Dynamics

Mass

Mass 质量 tells you how hard it is to change an object's motion. The larger the mass, the larger the force 力 needed to give it a certain acceleration 加速度. Mass is measured in kilograms ($\text{kg}$) and is a scalar 标量.

Momentum

Linear momentum 动量 is the product of mass and velocity:

Momentum is a vector 矢量 — it points the same way as the velocity 速度. Its unit is $\text{kg m s}^{-1}$, which is the same as $\text{N s}$.

Force as the rate of change of momentum

Newton's second law, in its general form: the resultant force 合力 on an object equals the rate of change of its momentum.

Impulse is the area under a force-time graph, equal to the change in momentum

When the mass is constant this becomes $F = ma$, because $\Delta p = m\,\Delta v$ and $\Delta v / \Delta t = a$. Cambridge questions often want you to use $F = \Delta p / \Delta t$ directly for a collision 碰撞 or an impulse 冲量: the average force equals the change in momentum divided by the contact time.

A ball hits a wall with momentum $p_{1}$ and bounces back with momentum $p_{2}$ in the opposite direction. The change in momentum is $\Delta p = p_{2} - p_{1}$ (give each direction the correct sign). The average force is $\Delta p / \Delta t$, where $\Delta t$ is the contact time.

Worked example. A $0.20\ \text{kg}$ ball hits a wall at $8.0\ \text{m s}^{-1}$ and bounces straight back at $6.0\ \text{m s}^{-1}$. The contact lasts $0.050\ \text{s}$. Find the average force on the ball.

Take the rebound direction as positive, so $u = -8.0\ \text{m s}^{-1}$ and $v = +6.0\ \text{m s}^{-1}$:

When you know the momentum but not the speed, the change in kinetic energy 动能 is

This comes from $E_{\text{k}} = \tfrac{1}{2} m v^{2} = p^{2} / (2m)$.

A rocket pushes gas down; by Newton's third law the gas pushes the rocket up.

First law

An object stays at rest, or keeps moving at constant velocity in a straight line, unless a resultant external force 外力 acts on it. In short: zero resultant force means zero acceleration.

Second law

The resultant force on an object equals its rate of change of momentum, and acts in the same direction as that change. In SI units,

Acceleration and resultant force always point the same way.

Free-body diagram showing all forces on a block being pulled at an angle

Weight and normal contact force on a book resting on a table

Third law

When body A pushes on body B, body B pushes back on body A with an equal and opposite force. The two forces:

• act on different objects,
• are of the same type (both gravitational, both contact, both electrostatic 静电, and so on),
• have the same size and opposite direction.

A common trap: the weight 重力 of a block on a table and the normal contact force 支持力 from the table are not a third-law pair (they act on the same object and are different types). The third-law partner of the block's weight is the pull the block makes on the Earth. The third-law partner of the table's contact force is the push the block makes on the table.

For a rocket: the thrust 推力 on the rocket and the force on the gases are a third-law pair (the engine pushes the gas down, the gas pushes the engine up). Weight and air resistance are not part of this pair.

Newton's third-law pair: R on the book (up) and R′ on the table (down)

Weight is the force on an object from a gravitational field 重力场. Near the Earth's surface,

where $g \approx 9.81\ \text{m s}^{-2}$ is the acceleration of free fall. Weight is a vector that points towards the centre of the Earth. Do not mix it up with mass: mass is the same everywhere, but weight changes with place.

Friction and drag forces

A friction 摩擦力 force between two solid surfaces acts along the surface and opposes the sliding. A viscous 黏性 or drag 阻力 force is the resistive force from a fluid 流体 (a liquid or gas) on an object moving through it; air resistance is the case for air. You do not need to use any coefficient 系数 of friction or viscosity.

A simple model: the drag gets bigger as the speed gets bigger. At zero speed, the drag is zero.

Free-body diagram of a book being pulled on a table

An object falling through air

For an object dropped from rest and falling through air:

1. At first, only weight acts, so the object speeds up downwards at $g$.
2. As the speed grows, the upward drag grows. The resultant force gets smaller, so the acceleration gets smaller.
3. In the end, the drag equals the weight. The resultant force is zero, the acceleration is zero, and the speed stays constant — the terminal velocity 收尾速度.

On a velocity–time graph, the line starts straight with gradient $g$, then bends and flattens at the terminal velocity. This shape (fast start, then slowing acceleration, then constant speed) is how "falling with air resistance" differs from "free fall in a vacuum".

Velocity–time graph for an object falling through air

Forces on a falling object in a fluid

Why the spread-eagle pose? Spreading out gives the largest area, so the most drag. The bigger the drag, the sooner drag balances weight — and the lower the steady terminal velocity they fall at

Energy during a terminal-velocity fall

At terminal velocity, a parachutist 跳伞者 has constant kinetic energy. But the gravitational potential energy 重力势能 keeps falling as they go down. Where does it go? Almost all of it turns into thermal energy 热能 of the air around them. It does not become kinetic energy of the parachutist — that stays constant.

Cyclist or car at constant speed

A vehicle at constant speed on a flat road has zero resultant force. The forward driving force is equal and opposite to the total resistive force (friction, air resistance, and rolling resistance). At higher speed the drag is larger, so the driving force must be larger too — and so the power 功率 must be larger.

In a crash, a large force acts over a very short time to change momentum.

The principle

For a system with no resultant external force, the total momentum stays constant. This is conservation of momentum 动量守恒.

It always holds when there is no outside resultant force — in collisions, explosions 爆炸, and recoil 反冲. In two dimensions, momentum is conserved along each direction on its own.

Worked example. A $2.0\ \text{kg}$ trolley and a $3.0\ \text{kg}$ trolley are held together against a compressed spring, then released from rest. The $2.0\ \text{kg}$ trolley flies off at $6.0\ \text{m s}^{-1}$. Find the speed of the other trolley.

The total momentum stays zero (it started at rest), so

in the opposite direction.

Newton's third law in an isolated two-particle system: equal and opposite forces

Elastic and inelastic collisions

In every collision, momentum is conserved (if there is no outside resultant force).

An elastic collision 弹性碰撞 is one where the total kinetic energy is also conserved. A quick test: in an elastic collision, the relative speed 相对速率 of approach equals the relative speed of separation.

In an inelastic collision 非弹性碰撞, momentum is conserved but kinetic energy goes down — some becomes thermal, sound, or deformation 形变 energy. If the two objects stick together, the collision is perfectly inelastic.

Solving collision problems (one dimension)

Head-on collision: velocities before and after

For two objects with masses $m_{1}, m_{2}$ and starting velocities $u_{1}, u_{2}$ that hit head-on 正面, write

Use signed velocities (positive in one chosen direction). If the collision is elastic, add the relative-speed equation

or, the same thing, $\tfrac{1}{2} m_{1} u_{1}^{2} + \tfrac{1}{2} m_{2} u_{2}^{2} = \tfrac{1}{2} m_{1} v_{1}^{2} + \tfrac{1}{2} m_{2} v_{2}^{2}$. That gives two equations for two unknowns.

Worked example. A $1500\ \text{kg}$ car moving at $12\ \text{m s}^{-1}$ runs into a stationary $1000\ \text{kg}$ car and they lock together. Find their common velocity just after the collision.

Momentum is conserved (the cars stick, so $v_{1} = v_{2} = v$):

A useful result for a head-on elastic collision of mass $m$ with a stationary 静止 mass $M$:

Collisions in two dimensions

If the objects move in two dimensions, split the velocities into perpendicular 垂直 components 分量 and apply conservation of momentum along each direction on its own. For a collision where the objects hit at an angle, choose one axis along the first object's motion and one across it. The total momentum is conserved along each axis.

A glancing collision resolved along two perpendicular axes

Rocket / pushing out mass

A rocket pushes out gas at velocity $u$ (relative to itself) at a mass-flow rate 质量流率 $\dot m$ (kg per second). It feels a thrust

which comes from $F = \Delta p / \Delta t$. The momentum given to the gas each second equals the thrust on the rocket (Newton's third law: the rocket pushes the gas one way, the gas pushes the rocket the other way).

• Newton's second law is $F = \Delta p / \Delta t$ (rate of change of momentum); $F = ma$ is the special case for constant mass.
• Identify third-law pairs correctly: the same type of force, acting on two different bodies — not the balanced forces on one body.
• Momentum is conserved in every collision; kinetic energy is conserved only in an elastic collision.
• For terminal velocity, explain that drag rises with speed until drag $=$ weight, so the acceleration becomes zero.