Unit 4: Systems of Particles and Linear Momentum

Systems and the Center of Mass

In earlier mechanics, you often treat an object as a single particle: all of its mass is concentrated at one point, and you only track translation. That works when the object does not rotate significantly, its size is small compared to the distances involved, or you only care about how the whole object moves.

A system of particles is the next level up: you model an object (or multiple objects) as many masses moving together. The payoff is that you can describe complicated motion using two complementary viewpoints: (1) external motion of the system as a whole and (2) internal motion within the system (how parts move relative to each other, including collisions and explosions). The bridge between those viewpoints is the center of mass (COM): the point that moves as if all external forces were applied to a single particle of mass equal to the system’s total mass. In practice, the COM is the “dot” you can often use to represent an extended object in a free-body diagram.

For a homogeneous (uniform-density) object, the COM is at the geometric center (for example, a uniform sphere, cube, or rectangular box). When density is not uniform, the COM shifts toward the denser/heavier regions.

Defining the center of mass (discrete particles)

For particles of masses $m_i$ at position vectors $\vec r_i$, the COM position vector $\vec r_{cm}$ is

$\vec r_{cm} = \frac{1}{M}\sum_i m_i\vec r_i$

where the total mass is

$M = \sum_i m_i$

In components (often how you compute it on exams):

$x_{cm} = \frac{1}{M}\sum_i m_i x_i$

$y_{cm} = \frac{1}{M}\sum_i m_i y_i$

$z_{cm} = \frac{1}{M}\sum_i m_i z_i$

Conceptually, COM is a weighted average of position. A heavier particle “pulls” the COM closer to itself.

A reliable computation routine (especially in 1D) is: choose an origin and axis, list each position, compute $\sum m_i x_i$, and divide by $\sum m_i$.

Center of mass for continuous mass distributions

For an extended object with continuously distributed mass, replace the sum with an integral. If $dm$ is a small mass element located at $\vec r$, then

$\vec r_{cm} = \frac{1}{M}\int \vec r\,dm$

Common special cases (useful when density is uniform):

• For a 1D rod along $x$ with linear mass density $\lambda$,

$dm = \lambda\,dx$

• For a 2D lamina in the $xy$ plane with surface mass density $\sigma$,

$dm = \sigma\,dA$

• For a 3D object with volume mass density $\rho$,

$dm = \rho\,dV$

In AP Physics C: Mechanics, you are most often expected to compute COM for discrete masses, or use symmetry arguments (for standard shapes, the COM is at the geometric center if density is uniform).

Motion of the center of mass (the “system-level” Newton’s 2nd law)

Differentiate the COM definition with respect to time. The COM velocity is

$\vec v_{cm} = \frac{d\vec r_{cm}}{dt} = \frac{1}{M}\sum_i m_i\vec v_i$

Differentiate again to get COM acceleration:

$\vec a_{cm} = \frac{d\vec v_{cm}}{dt} = \frac{1}{M}\sum_i m_i\vec a_i$

Apply Newton’s 2nd law to each particle:

$m_i\vec a_i = \vec F_{i,ext} + \sum_{j\neq i} \vec F_{ij}$

Summing over all particles makes the internal forces cancel in pairs (Newton’s 3rd law), leaving only external forces:

$M\vec a_{cm} = \vec F_{ext,net}$

Interpretation: the COM accelerates exactly as if the entire mass $M$ were a single particle experiencing only the net external force.

This immediately gives useful qualitative statements:

• If there is no net external force on the system, then $\vec a_{cm} = \vec 0$, so the COM remains at rest or moves with constant velocity.
• If there is a net external force, the COM accelerates in the direction of the net force.

A common misunderstanding is thinking internal forces can change COM motion. They cannot. Internal forces can change how parts move relative to each other, but by themselves they cannot change the system’s total momentum or COM motion.

Stability idea (COM and potential energy)

The COM concept is also useful for reasoning about stability. A system tends to be stable when its COM is at a low point in its potential energy and unstable when its COM is at a high point in its potential energy.

Center of gravity vs. center of mass

The center of mass depends only on mass distribution. The center of gravity is the point where the gravitational force (weight) can be considered to act; it depends on both the mass distribution and the gravitational field.

• Center of gravity: where weight is “evenly distributed,” where the object balances in a gravitational field, depends on the gravitational field and can change if the field changes or if the object’s orientation/position in a nonuniform field changes.
• Center of mass: where mass is “evenly distributed,” where the object balances if the gravitational field is uniform, depends only on mass distribution and is fixed relative to the object.

In many AP contexts with a uniform gravitational field near Earth’s surface, center of gravity and center of mass coincide, but the conceptual distinction is still important.

Example 1: COM of two masses on a line

Two masses lie on the $x$ axis: $m_1$ at $x_1 = 0$ and $m_2$ at $x_2 = L$. The center of mass is

$x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1+m_2} = \frac{m_2}{m_1+m_2}L$

Checks that build intuition:

• If $m_2$ is much larger than $m_1$, then $x_{cm}$ approaches $L$.
• If $m_1 = m_2$, then $x_{cm} = \frac{L}{2}$.

Example 2: COM motion through an explosion (qualitative constraint)

A projectile reaches the peak of its trajectory and explodes into two pieces. Neglect air resistance. At the peak and after the explosion, the external force is gravity only, so

$M\vec a_{cm} = \vec F_{ext,net} = M\vec g$

Internal forces during the explosion can give pieces different velocities, but the COM continues along the same projectile trajectory it would have followed if the explosion never occurred.

Sample Problem: Nonuniform rod with linear density

A bar with length $0.30\ \text{m}$ has linear density

$\lambda(x) = 10 + 6x$

where $x$ is in meters and $\lambda$ is in $\text{kg/m}$. Determine the mass of the bar and the COM position measured from the $x=0$ end.

Solution: The mass is

$M = \int_0^{0.30} \lambda(x)\,dx = \int_0^{0.30} (10+6x)\,dx = 10(0.30) + 3(0.30)^2 = 3.27\ \text{kg}$

The COM is

$x_{cm} = \frac{1}{M}\int_0^{0.30} x\lambda(x)\,dx = \frac{1}{3.27}\int_0^{0.30} (10x+6x^2)\,dx$

$x_{cm} = \frac{1}{3.27}\left(5(0.30)^2 + 2(0.30)^3\right) = \frac{0.504}{3.27} \approx 0.154\ \text{m}$

So the COM is about $15.4\ \text{cm}$ from the $x=0$ end.

Exam Focus

• Typical question patterns:
  • Compute $x_{cm}$ and $y_{cm}$ for a set of point masses; sometimes one mass moves and you track how COM shifts.
  • Use $M\vec a_{cm} = \vec F_{ext,net}$ to argue that COM follows a predictable trajectory during an explosion or separation.
  • Use symmetry to locate COM for uniform objects.
  • (Occasionally) compute COM for a 1D object with nonuniform density using integrals.
• Common mistakes:
  • Forgetting to divide by total mass $M$ (treating COM as a simple average rather than weighted average).
  • Treating internal forces as contributing to $\vec F_{ext,net}$.
  • Mixing up geometric center with COM when masses are unequal or density is nonuniform.
  • Confusing center of mass with center of gravity in cases where the gravitational field is not effectively uniform.

Linear Momentum: Single Particles and Systems

Momentum is the quantity that most directly tracks how motion is “carried” through interactions like pushes, explosions, and collisions.

Defining linear momentum

For a particle of mass $m$ moving with velocity $\vec v$, the linear momentum $\vec p$ is

$\vec p = m\vec v$

Momentum is a vector and points in the direction of velocity.

Momentum for a system of particles and the COM connection

For a system, the total linear momentum $\vec P$ is

$\vec P = \sum_i \vec p_i = \sum_i m_i\vec v_i$

Compare this to the COM velocity formula:

$\vec v_{cm} = \frac{1}{M}\sum_i m_i\vec v_i$

Therefore, the “glue equation” linking system momentum and COM motion is

$\vec P = M\vec v_{cm}$

So the total momentum of the system equals the momentum of a single particle of mass $M$ moving with the COM velocity.

Newton’s 2nd law in momentum form (single particle and system)

For a single particle with constant mass,

$\vec F_{net} = \frac{d\vec p}{dt}$

For a system, the parallel statement is that net external force changes total momentum:

$\vec F_{ext,net} = \frac{d\vec P}{dt}$

Internal forces can redistribute momentum among particles, but only external forces change the total.

Impulse: how forces change momentum over time

Collisions and other short interactions are hard to analyze with kinematics because forces can spike and vary rapidly with time. The tool you use instead is impulse.

Impulse $\vec J$ delivered over a time interval is

$\vec J = \int_{t_i}^{t_f} \vec F\,dt$

If the net force is constant (or you use average force), this simplifies to

$\vec J = \vec F_{avg}\Delta t$

Impulse has the same units as momentum.

Impulse-momentum theorem

Combining $\vec F_{net} = \frac{d\vec p}{dt}$ with the definition of impulse gives

$\vec J_{net} = \Delta \vec p = \vec p_f - \vec p_i$

This relation is often explicitly provided on the AP free-response equation sheet in the forms

$\vec J = \int \vec F\,dt$

and

$\vec J = \Delta \vec p$

Force-time graphs and area

If you are given a graph of $F$ vs. $t$, the impulse is the signed area under the curve between the relevant times:

• Positive area increases momentum in the positive direction.
• Negative area decreases it.

A frequent exam skill is computing areas of rectangles and triangles from piecewise force graphs.

Example 1: Average force during a collision

A cart of mass $m$ moves along +x with speed $v_i$ and hits a bumper, rebounding with speed $v_f$ in the -x direction. If contact lasts time $\Delta t$, the average force on the cart follows from

$\Delta p = p_f - p_i = m(-v_f) - mv_i = -m(v_f+v_i)$

and

$F_{avg} = \frac{\Delta p}{\Delta t} = \frac{-m(v_f+v_i)}{\Delta t}$

The negative sign indicates the force points in -x.

Example 2: Impulse from a triangular force pulse

A force in the +x direction rises linearly from 0 to $F_{max}$ over time $T$ and then returns linearly to 0 by time $2T$, forming a triangle on the $F$ vs. $t$ graph. The impulse is the triangle area:

$J = \frac{1}{2}(2T)F_{max} = TF_{max}$

For mass $m$ initially at rest,

$v_f = \frac{J}{m} = \frac{TF_{max}}{m}$

Sample Problem: Impulse and average force on a football

A kicker punts a football of mass $0.4\ \text{kg}$, giving it launch speed $30\ \text{m/s}$. The impact time is $8\ \text{ms}$. Find the impulse delivered and the average force exerted.

Solution: Taking the ball to start from rest, the impulse equals the change in momentum:

$J = \Delta p = m\Delta v = 0.4(30) = 12\ \text{N s}$

Average force magnitude is

$F_{avg} = \frac{J}{\Delta t} = \frac{12}{0.008} = 1500\ \text{N}$

Exam Focus

• Typical question patterns:
  • Use $\vec J = \Delta \vec p$ to relate a collision force (or average force) to velocity change.
  • Compute impulse as area under an $F(t)$ graph; then find $\Delta v$.
  • Use $\vec P = M\vec v_{cm}$ to connect system momentum to COM motion.
• Common mistakes:
  • Using scalar speeds when momentum is vector; sign errors are extremely common.
  • Confusing force with impulse (force depends on time profile; impulse accounts for duration).
  • Forgetting that impulse uses the net force (or for systems, net external force).

Conservation of Linear Momentum (and When It Applies)

Conservation laws are powerful because they let you jump from “before” to “after” without tracking the details in between.

The core idea

From the system momentum form of Newton’s 2nd law,

$\vec F_{ext,net} = \frac{d\vec P}{dt}$

Integrate over time:

$\int_{t_i}^{t_f} \vec F_{ext,net}\,dt = \vec P_f - \vec P_i$

So

$\vec J_{ext} = \Delta \vec P$

If the net external impulse is zero (or negligible), then

$\vec P_f = \vec P_i$

This is conservation of linear momentum. In words: the total momentum of a closed system remains constant if no external forces (more precisely, no net external impulse) act on the system.

What counts as “external” depends on your chosen system

Choosing the system boundary wisely is often the hardest part.

• If your system is “two colliding carts,” the contact forces between carts are internal and cancel in the total momentum equation.
• The track’s normal forces and gravity are external, but often vertical and irrelevant to horizontal momentum.
• Friction with the track is external and may break momentum conservation if it produces significant impulse during the collision.

Rule of thumb: momentum is conserved in a direction if the net external impulse in that direction is negligible.

Momentum conservation in components

Because momentum is a vector, conservation applies component-by-component:

$P_{x,f} = P_{x,i}$

$P_{y,f} = P_{y,i}$

when external impulses in those directions are negligible.

Center of mass viewpoint

Since

$\vec P = M\vec v_{cm}$

conservation of momentum implies constant COM velocity. In an isolated system, the COM moves in a straight line at constant speed.

Recoil and explosions

An explosion is an internal interaction that converts internal energy into kinetic energy of parts. Momentum conservation still holds (if external impulse is negligible), so pieces must recoil so that total momentum stays the same. If an object at rest explodes into two parts,

$\vec p_{1,f} + \vec p_{2,f} = \vec 0$

They fly apart with equal and opposite momentum vectors (not necessarily equal speeds).

Example 1: Two carts push apart (explosion-like)

Two carts of masses $m_1$ and $m_2$ are initially at rest on a frictionless track. A spring between them is released, pushing them apart. Conservation of momentum gives

$m_1 v_1 + m_2 v_2 = 0$

so

$v_2 = -\frac{m_1}{m_2}v_1$

Momentum alone gives the ratio and opposite directions; numerical speeds require additional information (like spring energy).

Example 2: Momentum conserved horizontally in a brief collision despite gravity

Two pucks collide on an air table. During the collision, gravity and normal forces act, but they are vertical. If friction is negligible and the collision time is short, the external impulse in the horizontal plane is approximately zero, so horizontal momentum is conserved in both $x$ and $y$.

Sample Problem: Astronaut recoil by throwing a tool

An astronaut floats in space near her shuttle. Her mass (body + suit + equipment) is $89\ \text{kg}$. She throws a $1\ \text{kg}$ tool directly away from the ship with speed $9\ \text{m/s}$. If the ship is $10\ \text{m}$ away, how long will it take her to reach it?

Solution: Take “away from the ship” as positive. The astronaut + tool system starts at rest, so total momentum is initially zero. After the throw,

$0 = (1)(9) + (89)v_{a}$

so

$v_{a} = -\frac{9}{89}\ \text{m/s}$

Her speed toward the ship is

$|v_a| = \frac{9}{89}\ \text{m/s}$

Assuming this speed stays constant,

$t = \frac{10}{9/89} \approx 98.9\ \text{s}$

So it takes about $99\ \text{s}$.

Collisions and energy: an important separation

Collisions are commonly classified as elastic or inelastic. Momentum is conserved for an isolated system in both cases, but kinetic energy is conserved only in elastic collisions.

A subtle but crucial point: momentum conservation does not imply kinetic energy conservation. Momentum conservation comes from Newton’s laws and negligible external impulse; kinetic energy conservation requires a special kind of interaction (elastic).

Exam Focus

• Typical question patterns:
  • Decide whether momentum is conserved in a chosen direction based on external forces/impulses.
  • Use $\vec P_i = \vec P_f$ for collisions, recoil, explosions; often in 1D first.
  • Use $\vec P = M\vec v_{cm}$ to reason about COM speed/direction.
• Common mistakes:
  • Conserving momentum when external impulse is not negligible (for example, significant friction during the interaction).
  • Conserving total momentum but then treating it as scalar (dropping vector signs/components).
  • Assuming kinetic energy is conserved in all collisions.

Collisions in One Dimension: Elastic Collisions

A collision is a short interaction where forces between objects are large and act over a short time. That “short time” is exactly why impulse and momentum are so useful.

Classifying collisions by kinetic energy behavior

Assuming net external impulse is negligible, momentum is conserved in all collisions, but kinetic energy may or may not be.

• Elastic collision: total kinetic energy is conserved.
• Inelastic collision: total kinetic energy decreases (converted to thermal energy, deformation, sound, etc.).
• Perfectly inelastic collision: objects stick together and move with a common final velocity.

Setting up a 1D elastic collision mathematically

Let masses $m_1$ and $m_2$ have initial velocities $v_{1i}$ and $v_{2i}$. After collision, velocities are $v_{1f}$ and $v_{2f}$.

Momentum conservation:

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

Kinetic energy conservation (elastic only):

$\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2$

Solve these two equations for the two unknown final velocities.

A very useful derived relationship (relative speed reversal)

For 1D elastic collisions,

$v_{1i} - v_{2i} = -(v_{1f} - v_{2f})$

Equivalently,

$v_{1i} - v_{2i} = v_{2f} - v_{1f}$

Interpretation: relative speed of approach equals relative speed of separation. This often reduces algebra and provides a powerful check.

Example 1: Elastic collision with a stationary target

A mass $m_1$ moving with speed $v_0$ strikes mass $m_2$ initially at rest. Using momentum plus relative speed reversal:

$m_1 v_0 = m_1 v_{1f} + m_2 v_{2f}$

$v_0 = v_{2f} - v_{1f}$

Solving gives

$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_0$

$v_{2f} = \frac{2m_1}{m_1+m_2}v_0$

Sanity checks:

• If $m_1 = m_2$, then $v_{1f} = 0$ and $v_{2f} = v_0$.
• If $m_2$ is very large, $v_{1f}$ approaches $-v_0$ and $v_{2f}$ is tiny.

Example 2: Using the COM frame idea (conceptual)

In an elastic collision, it is often helpful to imagine the COM frame, where total momentum is zero. In that frame, the two objects approach each other with equal and opposite momenta and separate with equal and opposite momenta. Transforming back to the lab frame helps you reason qualitatively about symmetry and speed relationships.

Exam Focus

• Typical question patterns:
  • “Two carts collide elastically in 1D” where one is initially at rest; derive or compute final speeds.
  • Conceptual questions using special cases (equal masses, very massive target) to predict outcomes.
  • Combine momentum with energy conservation specifically labeled as elastic.
• Common mistakes:
  • Applying kinetic energy conservation to a collision that is not stated to be elastic.
  • Losing track of signs when an object reverses direction.
  • Solving momentum and energy equations but choosing the nonphysical root (for example, the trivial solution $v_{1f}=v_{1i}$, $v_{2f}=v_{2i}$ corresponds to “no interaction”).

Collisions in One Dimension: Inelastic and Perfectly Inelastic Collisions

In many real collisions, some kinetic energy turns into deformation, heating, or sound. Momentum conservation still holds for an isolated system, but kinetic energy does not.

What “kinetic energy is not conserved” really means

When kinetic energy decreases, it has not disappeared. It has been transferred into internal energy (microscopic motion, deformation) and possibly energy carried away by sound.

Perfectly inelastic collisions (objects stick)

If two masses stick together after colliding, they share a common final velocity $v_f$. Momentum conservation gives

$m_1 v_{1i} + m_2 v_{2i} = (m_1+m_2)v_f$

so

$v_f = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1+m_2}$

This is a weighted average of the initial velocities.

Kinetic energy loss in a perfectly inelastic collision

Compute and compare:

$K_i = \frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2$

$K_f = \frac{1}{2}(m_1+m_2)v_f^2$

The difference $K_i - K_f$ is converted into internal energy.

A common misconception is that “perfectly inelastic means all kinetic energy is lost.” Not true: some kinetic energy generally remains because the combined mass can still be moving.

Inelastic but not perfectly inelastic

If objects do not stick, you still use momentum conservation, but you need more information to determine both final velocities (because kinetic energy conservation is not available). The extra information could be a stated final speed, a coefficient of restitution (sometimes used), or a second stage analyzed with energy (like ballistic pendulum setups).

Ballistic pendulum: a classic momentum-then-energy two-step

A ballistic pendulum combines a perfectly inelastic collision with subsequent conversion between kinetic and potential energy.

• Stage 1 (collision, short time): use momentum conservation (external impulse often negligible during the collision).
• Stage 2 (swing upward): use mechanical energy conservation for the pendulum motion (if nonconservative losses are negligible), because gravity is conservative and tension does no work.

Example 1: Perfectly inelastic collision on a frictionless track

A cart of mass $m_1$ moving at $v_0$ hits a stationary cart of mass $m_2$ and they stick. Momentum conservation:

$m_1 v_0 = (m_1+m_2)v_f$

so

$v_f = \frac{m_1}{m_1+m_2}v_0$

Kinetic energy loss:

$K_i = \frac{1}{2}m_1 v_0^2$

$K_f = \frac{1}{2}(m_1+m_2)\left(\frac{m_1}{m_1+m_2}v_0\right)^2 = \frac{1}{2}\frac{m_1^2}{m_1+m_2}v_0^2$

so

$K_i - K_f = \frac{1}{2}\frac{m_1 m_2}{m_1+m_2}v_0^2$

Example 2: Ballistic pendulum (momentum then energy)

A bullet of mass $m$ traveling at speed $v$ embeds in a block of mass $M$ suspended as a pendulum. After the collision, the block+bullet rise to maximum height $h$. Let $V$ be the speed just after impact.

Stage 1 (collision):

$mv = (M+m)V$

Stage 2 (rise):

$\frac{1}{2}(M+m)V^2 = (M+m)gh$

so

$V = \sqrt{2gh}$

Combine:

$v = \frac{M+m}{m}\sqrt{2gh}$

Common pitfall: trying to use energy conservation during the collision itself. Embedding is highly nonconservative, so energy is not conserved in Stage 1.

Sample Problem: Head-on collision of two rolling balls

Two balls roll toward each other. A red ball has mass $0.5\ \text{kg}$ and speed $4\ \text{m/s}$ just before impact. A green ball has mass $0.3\ \text{kg}$ and speed $2\ \text{m/s}$. After the head-on collision, the red ball continues forward with speed $1.7\ \text{m/s}$. Find the green ball’s speed after the collision. Was the collision elastic?

Solution: Momentum is a vector; direction matters. Let the red ball’s initial direction be positive:

$v_{r,i} = 4$

$v_{g,i} = -2$

After collision:

$v_{r,f} = 1.7$

Conservation of momentum gives

$0.5(4) + 0.3(-2) = 0.5(1.7) + 0.3 v_{g,f}$

Solving:

$v_{g,f} = 1.83\ \text{m/s}$

The positive sign shows the green ball reverses direction, which is common when a lighter object collides with a heavier one.

To test elasticity, compare kinetic energy before and after. In this case, an explicit calculation is not strictly necessary because both balls’ speeds decreased, so total kinetic energy must have decreased and the collision is inelastic (this is typical for macroscopic collisions). If you do compute, you find kinetic energy is not conserved, with most of the lost energy transferred as heat and deformation.

Exam Focus

• Typical question patterns:
  • Perfectly inelastic collision: find common final velocity and compare energies.
  • Multi-stage problems like ballistic pendulum: momentum for collision, then energy for subsequent motion.
  • Given one final speed in an inelastic collision: use momentum to solve for the other.
  • Conceptual questions asking which quantities are conserved in which stage.
• Common mistakes:
  • Using energy conservation across an inelastic collision.
  • Conserving momentum in a direction where external impulse is significant (for example, friction during contact).
  • Forgetting to treat velocity as signed in 1D; “sticking” does not mean the final velocity is automatically positive.

Momentum and Collisions in Two Dimensions (Vectors)

Real collisions are often not head-on. Once motion can occur in both $x$ and $y$, momentum conservation becomes a vector equation. The ideas are the same: justify that external impulse is negligible (in the relevant directions), then conserve momentum component-by-component.

The essential shift: treat momentum as a vector, not a scalar

In 2D, total momentum is

$\vec P = \sum_i m_i\vec v_i$

Conservation of momentum gives one vector equation, which becomes two component equations:

$\sum_i m_i v_{ix,i} = \sum_i m_i v_{ix,f}$

$\sum_i m_i v_{iy,i} = \sum_i m_i v_{iy,f}$

You generally solve these simultaneously.

When is momentum conserved in 2D problems?

Momentum is conserved in a direction when external impulse in that direction is negligible. Common AP setups where this is reasonable include pucks colliding on an air table (negligible friction) and explosions/fragmentation over a very short time interval.

Two-dimensional explosions: vector balancing

If an object initially at rest breaks into multiple pieces,

$\vec P_i = \vec 0$

so

$\sum_k \vec p_{k,f} = \vec 0$

Graphically, the final momentum vectors can be arranged tip-to-tail to form a closed polygon.

Two-dimensional collisions: elastic vs. inelastic

The classification is the same:

• Momentum is conserved (isolated system).
• Kinetic energy is conserved only if elastic.

A practical point: 2D elastic collisions can involve more unknowns than equations unless additional information is provided (angles, one object initially at rest, symmetry, etc.). AP problems are typically posed with enough constraints to solve.

Example 1: Two pucks stick together (perfectly inelastic) in 2D

Puck 1 of mass $m_1$ moves with velocity $\vec v_{1i}$. Puck 2 of mass $m_2$ is initially at rest. They collide and stick.

$m_1\vec v_{1i} = (m_1+m_2)\vec v_f$

so

$\vec v_f = \frac{m_1}{m_1+m_2}\vec v_{1i}$

If both pucks move initially,

$\vec v_f = \frac{m_1\vec v_{1i} + m_2\vec v_{2i}}{m_1+m_2}$

Example 2: 2D elastic collision with angles given (setup)

A puck of mass $m$ moves at speed $v_0$ in the +x direction and collides elastically with an identical puck initially at rest. After the collision, puck A moves at angle $\theta$ above +x with speed $v_A$, and puck B moves at angle $\phi$ below +x with speed $v_B$.

Momentum components:

$v_0 = v_A\cos\theta + v_B\cos\phi$

$0 = v_A\sin\theta - v_B\sin\phi$

Elastic energy (equal masses):

$v_0^2 = v_A^2 + v_B^2$

From the $y$ equation:

$v_A\sin\theta = v_B\sin\phi$

This system can be solved for unknown speeds given angles (or vice versa). A famous special case: for equal masses with one initially at rest, the two final velocity vectors are perpendicular in an elastic collision.

Example 3: Explosion in midair conserving momentum during the split

A projectile of total mass $M$ has velocity $\vec v$ and breaks into two pieces of masses $m_1$ and $m_2$ with velocities $\vec v_1$ and $\vec v_2$ immediately after. Neglect external impulse during the brief explosion:

$M\vec v = m_1\vec v_1 + m_2\vec v_2$

Solve for an unknown velocity vector if needed:

$\vec v_2 = \frac{M\vec v - m_1\vec v_1}{m_2}$

Common “vector discipline” habits that prevent errors

A reliable approach is to (1) draw axes and label angles relative to axes, (2) write momentum conservation as two separate component equations, (3) be consistent about sines and cosines based on how angles are defined, and (4) only then plug in numbers.

Exam Focus

• Typical question patterns:
  • 2D inelastic collision (often sticking): compute final velocity vector via component momentum conservation.
  • 2D elastic collision with one object initially at rest: use momentum components plus kinetic energy conservation.
  • Explosion/fragmentation in 2D: solve for an unknown velocity vector from the vector momentum equation.
• Common mistakes:
  • Conserving “speed” rather than conserving momentum components.
  • Using the wrong trig component (mixing sine and cosine because the angle reference is unclear).
  • Forgetting that momentum conservation gives two equations in 2D, but you may still need an additional constraint (elastic energy conservation or given angles) to fully solve.