Revise Forces and momentum for IB Physics (A.2)

Newton's laws are the foundation of the whole topic. The first law states that an object stays at rest or moves at constant velocity unless a resultant external force acts on it; this is the principle of inertia, and it tells you that constant velocity (including rest) always means the forces are balanced. The second law states that the resultant force equals the rate of change of momentum, $F = \Delta p / \Delta t$, which for constant mass becomes $F = ma$; force and acceleration are vectors pointing the same way. The third law states that if body A pushes on body B, then B pushes back on A with an equal and opposite force — the two forces are the same type, equal in magnitude, opposite in direction, and act on different bodies.

A free-body diagram represents a single chosen object as a dot or box, with one labelled arrow for every force acting ON it — and nothing else. Getting this diagram right is the most reliable way to score method marks, because every later equation reads off it. The forces you meet are of two kinds. Field forces act at a distance: weight $W = mg$ (always vertically down). Contact forces act where surfaces or media touch: the normal force $N$ (perpendicular to a surface), tension $T$ (a pull along a string, the same magnitude throughout an ideal light string), friction (along the surface, opposing relative motion or its tendency), drag (a resistive fluid force opposite to motion that grows with speed), buoyancy or upthrust (the upward force of a fluid on a submerged body), and the spring force.

Friction comes in two regimes. Static friction acts on a stationary object and adjusts itself, up to a maximum, to prevent sliding; kinetic friction acts on a sliding object and is roughly constant. The maximum friction on a surface is $f = \mu N$, where $\mu$ is the coefficient of friction and $N$ the normal force — so friction depends on how hard the surfaces are pressed together, not on the contact area. For a spring, Hooke's law gives a restoring force proportional to the extension.

F = -kx

where $F$ is the spring force, $k$ is the spring constant (N m⁻¹) and $x$ is the extension or compression from the natural length. The minus sign records that the force always points back toward the equilibrium position, opposing the displacement.

An object is in translational equilibrium when the resultant force on it is zero, $\Sigma F = 0$. By the first law it is then at rest or moving at constant velocity — its acceleration is zero. The working method is to resolve every force into perpendicular components (usually horizontal and vertical, or along and perpendicular to a slope) and set the total in each direction to zero. Equilibrium problems are really two simultaneous equations, one per direction.

Momentum is the quantity of motion an object carries: $p = mv$, a vector pointing along the velocity, measured in kg m s⁻¹. Newton actually framed his second law in terms of momentum — the resultant force is the rate of change of momentum, $F = \Delta p / \Delta t$ — and for a constant mass this becomes the familiar $F = ma$. Rearranging gives impulse: $J = F\Delta t = \Delta p$. The impulse of a force is its size multiplied by the time it acts, and it equals the change in momentum produced. On a force–time graph, impulse is the area under the curve. This is why a longer contact time means a smaller force for the same change in momentum — the physics behind airbags, crumple zones and bending your knees when you land.

$F = \frac{\Delta p}{\Delta t} \qquad p = mv \qquad J = F\Delta t = \Delta p$

When no external resultant force acts on a system, its total momentum stays constant: $\Sigma p_{\text{before}} = \Sigma p_{\text{after}}$. This follows directly from Newton's third law — the internal forces two colliding bodies exert on each other are equal and opposite, so they change the two momenta by equal and opposite amounts, leaving the total unchanged. Because momentum is a vector, you must choose a positive direction and give each velocity the correct sign. Conservation of momentum applies to every collision and every explosion, and it is the single most powerful tool in this topic.

Momentum is conserved in every collision, but kinetic energy is not. In a perfectly elastic collision the total kinetic energy is the same before and after — a good model for collisions between hard spheres or between atoms. In an inelastic collision some kinetic energy is transferred into heat, sound and permanent deformation, so the kinetic energy after is less than before. The extreme case is a perfectly inelastic collision, where the objects stick together and move off with a common velocity; this loses the most kinetic energy consistent with conserving momentum. The test is simple: calculate the total kinetic energy before and after — if it is unchanged the collision is elastic, and if it has fallen the collision is inelastic.

An object moving in a circle at constant speed is still accelerating, because its velocity — a vector — is continually changing direction. This acceleration points toward the centre of the circle and has magnitude $a = v^2/r = \omega^2 r$, where $\omega$ is the angular speed. By Newton's second law a net inward force is therefore required, called the centripetal force: $F = mv^2/r = m\omega^2 r$, always directed toward the centre. It is essential to understand that centripetal force is not a new or extra force — it is the name for the resultant of the ordinary forces (tension, friction, gravity, normal force) resolved toward the centre. There is no outward 'centrifugal' force on the object; the sensation of being flung outward is simply inertia, the body's tendency to continue in a straight line.

$a = \frac{v^2}{r} \qquad F = \frac{mv^2}{r} = m\omega^2 r$

This is the showcase for a calculation topic. In Paper 2 the marks are analytic: each is tied to a specific piece of working — a method mark (M) for a correct approach, or an answer mark (A) for a correct result — and, crucially, method marks and error-carried-forward (ECF) mean a wrong number early on does not cost you every mark that follows. That protection only exists if your method is written down. Study how each mark below is earned by a specific line.

These ideas underpin the rest of mechanics and beyond. The work–energy topic (A.3) recasts forces acting over distances as energy transfers; gravitational and orbital motion set the gravitational force equal to $mv^2/r$ to find orbital speeds; and in later topics the same momentum bookkeeping reappears for photons, nuclear reactions and rocket propulsion. Master the core habits here — draw the free-body diagram, apply $\Sigma F = ma$ or conservation of momentum, and resolve toward the centre for circular motion — and the rest of physics becomes variations on methods you already own.