3. Dynamics

Mass is inertia: for a given resultant force, a larger mass produces a smaller acceleration through F=ma. It is a scalar measured in kilograms.

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.

Mass does not depend on local g; an object’s weight can change while its inertia remains the same.

Newton’s second law states ΣF=ma. The acceleration vector has the same direction as the resultant and magnitude proportional to it for fixed mass.

Draw a free-body diagram, resolve components and include only forces acting on the chosen body. A negative component records the chosen sign convention.

A 3 kg trolley with resultant force (6,0) N accelerates at (2,0) m s⁻².

Using one applied force instead of the resultant gives the wrong acceleration when several forces act.

Momentum p=mv is a vector in the direction of velocity, measured in kg m s⁻¹. It is conserved for an isolated system when external impulse is negligible.

Use signed velocities on one axis and define the system before writing conservation. Momentum is distinct from kinetic energy, which depends on speed squared.

A 2 kg object moving at −3 m s⁻¹ has momentum −6 kg m s⁻¹ in the chosen positive direction.

Equal and opposite momenta can sum to zero while individual objects still move.

The general law is F=dp/dt. For constant mass it becomes F=ma, but variable-mass systems require momentum flux to be considered explicitly.

Choose the system boundary and direction before differentiating momentum. The force is the resultant external force on that system.

A constant 12 N force acting on a 3 kg object produces dp/dt=12 kg m s⁻² and, at fixed mass, acceleration 4 m s⁻².

F=ma is a special constant-mass form; force is not simply momentum divided by time unless the change is defined correctly.

First law defines inertial motion when resultant force is zero; second law gives ΣF=ma; third law pairs equal opposite forces acting on different bodies.

Apply one law to one chosen body at a time. Identify the frame, draw forces and distinguish acceleration from velocity.

A stationary book has weight and normal reaction balanced; the Earth–book gravitational interaction has a third-law partner acting on Earth.

Third-law forces do not cancel in one free-body diagram because they act on different objects.

Weight W=mg is the force exerted by a gravitational field on mass m. Its direction is toward the field source and its value depends on local g.

Use mass in inertial equations and weight as a force in free-body diagrams. In a non-uniform field, g can vary with position.

A 5 kg mass near Earth has weight about 49 N downward when g=9.8 m s⁻².

Weight is not mass, and apparent weight measured by a scale can differ from mg during acceleration.

Dry friction acts at contact and is limited by μR; viscous or drag forces depend on motion through a fluid and may vary with speed, often as kv or kv² in ideal models.

Identify the regime and direction before choosing a formula. Friction can be static up to a limit, while drag is zero when relative speed is zero.

A falling object experiences weight downward and air resistance upward once it moves; a block at rest can have static friction smaller than μR.

“Resistance” is not always μR, and drag does not have a fixed value independent of speed.

For a falling object, resultant force is mg minus upward resistance. As speed grows, resistance can grow and reduce the downward acceleration.

Write m dv/dt=mg−R(v), state the drag model and use signs consistently. The no-drag constant-g solution is only an ideal comparison.

With linear drag R=kv, acceleration decreases from g at release as v increases.

Gravity does not switch off when drag appears; drag changes the resultant and therefore acceleration.

Terminal velocity is reached when acceleration becomes zero, so the upward resistive force equals the downward weight and velocity becomes constant.

Find the terminal condition from ΣF=0, not from v=0. The object may approach terminal speed asymptotically rather than reach it in finite time.

For linear resistance kv, terminal speed is mg/k; doubling mass doubles terminal speed if k is unchanged.

Terminal velocity is not the same as zero velocity, and constant velocity means zero resultant force, not zero forces.

For a closed system with negligible external impulse, total momentum before an interaction equals total momentum after it: Σp_before=Σp_after.

Define the system and a positive direction first. Momentum is a vector, so opposite directions must carry opposite signs.

Two carts initially at rest apart can exchange equal and opposite momentum; for a collision, the signed total before and after must match.

Momentum is not automatically conserved for every chosen object: an external force can change that object’s momentum.

In one dimension, write m₁u₁+m₂u₂=m₁v₁+m₂v₂, with velocities signed along one chosen axis.

List known masses and velocities, keep units consistent, then solve for the unknown. A negative result means motion opposite to the chosen positive direction.

A 2 kg cart at 3 m s⁻¹ sticks to a 1 kg cart at rest: 6=(3)v, so the joined speed is 2 m s⁻¹ in the original direction.

Do not replace momentum with speed, and do not drop the sign of a cart moving in the opposite direction.

An elastic collision satisfies momentum conservation and also conserves total kinetic energy: KE_before=KE_after.

Use momentum and kinetic-energy equations together only when the collision is identified as elastic; otherwise kinetic energy may become thermal, sound or deformation energy.

For two identical smooth balls in a head-on elastic collision, exchanging their velocities satisfies both conservation laws.

Momentum conservation alone does not prove a collision is elastic; most real collisions are not perfectly elastic.

If the system is isolated, momentum remains conserved in an inelastic interaction, but total kinetic energy after is smaller because energy is transferred to deformation, heat or sound.

Check whether objects separate or stick, then use the momentum equation. Track the lost kinetic energy as a change of energy store, not as lost total energy.

A lump of clay hitting a stationary block and sticking has one common final velocity; the difference in kinetic energy becomes internal energy.

“Kinetic energy is not conserved” does not mean energy disappeared, and inelastic does not mean momentum is unbalanced.