Errors in IB Physics forces and momentum questions usually come from the method rather than the final calculation. Students omit forces, use inconsistent directions, confuse Newton's third-law pairs, or apply momentum conservation without defining a suitable system. These problems are fixable by comparing your work with a step-by-step solution and identifying the first point where your reasoning diverged.
In the current course, A.2 Forces and momentum sits within Theme A, Space, time and motion. The official IB Physics subject brief identifies it as common SL and HL content. Questions can appear in Paper 1A multiple-choice, Paper 1B data-based work, and Paper 2 short-answer or extended-response tasks, so you need both conceptual accuracy and a reliable written method.
The most common mistakes and their fixes
| Common mistake | Why it loses marks | Practical fix |
|---|---|---|
| Treating every listed force as positive | Force is a vector, so direction determines sign | Choose positive axes before writing equations |
| Drawing forces that do not act on the selected body | The free-body diagram no longer represents one system | Isolate one object and draw only external forces on it |
| Assuming the normal force always equals weight | This is true only in particular situations | Resolve forces perpendicular to the surface and calculate the normal force |
| Treating Newton's third-law forces as balanced forces | Third-law partners act on different bodies | Name both interacting bodies for every action-reaction pair |
| Using momentum as a scalar | Opposite velocities then receive incorrect signs | Write velocity and momentum with direction or signed components |
| Assuming kinetic energy is conserved in every collision | Momentum and kinetic energy have different conservation conditions | Conserve momentum first, then test whether kinetic energy is conserved |
| Using force multiplied by time for a changing force without qualification | The equation requires an average force | Use average resultant force or the area under a force-time graph |
| Substituting immediately into formulas | The physical model remains unclear and sign errors are hidden | Draw, define, write the symbolic equation, and then substitute |
Mistake 1: Starting without a free-body diagram
A free-body diagram represents all external forces acting on one selected body or system. It should not contain velocity arrows, acceleration arrows, or forces exerted by the selected body on something else. Common forces include weight, normal contact force, tension, friction, drag, buoyancy and applied forces.
Suppose a block slides down a rough slope of angle θ. Its weight is vertically downward, not parallel to the slope. Resolve it into mg sin θ down the slope and mg cos θ perpendicular to the slope; if there is no perpendicular acceleration and no other perpendicular force, the normal force is mg cos θ rather than mg.
Fix: Pause before using F = ma. Identify the object, choose axes that simplify the motion, draw every external force, and resolve only the forces that are not already parallel to those axes. The RevisionDojo A.2 Forces and Momentum notes can be used to check the force types and Newton's-law relationships before attempting questions.
Mistake 2: Confusing resultant force with an individual force
Newton's second law concerns the resultant force:
ΣF = ma for constant mass, or more generally F = Δp/Δt.
If a car experiences a 2400 N driving force and 700 N of resistance, the force producing its acceleration is 1700 N, not 2400 N. Similarly, zero resultant force means zero acceleration, not zero velocity. An object can move at constant non-zero velocity while the forces on it are balanced.
Fix: Write a component equation such as ΣFx = driving force − resistance = ma. This makes it clear which forces are being combined and why one has a negative sign. It also prevents the common error of setting every force separately equal to ma.
Mistake 3: Mishandling Newton's third law
Newton's third law states that interacting bodies exert forces on each other that are equal in magnitude and opposite in direction. These forces act on different bodies, so they do not cancel on one free-body diagram.
For a book resting on a table, the upward normal force on the book and the downward weight of the book may be equal, but they are not a third-law pair. Both act on the book. The partner to the table's normal force on the book is the book's normal force on the table, while the partner to Earth's gravitational force on the book is the book's gravitational force on Earth.
Fix: Name each force in the form “force exerted by A on B.” A valid third-law pair reverses A and B, has the same force type, and acts on separate objects.
Mistake 4: Treating friction as automatically equal to μN
Static friction adjusts to the value needed to prevent relative motion, up to a maximum. It is therefore represented by Ff ≤ μsN. Dynamic friction for a sliding surface is modelled as Ff = μdN in the standard model.
A stationary block on a shallow slope does not necessarily experience the maximum possible static friction. If equilibrium requires mg sin θ, the friction force has that value, provided it does not exceed μsN.
Fix: First determine whether the surfaces are stationary relative to each other or sliding. For static equilibrium, calculate the friction required and then compare it with the limiting value. Do not insert μN automatically merely because a coefficient is supplied.
Mistake 5: Ignoring direction in momentum calculations
Linear momentum is the vector p = mv. A consistent positive direction is therefore essential in collisions, explosions and rebound questions.
Consider a 0.20 kg ball moving right at 6.0 m s⁻¹ before rebounding left at 4.0 m s⁻¹. Taking right as positive gives:
- Initial momentum = 0.20 × 6.0 = +1.2 kg m s⁻¹
- Final momentum = 0.20 × (−4.0) = −0.80 kg m s⁻¹
- Change in momentum = −0.80 − 1.2 = −2.0 kg m s⁻¹
Using speed alone would incorrectly produce a much smaller change. The negative answer shows that the impulse on the ball is directed left.
Fix: Draw separate before-and-after diagrams, mark the positive direction, and assign a sign to every velocity before calculating momentum. The RevisionDojo linear momentum and impulse notes provide further worked examples of this setup.
Mistake 6: Conserving momentum without defining the system
Total momentum remains constant when the resultant external impulse on the chosen system is zero or negligible during the interaction. Internal forces between bodies occur in equal and opposite pairs and do not change the system's total momentum.
In a short collision between two carts, friction may be negligible over the collision time, making the two-cart system approximately isolated. For one cart alone, however, the force from the other cart is external, so that cart's individual momentum changes.
Fix: State the system explicitly and ask which forces are external to it. Then write the complete vector equation Σp before = Σp after. Avoid unexplained statements such as “momentum is always conserved,” because the conservation claim depends on the selected system and external impulse.
Mistake 7: Confusing momentum conservation with kinetic-energy conservation
Momentum is conserved in an isolated system during both elastic and inelastic collisions. Kinetic energy is conserved only in an elastic collision. In an inelastic collision, some kinetic energy is transferred into internal energy, deformation, sound or other forms.
| Interaction | Momentum of isolated system | Total kinetic energy |
|---|---|---|
| Elastic collision | Conserved | Conserved |
| Inelastic collision | Conserved | Not conserved |
| Perfectly inelastic collision | Conserved | Not conserved; objects stick together |
| Explosion in an isolated system | Conserved | May increase as stored energy is released |
Fix: Use momentum conservation independently of the energy classification. Only write a kinetic-energy conservation equation when the question states or establishes that the collision is elastic. If asked to determine the collision type, calculate total kinetic energy before and after rather than relying on whether the objects separate.
Mistake 8: Misreading impulse and force-time graphs
Impulse is the change in momentum:
J = Δp = Favg Δt.
For a variable force, impulse is the signed area under the force-time graph, not necessarily the peak force multiplied by the total time. For a triangular pulse, for example, the impulse is ½ × base × height. Areas below the time axis represent impulse in the negative direction.
Fix: Identify the graph's geometric sections, calculate each signed area, and add them. If average force is required, divide the total impulse by the contact time. Keep N s and kg m s⁻¹ available as equivalent impulse units.
How to review worked video solutions effectively
Watching a solution passively rarely changes exam performance. Use a question from the A.2 Forces and Momentum Questionbank, attempt it without support, and then compare your method with the available step-by-step explanation or video solution. For broader demonstrations, use the IB Physics video library and the focused linear momentum and impulse videos.
When reviewing a worked or past-paper video solution, record four things:
- Setup: What system, axes and positive direction did the solution choose?
- Principle: Was the key idea Newton's second law, impulse, momentum conservation or energy?
- Equation: What symbolic equation appeared before numerical substitution?
- Communication: How were units, vector direction and assumptions stated?
Stop the video before each major step and predict what should happen next. If your prediction differs, identify the first difference rather than copying the final answer. Jojo AI can then help explain why a force, sign or conservation equation was inappropriate, but you should rewrite the complete solution yourself afterwards.
For past papers, use copies supplied legally by your school or another licensed source. Pair each attempted question with RevisionDojo's per-question worked or video solutions where available, then practise a similar question from the Questionbank. The official IB specimen papers for first examinations in 2025 are also useful for understanding the current question formats.
A reliable exam method
Use the following sequence whenever a question involves forces, collisions or impulse:
- Define the body or system.
- Draw the situation, including before-and-after diagrams for collisions.
- Choose axes and a positive direction.
- Draw a free-body diagram if forces are involved.
- Select the governing principle, rather than searching randomly for a formula.
- Write a symbolic vector or component equation.
- Substitute values with consistent SI units.
- Check sign, direction, units and physical plausibility.
The RevisionDojo A.2 topic hub combines notes, lessons, flashcards and exam-style practice. Use it to alternate between concept review and calculation rather than completing large numbers of similar questions without analysing mistakes.
Conclusion
The most damaging forces and momentum errors are systematic: incomplete diagrams, incorrect force pairs, unjustified friction formulas, inconsistent momentum signs, undefined systems and confusion between momentum and kinetic energy. A structured method fixes these weaknesses because it makes the physics visible before the arithmetic begins.
After each practice set, classify errors by cause and review the corresponding worked method. RevisionDojo's A.2 Questionbank, focused notes, Jojo AI feedback and past-paper video solutions where available are most useful when you pause, predict each step and then redo the question independently.
Sources and referenced URLs
- Official IB Physics subject brief
- Official IB Physics specimen papers for first examinations in 2025
- RevisionDojo IB Physics A.2 topic hub
- RevisionDojo A.2 Forces and Momentum notes
- RevisionDojo A.2 Forces and Momentum Questionbank
- RevisionDojo linear momentum and impulse notes
- RevisionDojo linear momentum and impulse videos
- RevisionDojo IB Physics video library
