Forces and Momentum

Newton's laws are the rulebook for all of mechanics. The first law says that in an inertial frame a body keeps a constant velocity (including zero) unless a net external force acts on it: $\Sigma F = 0 \implies v = \text{constant}$.

The second law is best stated in terms of momentum, given by:

$$F_{\text{net}} = \frac{\Delta p}{\Delta t}$$

This becomes the familiar $F = ma$ only when mass stays constant. The third law says forces come in pairs: if body $A$ pushes on body $B$, then $B$ pushes back on $A$ with a force equal in magnitude and opposite in direction.

Students often misidentify action-reaction pairs because they look at a single body in equilibrium. A genuine third-law pair must involve two different bodies acting on each other, and the two forces must be the same type (for example, the gravitational pull of Earth on a book and the gravitational pull of the book on Earth).

The upward normal force and the downward weight on a book resting on a table are *not* a third-law pair: they act on the same body and arise from different interactions (contact electromagnetism versus gravity).

You must be able to state all three laws, pick out correct force pairs, and apply $F = ma$ to single bodies and connected systems such as a car towing a trailer.

A free-body diagram strips a problem down to one body and the forces acting on it:

• weight $mg$ acting from the centre of mass
• the normal force perpendicular to a surface
• tension along a string
• friction along a surface
• where relevant, drag or buoyancy

Forces exerted *by* the body on other things never appear on its own diagram.

Once the diagram is drawn, choose perpendicular axes (often along and perpendicular to an incline) and resolve each force into components. On each axis apply:

$$\Sigma F = ma$$

If the body is in translational equilibrium, $\Sigma F = 0$ on every axis. For a block at rest on a slope of angle $\theta$, resolving gives $F_N = mg\cos\theta$ perpendicular to the surface and a friction force balancing $mg\sin\theta$ along it.

In exams you must draw arrows starting on the body, label each with its proper physical name, and make sure no "net force" or "centripetal force" arrow is added as if it were an extra interaction.

A clean, correctly labelled diagram is usually the first mark in any dynamics question, and most algebra errors trace back to a missing or misdirected force at this stage.

When two surfaces touch, the contact interaction has two components: the normal force $F_N$, perpendicular to the surfaces, and friction, parallel to them.

Static friction acts when there is no relative sliding; it adjusts itself to oppose the *tendency* of motion and can take any value up to a maximum:

$$F_f \leq \mu_s F_N$$

Once sliding actually begins, dynamic (kinetic) friction takes over, always directed against the relative motion, with the roughly constant value:

$$F_f = \mu_d F_N$$

Because $\mu_s$ is usually larger than $\mu_d$, a crate is harder to start moving than to keep moving: the push needed jumps up to $\mu_s F_N$, then drops once the crate slides.

The coefficients $\mu_s$ and $\mu_d$ are dimensionless and depend on the pair of materials, not on the contact area. On an incline of angle $\theta$, remember that $F_N = mg\cos\theta$, not $mg$, so the available friction is reduced.

You must be able to:

• decide whether a body remains static (compare the required force with $\mu_s F_N$)
• calculate the friction force in each regime
• include friction correctly in $\Sigma F = ma$ problems such as braking cars or blocks on slopes

A body in uniform circular motion moves at constant speed, but its velocity vector continuously changes direction, so it is accelerating. This centripetal acceleration points toward the centre of the circle with magnitude:

$$a = \frac{v^2}{r} = \omega^2 r$$

where $\omega = \frac{2\pi}{T}$ is the angular speed and $v = \omega r$.

By Newton's second law, a net inward force is required:

$$F_c = \frac{mv^2}{r}$$

It is always supplied by real physical forces:

• tension for a ball whirled on a string
• friction for a car turning on a flat road
• gravity for a satellite in orbit
• the horizontal component of the normal force on a banked track

Never draw "centripetal force" as an extra arrow on a free-body diagram; instead identify which actual force (or combination) provides the inward resultant and set it equal to $\frac{mv^2}{r}$.

A useful extension (needed for HL rigid-body work): if the speed changes, a tangential acceleration $a_t = \frac{\Delta v}{\Delta t}$ appears alongside the radial $a$, and the total acceleration is the vector sum of the two. The net force then no longer points at the centre but is angled forward when speeding up and backward when slowing down.

You must be able to convert between $v$, $\omega$ and $T$, and solve for speeds, radii and forces in horizontal and vertical circles.

Linear momentum, $p = mv$, is a vector that measures how hard it is to stop a moving body. Newton's second law in its general form says the net force equals the rate of change of momentum, $F = \frac{\Delta p}{\Delta t}$.

Rearranging gives the impulse, the product of the average net force and the time for which it acts:

$$J = F\Delta t = \Delta p$$

On a force-time graph, the area under the curve equals the impulse and therefore the change in momentum; this works even when the force varies, as in a bat striking a ball.

Impulse explains most safety engineering. In a car crash the change in momentum of a passenger is fixed by the speeds involved, but crumple zones, seat belts and airbags stretch out the stopping time $\Delta t$, so the average force $F = \frac{\Delta p}{\Delta t}$ on the body is much smaller. The same idea explains bending your knees on landing or the follow-through in catching a ball.

You must treat momentum with signs in one dimension, read impulse off $F$-$t$ graphs, and compute average forces during impacts, taking care that a rebound reverses the sign of the velocity so $\Delta p$ can exceed the initial momentum.

In a closed, isolated system the total momentum is constant, because every internal force belongs to a Newton's third-law pair and the impulses cancel.

For two bodies, applied with a clearly defined positive direction so velocities carry the correct signs, this reads:

$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$

Collisions are classified by what happens to kinetic energy:

• In an elastic collision both momentum and total kinetic energy are conserved (a good approximation for colliding gas molecules or steel balls).
• In an inelastic collision momentum is conserved but some kinetic energy is transferred to internal energy, sound and deformation.
• In a perfectly inelastic collision the bodies stick together and move with a common velocity, giving the maximum possible kinetic-energy loss consistent with momentum conservation.
• An explosion runs the logic in reverse: a system initially at rest has zero total momentum, so the fragments must carry momenta that sum to zero, like a cannon recoiling backward as the shell flies forward.

You must be able to apply conservation of momentum to find unknown velocities, then separately compare initial and final kinetic energies to decide what kind of collision occurred; never assume kinetic energy is conserved just because momentum is.