IB Physics SL A.2 Forces and momentum Notes

Newton’s laws answer different questions. First law is about zero resultant force. Second law links non-zero resultant force to acceleration. Third law identifies paired forces in an interaction, but those paired forces never act on the same body.

State Newton’s third law and explain why the two forces in a third-law pair do not cancel each other.

Forces do not appear alone. A force on one body is exerted by another body. Good force labels say “force on object by agent”, such as weight on book by Earth or normal force on block by table. This prevents adding forces from the wrong system.

Explain why the phrase “the object has a force” is less precise than “the table exerts a normal force on the object”.

A clean free-body diagram starts by choosing the body. Then list every external force acting on that body and nothing else. Third-law partner forces are on other bodies, so they belong on different diagrams.

A box slides across a rough horizontal floor. List the real forces that should appear on the free-body diagram of the box.

After drawing the free-body diagram, choose axes and add force components. The resultant is not a new interaction; it is the net effect of all real forces on the object.

A 5.0 kg object has horizontal forces of 18 N to the right and 8 N to the left. Calculate the resultant force and acceleration.

Classifying forces helps choose directions in a free-body diagram. Some forces need contact and a surface direction; others are field forces and can act through space. The name of the force should come with a direction rule.

Classify weight, normal force, friction and magnetic force as contact or field forces, and state one direction rule.

Normal force FN acts perpendicular to the contact surface.

In an IB-style response, define the quantity and state whether direction or an interval matters.

First decide the motion state at the contact. If the surfaces are not sliding, the friction is static: it can take whatever value is needed for equilibrium, but only up to F_f,max = μ_s F_N. If the object is just about to slip, static friction is at this limiting value. Once the surfaces slide, use dynamic friction, F_f = μ_d F_N, directed opposite the relative motion. In both cases the size depends on the normal reaction F_N, not directly on mass unless F_N has first been found from the force balance.

A horizontal force is applied to a block on a rough surface. The block may remain at rest, be just about to slip, or slide. State which friction model applies and justify the equation or inequality used.

Start from the selected object, not from the string as a whole. If a taut string is attached to that object, draw the tension force along the string, pointing away from the object at the point of attachment. In IB force problems the phrase ideal string usually means massless and inextensible, and an ideal pulley is frictionless and massless; under those assumptions one continuous string has the same tension magnitude throughout. The pulley changes the direction of the tension force but not its magnitude. If the string is slack, cut, massive, or passes over a non-ideal pulley, that shortcut no longer automatically applies.

For a mass attached to a taut string, draw or describe the tension force on the mass and state when the same value of T can be used elsewhere in the system.

For a spring or elastic material in its Hooke’s-law region, the restoring force increases in direct proportion to the extension or compression from the natural length. In vector form this is F_H = -kx: x is measured from equilibrium and the negative sign shows that the force points back toward equilibrium. The magnitude is often written F = kx. Once the force-extension graph stops being a straight line through the origin, Hooke’s law no longer applies, so do not use one constant k beyond the limit of proportionality. In graph questions, check the axes before taking a gradient: force against extension gives k, while extension against force gives 1/k.

A spring is stretched and a force-extension graph is provided. Use Hooke’s law to find the spring constant or restoring force, and state the condition for using the law.

Use Stokes’ law only after checking the physical model: a small spherical object moving slowly through a viscous fluid with smooth/laminar flow. The drag force acts opposite the velocity relative to the fluid, so increasing speed increases the resistive force in this model. The symbols must be read carefully: η is viscosity in Pa s, r is radius in metres, and v is speed relative to the fluid. In falling-sphere questions, the drag force grows as the sphere speeds up until the forces balance at terminal velocity; at that point the acceleration is zero, not the speed.

A small sphere falls through a viscous fluid. Use Stokes’ law to calculate or explain the drag force, and state the condition for terminal velocity.

Buoyancy comes from the fluid pressure being larger at the bottom of the object than at the top, giving a net upward force. The useful IB calculation is F_b = ρ_fluid V_displaced g: multiply the density of the fluid by the volume of fluid displaced and by g. Be careful with the volume. A fully submerged sphere displaces its full volume, but a floating object displaces only the volume below the fluid surface. Do not automatically set buoyancy equal to weight; that equality is a force-balance condition for floating, hovering, or terminal-motion cases, not the definition of buoyancy.

An object floats or is submerged in a fluid. Determine the buoyant force or the submerged fraction using Archimedes’ principle.

Classify a force by the interaction mechanism, not by where the arrow happens to be drawn. Field forces do not require contact: Earth can pull a mass gravitationally, an electric field can push or pull a charge, and a magnetic field can exert a force on a current, magnet, or moving charge. Contact forces require touching surfaces or a material connection, such as normal reaction, friction, tension, elastic force, viscous drag, and buoyancy. On a free-body diagram you still draw only the force acting on the selected object; the source of the field is not drawn as an extra force on that same object.

Classify the forces acting on an object as contact or field forces, and identify the field-force family when relevant.

In A.2, weight is the gravitational field force on the selected object. Its magnitude is F_g = mg, where m is the object’s mass and g is the local gravitational field strength. This means a 5 kg object has mass 5 kg but weight about 49 N near Earth. On a free-body diagram, weight acts downward near Earth because the gravitational field points toward Earth’s centre. Do not replace every vertical force with mg: the normal reaction is a contact force from a surface and equals mg only under specific force-balance conditions, such as a stationary object on a horizontal surface with no other vertical forces.

Calculate or identify the weight of an object in a gravitational field and draw it correctly on a free-body diagram.

Treat electric force as a field-force option in A.2 force diagrams. First ask whether the selected object has charge, or whether the question explicitly says it is in an electric field. If the charge is positive, the electric force is along the electric field direction; if the charge is negative, the force is opposite the field. For pairs of charges, same signs repel and opposite signs attract. Avoid drawing an electric force just because electric field lines are shown somewhere else; it must act on the selected object.

A charged particle is shown in an electric field. State whether an electric force acts on it and give the direction of the force.

In A.2, the main skill is recognizing magnetic force as one of the field-force families. It belongs on the selected object only when the object is a magnet, carries current, or is a charge moving through a magnetic field. A charge at rest in a magnetic field does not feel a magnetic force just because the field is present. When the object is a moving charge or current-carrying wire, the force direction is perpendicular to both the magnetic field and the motion/current direction; detailed hand-rule calculations can be handled later, but the perpendicular nature is important for force diagrams.

A particle, wire, or magnet is placed in a magnetic field. State whether a magnetic force acts and identify the basic direction condition.

Momentum conservation is not a magic rule for every collision. First draw the boundary around the system you are analysing. Forces between objects inside that boundary are internal and cancel in pairs for the total system momentum. Forces from outside the boundary are external; if their resultant is zero or negligible during the interaction, then total momentum before equals total momentum after. If an external impulse such as friction, a wall, a hand push, or gravity on a non-isolated vertical system is significant, the system momentum changes in that direction.

For a collision or explosion, define the system and state whether momentum is conserved, using the resultant external force condition.

Impulse connects forces during an interaction to the change in momentum they produce. If the resultant external force is constant, use J = FΔt; more generally, use average resultant force, J = F_avg Δt. If a force-time graph is given, the impulse is the area under the graph, including sign if force direction is shown. Because J = Δp, impulse has units N s, which are equivalent to kg m s^-1. The direction of impulse is not optional: it is the direction in which the object’s momentum changes.

A force-time graph is given for an interaction. Determine the impulse and explain how it changes the object’s momentum.

This card turns the impulse idea into a calculation routine. Choose a positive direction, make initial and final velocities signed quantities, then calculate p_initial = mu and p_final = mv. The momentum change is Δp = p_final - p_initial = m(v - u), so the impulse has the same value and sign. If the time interval is known, divide by time to find the average resultant force. The sign of the answer tells you direction; do not remove it just because the magnitude is positive.

An object changes velocity during a collision. Calculate the impulse and average resultant force using a stated sign convention.

The momentum form is the more general statement: the resultant external force on an object or system equals the rate at which its momentum changes. Over a finite interval this is F_avg = Δp/Δt. If the mass is constant, Δp = mΔv, so the equation reduces to F_resultant = ma. This is why F = ma is powerful but conditional: it uses resultant force, constant mass, and consistent vector directions. If a problem gives momentum change and time directly, the momentum-rate form is often cleaner than forcing an acceleration calculation.

A problem gives either acceleration or a momentum change over time. Choose the appropriate Newton’s second law form and justify it.

The word elastic is not about whether objects bounce visually; it is about kinetic energy. If the system has no resultant external impulse, total momentum is conserved for both elastic and inelastic collisions. The difference is kinetic energy: in an elastic collision, total kinetic energy before equals total kinetic energy after; in an inelastic collision, some kinetic energy is transferred into other forms such as deformation, thermal energy, or sound. If the objects stick together, the collision is perfectly inelastic.

A collision is described with before-and-after velocities. State whether it is elastic or inelastic and justify using momentum and kinetic energy.

An explosion is the reverse style of a collision: one system separates into parts because internal energy becomes kinetic energy. Momentum conservation still depends on the same condition: no significant resultant external impulse on the chosen system during the short event. Write the total momentum before, then the vector sum of fragment momenta after. If the original object was at rest, the final fragment momenta must add to zero, so the fragments move in opposite directions with equal magnitude momenta. Different masses then have different speeds.

An object explodes into two fragments. Use conservation of momentum to find an unknown fragment speed or direction.

Use momentum and kinetic energy differently. Momentum needs signs and directions; kinetic energy uses speed squared and is always non-negative. To track energy, calculate the total kinetic energy before and after: Σ(1/2mv^2). If the totals are equal, the collision is elastic. If the final kinetic energy is lower, the collision is inelastic and the difference has gone into internal energy stores such as deformation, heating, and sound. In an explosion, internal or chemical energy can become kinetic energy, so final kinetic energy may be greater than initial kinetic energy while momentum is still conserved for an isolated system.

Calculate total kinetic energy before and after a collision or explosion, then classify or explain the event.

An object moving around a circle at constant speed is still accelerating because velocity is a vector and its direction is changing. The acceleration needed for this direction change points toward the centre of the circle at every instant. The velocity arrow is tangent to the path, while the acceleration arrow is radial and inward. Use a = v^2/r when linear speed and radius are known, a = ω^2r when angular speed is known, or a = 4π^2r/T^2 when period is known.

For an object moving at constant speed in a circle, draw the velocity and acceleration directions and calculate the centripetal acceleration.

Centripetal force is a role played by the resultant force, not a new category of force. In a ball-on-string problem, tension may provide the inward resultant. In a car turning on a flat road, friction may provide it. In an orbit, gravity provides it. The free-body diagram should show real forces; then resolve them toward the centre and set the inward resultant equal to mv^2/r. If you draw both “tension” and a separate “centripetal force” for the same cause, you have double-counted.

Explore the force that keeps the object turning

For an object in circular motion, identify the real force providing the centripetal force and calculate the required inward resultant.

The phrase “constant speed” can hide the key vector idea. In circular motion the velocity arrow is always tangent to the path, so its direction changes as the object moves. The acceleration and resultant force point inward, perpendicular to the instantaneous velocity. Because the resultant force has no tangential component in uniform circular motion, it changes the direction of velocity without changing its magnitude. A tangential force component would change the speed.

Explain why an object moving at constant speed in a circle is accelerating, and draw the velocity and resultant-force directions.

Angular speed describes how quickly the angle changes; linear speed describes how quickly a point moves around the circumference. One full revolution is 2π radians, so ω = 2π/T or 2πf. A point at radius r travels one circumference, 2πr, every period T, giving v = 2πr/T. Combining these gives v = ωr. This means two points on the same spinning disc can share the same angular speed but have different linear speeds because their radii are different.

A point on a rotating object has a given period, frequency, angular speed, or radius. Calculate its angular or linear speed.

A.2 is a model-selection topic, not a formula list. First decide whether the question is about a single object’s resultant force, a system’s momentum, an interaction over time, a collision/explosion energy pathway, or circular motion. Then choose the model condition: real forces on a free-body diagram, no resultant external force for momentum conservation, graph area for impulse, kinetic-energy comparison for collision type, and inward resultant force for circular motion. The best answers state the physical condition before substituting numbers.

A mixed A.2 problem combines forces, momentum, and circular motion. State the model choice and condition before starting calculations.