², the scale's normal force is N = 60(9.81 + 1.2) = 660.6 N, or 661 N to three significant figures. The scale force exceeds the person's weight because the resultant force is upwards.
Now suppose the lift is travelling upwards but slowing down. Its acceleration is downwards, so N is less than mg. The direction of travel alone does not determine the force balance. Label acceleration separately on your sketch; this prevents confusing 'moving upwards' with 'accelerating upwards' when a question describes braking or stopping.
Use energy when comparing starting and finishing states
A 0.50 kg block slides from rest down a smooth slope through a vertical height of 0.80 m. Taking gravitational potential energy lost to equal kinetic energy gained gives mgh = ½mv², so v = √(2gh) = 3.96 m s⁻¹. The mass cancels, and the slope length is unnecessary because the energy change depends on vertical height.
The result assumes no energy is transferred into heating or other stores. On a rough slope, subtract the work done against friction before calculating kinetic energy. If friction is constant, that transfer is frictional force multiplied by distance along the slope, not by vertical height. Writing a short energy-accounting sentence helps keep these different distances in the right places.
Test constant acceleration before using SUVAT
A car slowing from 20 m s⁻¹ at a constant acceleration of −4.0 m s⁻² stops after 5.0 s. Using v² = u² + 2as gives a stopping distance of 50 m. Both answers follow from the stated constant-acceleration model; neither includes the distance travelled during a driver's reaction time before braking begins.
A falling object experiencing increasing air resistance generally has changing acceleration. Applying one constant value over the whole fall would need justification as an approximation. A velocity–time graph may instead allow displacement to be found from the area and instantaneous acceleration from a tangent. Recognising when a familiar equation is inappropriate is as valuable as rearranging it correctly.
Conserve momentum across a suitable interaction
A 0.40 kg trolley moving at 3.0 m s⁻¹ collides with a stationary 0.60 kg trolley and they stick together. If external impulse is negligible during the collision, momentum conservation gives 0.40 × 3.0 = (0.40 + 0.60)v, so v = 1.2 m s⁻¹ in the original direction. Treating the two trolleys together removes the internal collision forces from the momentum balance.
The initial kinetic energy is 1.8 J; the final kinetic energy is 0.72 J. The difference, 1.08 J, is transferred into other forms such as deformation and thermal energy. Momentum conservation does not imply kinetic-energy conservation. State the short interaction interval and negligible external impulse, rather than claiming that no forces act during the collision.
Practise switching models within one question
After the collision above, suppose the joined trolleys travel onto a rough horizontal section with a constant resisting force of 0.30 N. Find their stopping distance. Their combined mass is 1.0 kg, so acceleration is −0.30 m s⁻². From 0 = 1.2² + 2(−0.30)s, the distance is 2.4 m. Equivalently, 0.30s = 0.72 J gives the same result through work and energy.
Notice the model change: momentum was useful across the brief collision, while the external resisting force matters during the subsequent stopping phase. Do not conserve momentum over that whole phase. When revising, mark the boundary between events and write one governing principle for each. That habit turns a complicated story into a sequence of justified calculations.
Questions, explained
Choose a question for a direct answer, then explore the explanation and supporting resources. Each answer has its own link to save or share.
How do I choose between energy and SUVAT in mechanics?
Use energy when the question connects starting and finishing states through work or energy transfers, especially when time is not needed. SUVAT is useful for motion with constant acceleration when displacement, time or velocity is involved. Sometimes both work: a consistent alternative solution is a useful check. First test the assumptions, because a familiar set of supplied quantities does not guarantee constant acceleration.
How can momentum be conserved when large collision forces act?
The collision forces between the objects are internal to a system containing both objects. They transfer momentum between those objects but cancel in the total system balance. Total momentum remains constant when the net external impulse during the interval is negligible. The claim concerns the combined system, not the momentum of each object separately.
Does negative acceleration always mean an object is slowing down?
No. A negative sign describes direction relative to the chosen positive axis. If velocity is also negative, negative acceleration increases the speed. Slowing occurs when acceleration is opposite to velocity. A sketch with separate velocity and acceleration arrows often resolves this before any calculation, particularly for lifts, braking vehicles and objects thrown vertically upwards.