3. Dynamics

Keep mass separate from weight, which is mg, and use the chosen body’s mass in its equation of motion.

The same 10 N force gives 2 m s⁻² to a 5 kg object but 1 m s⁻² to a 10 kg object.

Choose a positive direction, draw only forces acting on the body, add their signed components, then divide the resultant by mass.

A 850 kg car has 1600 N forwards and 1200 N backwards. Resultant force = 400 N forwards, so a = 400/850 = 0.47 m s⁻² forwards.

Take right as positive. A 0.10 kg ball changes velocity from +20 m s⁻¹ to −15 m s⁻¹: initial p = +2.0 kg m s⁻¹, final p = −1.5 kg m s⁻¹, and change Δp = −3.5 kg m s⁻¹.

A 0.20 kg ball moving at +14.0 m s⁻¹ rebounds at −7.0 m s⁻¹ in 0.60 s. Δp = 0.20(−7.0 − 14.0) = −4.2 kg m s⁻¹, so average F = −4.2/0.60 = −7.0 N: 7.0 N opposite to its initial motion.

For a book resting on a table, weight and normal contact force balance on the book: this applies the first/second-law resultant. The third-law partner of the table's force on the book is the book's force on the table; the partner of Earth's gravitational force on the book is the book's gravitational force on Earth.

A 5.0 kg object has weight 5.0 × 9.8 = 49 N near Earth. Where g = 1.6 N kg⁻¹, its mass remains 5.0 kg but its weight is 8.0 N.

A changing drag force makes the resultant force change, so acceleration need not be constant. For example, as a falling object speeds up, upward air resistance increases while weight remains downward.

As downward speed increases, upward air resistance increases. Weight stays constant, so the downward resultant decreases; therefore downward acceleration decreases even while velocity continues to increase.

For an object moving upward, both weight and air resistance act downward. As the object slows, air resistance decreases; at the highest point its speed and air resistance are momentarily zero, but weight and downward acceleration remain.

For an object falling through air: it first accelerates downward; increasing speed produces increasing upward drag; the downward resultant and acceleration decrease; when drag equals weight, velocity becomes constant.

In a liquid, upthrust may also act upward. At terminal velocity the complete force balance is weight = drag + upthrust, not necessarily weight = drag alone.

Define which objects belong to the system, choose positive directions and add their vector momenta. Internal interaction forces can change each object's momentum but produce equal and opposite changes in the system total.

An initially stationary firework has total momentum zero. After it explodes into two fragments, their momenta are equal in magnitude and opposite in direction, so the total remains zero.

A 2.0 kg cart moving at +3.0 m s⁻¹ sticks to a stationary 1.0 kg cart. Before: total p = 6.0 kg m s⁻¹. After: combined mass = 3.0 kg, so v = 6.0/3.0 = +2.0 m s⁻¹.

Use the same momentum method for collisions, sticking interactions, recoil and explosions. Whether the interaction is elastic or inelastic changes the kinetic-energy analysis, not the isolated-system momentum equation.

For an isolated collision, momentum is also conserved. Use momentum together with either elastic condition to solve unknown velocities; do not replace signed velocities with speeds in the momentum equation.

In a head-on elastic collision of identical balls where one is initially stationary, the moving ball can stop and the other leave with its speed. Total momentum and kinetic energy are unchanged, and approach speed equals separation speed.

Momentum conservation follows from the absence of a resultant external force on the system. It does not require kinetic energy to remain constant. Total energy is still conserved when kinetic energy changes form.