Collisions in Systems of Particles (AP Physics C: Mechanics, Unit 4)

Elastic Collisions (1D and 2D)

What an elastic collision is

A collision is an interaction over a short time interval during which two objects exert large forces on each other, changing their velocities. In the momentum unit, the key idea is that during this short time, external impulses (from forces outside the two-object system) are often negligible compared with the internal collision forces. That lets you treat the pair of objects as an (approximately) isolated system for the duration of impact.

An elastic collision is a collision in which two things are conserved for the system:

1. Total linear momentum is conserved.
2. Total kinetic energy is conserved.

Momentum conservation comes from Newton’s laws when the net external impulse on the system is negligible. Kinetic energy conservation is an additional physical condition: it means no net kinetic energy is transformed into internal energy (thermal energy, sound, deformation) during the collision.

Why this matters: many real collisions (billiard balls, air-hockey pucks, some atomic collisions) are close to elastic, and “elastic” is also the cleanest model to practice the momentum method. It forces you to coordinate momentum (a vector) with kinetic energy (a scalar), and that combination is exactly what makes these problems interesting.

Momentum conservation (always your starting point)

For two objects with masses $m_1$ and $m_2$ and velocities before collision $\vec v_{1i}$, $\vec v_{2i}$ and after collision $\vec v_{1f}$, $\vec v_{2f}$, conservation of linear momentum is

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

This is a vector equation. In 1D, it’s one equation. In 2D, it represents two component equations (x and y).

Common pitfall: students sometimes treat momentum as a scalar even in 2D. Always decide on a coordinate system and write x and y component equations if motion is not purely along one line.

Kinetic energy conservation (what makes it elastic)

Elastic collisions also satisfy

$\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

Here $v$ is the speed (magnitude of velocity). This is a scalar equation and does not carry direction information.

A subtle but important point: kinetic energy can be conserved even though forces during the collision are huge and complicated, because those forces are internal to the system and act over very short times. Energy conservation is not automatic; it depends on whether the interaction is “lossless” in the sense of deformation/heat/sound.

1D elastic collisions: how they work and how to solve

In 1D, everything lies along one line (take it as the x-axis). Then momentum conservation becomes

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

and kinetic energy conservation is

$\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

You now have two equations for the two unknowns $v_{1f}$ and $v_{2f}$.

A powerful 1D elastic property: relative speed reverses

For a 1D elastic collision, the relative speed of approach equals the relative speed of separation (with opposite sign). Algebraically,

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

This is not a separate physical law; it’s equivalent to combining momentum and kinetic energy conservation for 1D elastic collisions. It’s often easier to use than the squared-energy equation because it avoids quadratics.

Why it helps: with momentum conservation plus the relative-speed relation, you can solve most 1D elastic collisions cleanly.

Closed-form results (useful, but don’t memorize blindly)

If you solve the momentum equation together with the relative-speed relation, you obtain

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

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

These are correct for any 1D elastic collision. Still, it’s safer on an exam to derive as needed or to rely on the two-equation method, because sign mistakes (choosing the positive direction) are more common than algebra mistakes.

Worked example (1D): cart hits a stationary cart (elastic)

A cart of mass $m_1 = 2.0\ \text{kg}$ moves right at $v_{1i} = 3.0\ \text{m/s}$ and collides elastically with a stationary cart of mass $m_2 = 1.0\ \text{kg}$ with $v_{2i} = 0$. Find $v_{1f}$ and $v_{2f}$.

Use momentum conservation:

$2(3) + 1(0) = 2v_{1f} + 1v_{2f}$

So

$6 = 2v_{1f} + v_{2f}$

Use the 1D elastic relative-speed relation:

$3 - 0 = -(v_{1f} - v_{2f})$

So

$3 = -v_{1f} + v_{2f}$

Now solve the system. From the second equation:

$v_{2f} = v_{1f} + 3$

Substitute into momentum equation:

$6 = 2v_{1f} + (v_{1f} + 3)$

$6 = 3v_{1f} + 3$

$v_{1f} = 1.0\ \text{m/s}$

Then

$v_{2f} = 1.0 + 3.0 = 4.0\ \text{m/s}$

Interpretation: the heavier cart keeps moving right but slows down; the lighter cart shoots forward faster.

Common mistake to avoid: forgetting that velocities carry sign. If left is negative, keep that consistent from start to finish.

2D elastic collisions: what changes in two dimensions

In 2D, you still conserve total momentum, but now it’s a vector:

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

Write components:

$m_1v_{1i,x} + m_2v_{2i,x} = m_1v_{1f,x} + m_2v_{2f,x}$

$m_1v_{1i,y} + m_2v_{2i,y} = m_1v_{1f,y} + m_2v_{2f,y}$

Kinetic energy conservation is still

$\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

Why 2D is harder

Count unknowns. In 2D, the final velocities have components, so you often have more unknowns than equations. Momentum gives two equations, kinetic energy gives one, for three total equations. But you may have four unknown components (two for each final velocity). That means you typically need additional information:

• A given scattering angle (direction of one object after collision)
• A constraint like “object 2 initially at rest” and “masses are equal,” which produces special geometry
• A known line of impact (common in problems that specify the collision is smooth and impulsive along a given direction)

A classic AP-style scenario: one object is initially at rest; after an elastic collision, you’re given one or both final directions and asked for speeds.

Special but important case: equal masses, target initially at rest

If $m_1 = m_2$ and object 2 starts at rest, an elastic collision has a very useful geometric result: the two final velocity vectors are perpendicular.

This comes from combining momentum and energy conservation:

• Momentum: $\vec v_{1i} = \vec v_{1f} + \vec v_{2f}$
• Energy: $v_{1i}^2 = v_{1f}^2 + v_{2f}^2$

If you square the momentum vector relation (dot both sides with itself), you get

$v_{1i}^2 = v_{1f}^2 + v_{2f}^2 + 2\vec v_{1f}\cdot\vec v_{2f}$

Comparing with the energy equation implies

$\vec v_{1f}\cdot\vec v_{2f} = 0$

which means the vectors are perpendicular.

Why it matters: it turns a messy 2D algebra problem into a geometry/trig problem.

Worked example (2D): billiard-ball style, equal masses

A puck (1) of mass $m$ moves in the +x direction at speed $v_{1i} = 5.0\ \text{m/s}$ and hits an identical puck (2) initially at rest. After an elastic collision, puck 1 moves at angle $30^\circ$ above +x. Find the speeds $v_{1f}$ and $v_{2f}$.

Because masses are equal and puck 2 starts at rest, the final velocities are perpendicular. If puck 1 is at $30^\circ$, puck 2 must be at $-60^\circ$ (so that the angle between them is $90^\circ$ and momentum still adds to a purely +x initial momentum).

Use momentum components. Initial momentum is entirely x:

$mv_{1i} = mv_{1f}\cos 30^\circ + mv_{2f}\cos 60^\circ$

Cancel $m$:

$5.0 = v_{1f}\cos 30^\circ + v_{2f}\cos 60^\circ$

In y:

$0 = v_{1f}\sin 30^\circ - v_{2f}\sin 60^\circ$

Solve y-equation for $v_{2f}$:

$v_{1f}\sin 30^\circ = v_{2f}\sin 60^\circ$

$v_{2f} = v_{1f}\frac{\sin 30^\circ}{\sin 60^\circ}$

Using $\sin 30^\circ = 1/2$ and $\sin 60^\circ = \sqrt{3}/2$ gives

$v_{2f} = \frac{v_{1f}}{\sqrt{3}}$

Substitute into x-equation. Use $\cos 30^\circ = \sqrt{3}/2$ and $\cos 60^\circ = 1/2$:

$5.0 = v_{1f}\frac{\sqrt{3}}{2} + \left(\frac{v_{1f}}{\sqrt{3}}\right)\frac{1}{2}$

$5.0 = v_{1f}\left(\frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}}\right)$

Combine:

$\frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}} = \frac{3 + 1}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$

So

$5.0 = v_{1f}\frac{2}{\sqrt{3}}$

$v_{1f} = 5.0\frac{\sqrt{3}}{2} = 4.33\ \text{m/s}$

Then

$v_{2f} = \frac{4.33}{\sqrt{3}} = 2.50\ \text{m/s}$

Quick reasonableness check with energy:

$v_{1f}^2 + v_{2f}^2 = (4.33)^2 + (2.50)^2 = 18.75 + 6.25 = 25.0 = v_{1i}^2$

Works.

Common mistake to avoid: assuming “perpendicular final velocities” for any elastic collision. That perpendicular result needs equal masses and the target initially at rest.

Notation reference (common symbols)

Exam Focus

• Typical question patterns
  • “Two objects collide elastically; find final speeds/velocities” using momentum + kinetic energy (often 1D).
  • 2D “puck collision” where one object is initially at rest and you’re given one scattering angle.
  • Conceptual comparisons: which collision is elastic vs inelastic based on kinetic energy change.
• Common mistakes
  • Using kinetic energy conservation for a collision that is not stated (or implied) to be elastic.
  • In 2D, writing only one momentum equation instead of x and y components.
  • Dropping minus signs by confusing “speed” with “velocity” in 1D.

Inelastic Collisions

What an inelastic collision is (and what is still conserved)

An inelastic collision is a collision in which total kinetic energy is not conserved for the colliding objects. Some kinetic energy is converted into other forms of energy: thermal energy from friction-like deformation, sound, permanent deformation, or internal vibrations.

The critical point for AP Physics C: momentum is still conserved (as long as the net external impulse on the system is negligible during the collision).

So for an inelastic collision you generally use:

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

but you do not set kinetic energies equal.

Why it matters: a huge fraction of real-world collisions are inelastic. If you mistakenly assume kinetic energy conservation, you’ll often get an answer that looks “nice” but is physically impossible.

Perfectly inelastic collisions (objects stick together)

A perfectly inelastic collision is the extreme case where the objects stick together after impact and move with a common final velocity $\vec v_f$.

Then momentum conservation becomes especially simple:

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

Solving for the final velocity:

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

This formula is fundamentally the same as a center-of-mass velocity: when objects stick, they move together at the system’s center-of-mass velocity (assuming no external impulse during the collision).

Energy change in a perfectly inelastic collision

Even though you typically do not need the energy loss to find velocities, AP questions sometimes ask for how much kinetic energy is transformed.

Initial kinetic energy:

$K_i = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2$

Final kinetic energy:

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

The change in kinetic energy of the two-object system is

$\Delta K = K_f - K_i$

For a perfectly inelastic collision, $\Delta K$ is negative (unless the objects already shared the same velocity, in which case no real “collision” effect occurs).

A frequent conceptual confusion: students think “inelastic means energy is not conserved,” then conclude “energy disappears.” Total energy is still conserved, but it leaves the translational kinetic energy account and goes into internal energy modes.

How to approach general inelastic collisions (objects don’t stick)

“Inelastic” does not automatically mean “stick.” Many collisions are partially inelastic: the objects bounce apart, but with less total kinetic energy than before.

In that general case:

• Momentum conservation still holds.
• You need additional information to determine the final state (because you no longer have kinetic energy conservation as a second equation).

What can provide the extra information?

• A given final speed or direction (common in problems)
• A stated relationship like “object 2 leaves at angle $\theta$” in 2D
• Sometimes a coefficient of restitution $e$ is provided in 1D or along a known impact direction. If used, it is defined (along the line of impact) by

$e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}}$

In 1D, that becomes

$e = \frac{v_{2f} - v_{1f}}{v_{1i} - v_{2i}}$

with the understanding that the numerator and denominator are taken along the same line, with consistent sign convention. For an elastic collision, $e = 1$; for a perfectly inelastic collision (sticking), $e = 0$.

If a problem does not mention $e$, you typically should not invent it; use only the relationships the problem gives.

Worked example (1D): perfectly inelastic, ball + block

A ball of mass $m_1 = 0.20\ \text{kg}$ travels right at $v_{1i} = 10\ \text{m/s}$ and hits a block of mass $m_2 = 0.80\ \text{kg}$ initially at rest on a frictionless surface. The ball embeds in the block (perfectly inelastic). Find the final speed and the kinetic energy lost.

Step 1: Momentum conservation to get final velocity.

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f$

$0.20(10) + 0.80(0) = (1.00)v_f$

$v_f = 2.0\ \text{m/s}$

Step 2: Compute kinetic energies to find the loss.

$K_i = \frac{1}{2}(0.20)(10^2) + \frac{1}{2}(0.80)(0^2)$

$K_i = 10\ \text{J}$

$K_f = \frac{1}{2}(1.00)(2.0^2)$

$K_f = 2.0\ \text{J}$

$\Delta K = K_f - K_i = 2.0 - 10 = -8.0\ \text{J}$

So 8.0 J of kinetic energy is converted into internal energy (deformation, heat, sound).

Common mistake to avoid: trying to conserve kinetic energy to solve for $v_f$. If you do that here, you’d incorrectly get $v_f = 10\ \text{m/s}$, which would violate momentum conservation.

Worked example (2D): sticking together with angled motion

A cart of mass $m_1 = 3.0\ \text{kg}$ moves east at $4.0\ \text{m/s}$ and collides with a cart of mass $m_2 = 1.0\ \text{kg}$ moving north at $6.0\ \text{m/s}$. They stick together. Find the final velocity vector.

Choose +x east, +y north.

Initial momentum components:

$p_{x,i} = m_1v_{1i,x} + m_2v_{2i,x} = 3.0(4.0) + 1.0(0) = 12\ \text{kg m/s}$

$p_{y,i} = m_1v_{1i,y} + m_2v_{2i,y} = 3.0(0) + 1.0(6.0) = 6\ \text{kg m/s}$

They stick, so total mass is $m_1 + m_2 = 4.0\ \text{kg}$ and

$v_{f,x} = \frac{p_{x,i}}{m_1 + m_2} = \frac{12}{4.0} = 3.0\ \text{m/s}$

$v_{f,y} = \frac{p_{y,i}}{m_1 + m_2} = \frac{6}{4.0} = 1.5\ \text{m/s}$

So

$\vec v_f = (3.0\ \text{m/s})\hat i + (1.5\ \text{m/s})\hat j$

Speed:

$v_f = \sqrt{3.0^2 + 1.5^2} = 3.35\ \text{m/s}$

Direction above east:

$\theta = \tan^{-1}\left(\frac{1.5}{3.0}\right) = 26.6^\circ$

The important habit here is treating momentum as a vector. Sticking collisions in 2D are often “easy points” if you reliably resolve components.

Real-world interpretation: why kinetic energy drops

In real collisions, objects deform. Deformation means internal forces act over internal displacements, converting some translational kinetic energy into microscopic energy. The more deformation and internal friction, the more inelastic the collision.

A helpful analogy: momentum conservation is like “accounting for motion” that must balance if there is no external impulse; kinetic energy is like “usable mechanical energy” that can be spent on internal changes during a collision.

A practical problem-solving workflow for inelastic collisions

1. Define the system (usually the two colliding objects).
2. Check external impulse during collision (frictionless surface, short time, etc.). If negligible, momentum is conserved.
3. Write momentum conservation in components if needed.
4. If objects stick, use the combined-mass form immediately.
5. Only compute kinetic energy changes after you have velocities.

Exam Focus

• Typical question patterns
  • Perfectly inelastic “stick together” problems in 1D or 2D asking for final speed/direction.
  • Energy accounting: “How much kinetic energy is lost?” after a sticking collision.
  • Multi-step setups: a collision followed by motion up a ramp or compression of a spring (momentum for the collision, then energy for the post-collision motion).
• Common mistakes
  • Assuming kinetic energy is conserved in any collision scenario.
  • Mixing conservation laws across time intervals: using momentum conservation during a long interval where external forces (like friction) matter.
  • In 2D, computing final speed from momentum magnitude alone without respecting x and y components (direction matters).