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4.1 Linear Momentum

Connect mass and velocity

An object's linear momentum is its mass multiplied by its velocity. Because mass is a positive scalar, momentum points in the same direction as velocity.

$\begin{gathered}\vec p=m\vec v\\ [p]=\text{kg}\,\text{m/s}\end{gathered}$

Keep direction in the result

Worked example: choose right as positive. A $2.0\,\text{kg}$ cart moves left at $3.0\,\text{m/s}$ , so $v=-3.0\,\text{m/s}$ .

$p=mv=(2.0\,\text{kg})(-3.0\,\text{m/s})=-6.0\,\text{kg}\,\text{m/s}$ .

The negative sign means the momentum is leftward. At fixed velocity, doubling mass doubles $|p|$ ; at fixed mass, doubling speed doubles $|p|$ .

Recognize interaction models

Model	Defining idea	What the model lets us focus on
Collision	Forces between the involved objects are much larger than the net external force during the brief interaction	Compare the objects' initial and final states using the object model
Explosion	Forces internal to the chosen system move its objects apart	Compare the system's states before and after the internal separation

Respect the model limits

A collision model does not mean external forces are literally zero; it means the internal interaction forces dominate during the collision interval. Unless stated otherwise in AP Physics 1, momentum means linear momentum. Impulse and momentum conservation are separate relationships developed in the following Topics.

4.2 Change in Momentum and Impulse

Connect force to momentum rate

A net external force changes an object's or system's momentum. The stronger the force or the longer it acts, the larger the momentum effect.

$\begin{gathered}\vec F_{\text{net}}=\frac{\Delta\vec p}{\Delta t}\\ \vec J=\vec F_{\text{avg}}\Delta t\end{gathered}$

Calculate a signed impulse

Signed pulse example: choose right as positive. A net average force of $-40\,\text{N}$ acts for $0.25\,\text{s}$ .

$J=F_{\text{avg}}\Delta t=(-40\,\text{N})(0.25\,\text{s})=-10\,\text{N}\,\text{s}$ .

The impulse is $10\,\text{N}\,\text{s}$ leftward, the same direction as the net force.

Map area and slope

Graph	What to read	Physical meaning
Net external force $F_{\text{net}}$ versus time $t$	Signed area under the curve	Impulse delivered during the interval
System momentum $p$ versus time $t$	Slope $\Delta p/\Delta t$	Net external force during the interval

Keep direction and graph roles

Impulse is a vector, not merely force magnitude times time. Area below the time axis gives impulse in the negative direction. On a momentum–time graph it is the slope, not the area, that represents net external force.

Subtract final and initial vectors

Change in momentum is a vector subtraction: final momentum minus initial momentum. Choose a positive direction before substituting one-dimensional velocities.

$\begin{gathered}\Delta\vec p=\vec p_f-\vec p_i\\ \vec J=\vec F_{\text{avg}}\Delta t=\Delta\vec p\end{gathered}$

Calculate a reversal

Direction-reversal example: choose right as positive. A $0.50\,\text{kg}$ object changes velocity from $+4.0\,\text{m/s}$ to $-2.0\,\text{m/s}$ .

$\Delta p=m(v_f-v_i)=(0.50\,\text{kg})(-2.0-4.0)\,\text{m/s}=-3.0\,\text{kg}\,\text{m/s}$ .

Therefore $J=-3.0\,\text{N}\,\text{s}$ : the impulse is leftward.

Recover Newton's second law

For a system of constant mass,

$\vec F_{\text{net}}=\frac{\Delta\vec p}{\Delta t}=m\frac{\Delta\vec v}{\Delta t}=m\vec a.$

Newton's second law therefore follows directly from the impulse–momentum theorem when mass does not change.

Keep the mass condition

When velocity reverses, subtract signed vectors—not speed magnitudes. AP Physics 1 does not require quantitative analysis of systems whose mass changes with time, so the step $\Delta\vec p=m\Delta\vec v$ is used here only for constant mass.

4.3 Conservation of Linear Momentum

Represent the system

Total system momentum is the vector sum of its parts. Dividing that total by total mass gives the center-of-mass velocity.

$\begin{aligned}\vec p_{\text{sys}}&=\sum_i\vec p_i=\sum_i m_i\vec v_i\\[4pt]\vec v_{\text{cm}}&=\frac{\sum_i\vec p_i}{\sum_i m_i}\end{aligned}$

Balance internal changes

If the net external force is zero, $\vec v_{\text{cm}}$ and $\vec p_{\text{sys}}$ stay constant. Internal forces come in Newton's-third-law pairs, so their impulses are equal and opposite: one part's momentum gain is balanced by another part's loss.

Apply conservation

One-dimensional explosion: a $3.0\,\text{kg}$ system starts at rest and separates into $1.0\,\text{kg}$ and $2.0\,\text{kg}$ parts. Choose right as positive. If the lighter part leaves at $+6.0\,\text{m/s}$ , then

$0=(1.0\,\text{kg})(+6.0\,\text{m/s})+(2.0\,\text{kg})v_2$ ,

so $v_2=-3.0\,\text{m/s}$ . The total momentum remains zero, although each part now has nonzero momentum.

Keep the system and course boundary

Conservation applies to the system total, not separately to every object. In two dimensions, conserve momentum by components. AP Physics 1 expects semiquantitative reasoning and correct equation setup, not solving a full simultaneous-equation system for an unknown final velocity.

Define the boundary

Choose the system boundary first. Forces between included objects are internal; forces from outside are external. Only net external impulse changes the selected system's momentum.

$\begin{aligned}\Delta\vec p_{\text{sys}}&=\vec J_{\text{ext}}\\[4pt]\vec F_{\text{ext,net}}=\vec 0&\Rightarrow\Delta\vec p_{\text{sys}}=\vec 0\end{aligned}$

Compare two selections

Selected system during a short cart collision	Force between carts	Momentum conclusion
Both carts	Internal	If other external impulse is negligible, total momentum of both carts is constant
Cart A only	Force from cart B is external	Cart A's momentum changes by the impulse from cart B

Track momentum transfer

Momentum is conserved in every interaction when the system and environment are considered together. If a chosen subsystem gains momentum, an equal-and-opposite amount is transferred elsewhere. A nonzero net external force therefore signals momentum crossing the selected boundary—not momentum being created or destroyed.

Test isolation

A collision does not automatically make the selected system isolated. Check the external impulse over the interaction interval. A force may be internal for a two-object system but external for either object considered alone.

4.4 Elastic and Inelastic Collisions

Choose the system-level test

Use the total kinetic energy of the colliding system immediately before and immediately after the collision. Momentum conservation alone cannot distinguish elastic from inelastic collisions.

$K_{\text{sys}}=\sum_i\frac{1}{2}m_i v_i^2$

Compare collision classes

Collision type	System kinetic-energy test	What happens after?
Elastic	$K_f=K_i$	Objects separate; each object's kinetic energy may still change
Inelastic	$K_f<K_i$	Some kinetic energy is transformed into other forms
Perfectly inelastic	$K_f<K_i$	Objects stick and share one final velocity

Classify from energy data

Classification example: the colliding objects have $K_i=18\,\text{J}$ before and $K_f=12\,\text{J}$ after.

$\Delta K_{\text{sys}}=K_f-K_i=12\,\text{J}-18\,\text{J}=-6\,\text{J}$ .

Because system kinetic energy decreases, the collision is inelastic. The missing $6\,\text{J}$ of kinetic energy is transformed into other forms. If the objects also stick, it is perfectly inelastic.

Keep totals and transformations clear

Elastic does not mean every object keeps its own kinetic energy; only the system total is unchanged. Inelastic does not mean energy disappears. It means some initial kinetic energy is not restored as kinetic energy after the interaction.

4 Linear Momentum

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