Two-Dimensional Collisions

AP Physics 1· AP Physics 1 CED — Momentum· 14 min read

1. Fundamentals of Two-Dimensional Collisions★★☆☆☆⏱ 2 min

Two-dimensional collisions are collisions where objects move at angles to each other before or after impact, rather than along a single straight line. This is the most common real-world collision type, from billiard balls to intersection crashes, and extends 1D collision concepts you already learned.

📘 Definition

Two-Dimensional Collision

A collision where at least one object moves along a direction not collinear with the initial motion of the system, requiring vector component analysis to solve.

By convention, we almost always align the x-axis along the direction of motion of the incoming (incident) object to simplify calculations, with the y-axis perpendicular to that direction. Any collision where final velocity vectors are not all along the original incident direction counts as a 2D collision.

2. Conservation of Momentum in Perpendicular Components★★★☆☆⏱ 4 min

The fundamental principle that governs all 2D collisions (with no net external force on the system) is that momentum is conserved separately in the x-direction and y-direction. Because momentum is a vector, each component of total momentum is conserved independently, turning a complex 2D problem into two separate 1D problems you already know how to solve.

\sum p_{x,\text{initial}} = \sum p_{x,\text{final}} \\ \sum p_{y,\text{initial}} = \sum p_{y,\text{final}}

For the common case where the incident object ( $m_{1}$ ) moves along the +x axis before collision, and the target ( $m_{2}$ ) is initially at rest, the equations simplify to:

m_1 v_{1i} = m_1 v_{1f} \cos\theta_1 + m_2 v_{2f} \cos\theta_2 \\ 0 = m_1 v_{1f} \sin\theta_1 + m_2 v_{2f} \sin\theta_2

where $θ_{1}$ is the angle of $m_{1}$ above the x-axis, and $θ_{2}$ is the angle of $m_{2}$ below the x-axis, so their sine terms have opposite signs to give a total final y-momentum of zero.

📐 Worked Example

A 0.16 kg cue ball moving at 2.5 m/s along the +x-axis strikes a stationary 0.16 kg 8-ball. After the collision, the cue ball moves at 1.5 m/s at 35° above the +x-axis. What is the speed of the 8-ball after the collision?

1
Set up coordinate system: total initial momentum is:
2
$p_{x,i} = m v_{i} = (0.16\ \text{kg})(2.5\ \text{m/s}) = 0.4\ \text{kg·m/s}, \quad p_{y,i} = 0$
3
Calculate x and y components of the cue ball's final momentum:
4
$p_{1x,f} = (0.16)(1.5)\cos(35^\circ) \approx 0.24 \cdot 0.819 \approx 0.197\ \text{kg·m/s} \\ p_{1y,f} = (0.16)(1.5)\sin(35^\circ) \approx 0.24 \cdot 0.574 \approx 0.138\ \text{kg·m/s}$
5
Apply conservation of momentum to find the 8-ball's momentum components:
6
$p_{2x,f} = p_{x,i} - p_{1x,f} = 0.4 - 0.197 = 0.203\ \text{kg·m/s} \\ p_{2y,f} = p_{y,i} - p_{1y,f} = 0 - 0.138 = -0.138\ \text{kg·m/s}$
7
Calculate total momentum and speed for the 8-ball:
8
$p_{2f} = \sqrt{(0.203)^2 + (-0.138)^2} \approx 0.245\ \text{kg·m/s} \\ v_{2f} = p_{2f}/m_2 = 0.245 / 0.16 \approx 1.5\ \text{m/s}$

Exam tip:

Always align your coordinate system with the initial motion of the incident object to set one initial momentum component to zero, cutting calculation work and reducing sign errors.

3. Perfectly Inelastic Two-Dimensional Collisions★★★☆☆⏱ 3 min

In a perfectly inelastic collision, the two objects stick together after impact, so they share the same final velocity vector. This simplifies the momentum equations dramatically, because we only have one final velocity (two unknowns: speed and direction) to solve for, instead of two separate velocities. Unlike elastic 2D collisions, we do not need kinetic energy conservation to solve for unknowns here, since the two momentum conservation equations are sufficient.

The most common perfectly inelastic 2D scenario tested on the AP exam is perpendicular collisions at intersections, where two moving objects collide and stick together. For this scenario, we just assign one direction of motion to the x-axis and the other to the y-axis, then conserve momentum in each component as usual.

📐 Worked Example

A 1200 kg car moving east at 15 m/s collides with a 1500 kg car moving north at 10 m/s at an intersection. The cars lock together and move as one after the collision. What is the magnitude of their combined velocity immediately after impact?

1
Assign coordinate system: +x = east, +y = north. Calculate total initial momentum in each direction:
2
$p_{x,i} = m_A v_A = (1200\ \text{kg})(15\ \text{m/s}) = 18{,}000\ \text{kg·m/s} \\ p_{y,i} = m_B v_B = (1500\ \text{kg})(10\ \text{m/s}) = 15{,}000\ \text{kg·m/s}$
3
Total mass after collision: $M = 1200 + 1500 = 2700 kg$ . By conservation of momentum, total final momentum equals total initial momentum.
4
Calculate final speed from total momentum:
5
$p_f = \sqrt{p_{x,f}^2 + p_{y,f}^2} = \sqrt{(18000)^2 + (15000)^2} \approx 23{,}430\ \text{kg·m/s} \\ v_f = p_f / M = 23430 / 2700 \approx 8.7\ \text{m/s}$

Exam tip:

For any question asking if kinetic energy is conserved in a perfectly inelastic collision, the answer is always no: kinetic energy is lost to deformation, heat, and sound.

4. Center of Mass Motion in Two-Dimensional Collisions★★★☆☆⏱ 3 min

Since total momentum of the system is conserved (no external forces), the velocity of the center of mass (CM) of the system is constant before, during, and after the collision. This is a powerful conceptual and problem-solving tool that is frequently tested in AP Physics 1 MCQs.

\vec{v}_{cm} = \frac{m_1 \vec{v}_{1i} + m_2 \vec{v}_{2i}}{m_1 + m_2} = \frac{m_1 \vec{v}_{1f} + m_2 \vec{v}_{2f}}{m_1 + m_2}

Because total momentum is the same before and after collision, $v_{c m}$ does not change. Even when the two colliding objects move off at angles after impact, the center of mass keeps moving in the same straight line at the same speed it had before the collision. This holds for all 2D collisions, elastic or inelastic, as long as there are no external forces acting on the system.

📐 Worked Example

For the 1200 kg eastbound car (15 m/s) and 1500 kg northbound car (10 m/s) collision described earlier, calculate the speed of the center of mass before collision, and compare it to the final speed of the combined cars after collision.

1
Calculate x and y components of $v_{c m}$ before collision:
2
$v_{cm,x} = \frac{m_A v_{A,x} + m_B v_{B,x}}{m_A + m_B} = \frac{(1200)(15) + (1500)(0)}{2700} \approx 6.67\ \text{m/s} \\ v_{cm,y} = \frac{(1200)(0) + (1500)(10)}{2700} \approx 5.56\ \text{m/s}$
3
Find the magnitude of $v_{c m}$ before collision:
4
$v_{cm} = \sqrt{(6.67)^2 + (5.56)^2} \approx 8.7\ \text{m/s}$
5
Compare to the final combined speed calculated earlier: $v_{f} = 8.7 m/s$ , which matches exactly. The center of mass speed is unchanged by the collision, as predicted by momentum conservation.

Exam tip:

For any conceptual question asking about the center of mass path after a collision, it continues moving in the same straight line at constant speed unless an external force acts.

5. AP-Style Concept Check★★★☆☆⏱ 3 min

✓ Quick check

Test your understanding of core 2D collision concepts:

A 2 kg object moving at 3 m/s along the +x-axis collides with a stationary 1 kg object. After the collision, the 2 kg object moves at 2 m/s at 30° above the +x-axis. What is the x-component of the momentum of the 1 kg object after the collision?
- 2.5 kg·m/s
- 6.0 kg·m/s
- 4.3 kg·m/s
- 2.0 kg·m/s

6. Common Pitfalls

Wrong move:

Adding total momentum as scalars (summing masses times speeds directly) instead of summing x and y components separately.

Why:

Students get used to 1D collisions where direction is just a sign, so they forget momentum is a vector in 2D.

Correct move:

Always split all momentum into x and y components, sum components separately, then combine to get the final momentum vector.

Wrong move:

Using positive signs for both sine terms when two objects scatter on opposite sides of the x-axis.

Why:

Students assume angles are always positive, so they get a non-zero total y-momentum when it should be zero.

Correct move:

Define signs based on your coordinate system before calculating, and explicitly add a negative sign for angles below the x-axis.

Wrong move:

Using kinetic energy conservation to solve for an unknown speed in a perfectly inelastic collision.

Why:

Students remember KE conservation for elastic collisions, so they incorrectly apply it to inelastic collisions.

Correct move:

Only use the two momentum conservation equations for perfectly inelastic 2D collisions; KE is not conserved, so it will give an incorrect answer.

Wrong move:

Claiming the center of mass changes direction after collision because the individual objects change direction.

Why:

Students confuse the motion of individual objects with the motion of the system's center of mass.

Correct move:

If there are no external forces, always assume center of mass velocity is constant before, during, and after collision.

Wrong move:

Dividing final momentum by only the mass of one object instead of the sum of both masses for perfectly inelastic collisions.

Why:

Students rush calculations and forget the masses stick together.

Correct move:

Write down the total combined mass $M = m_{1} + m_{2}$ at the start of any perfectly inelastic collision problem.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Conservation of x-momentum	\sum p_{x,\text{initial}} = \sum p_{x,\text{final}}	Applies to all 2D collisions with no external forces
Conservation of y-momentum	\sum p_{y,\text{initial}} = \sum p_{y,\text{final}}	Applies to all 2D collisions with no external forces
Incident along x, target at rest	m_1 v_{1i} = m_1 v_{1f} \cos\theta_1 + m_2 v_{2f} \cos\theta_2 \ 0 = m_1 v_{1f} \sin\theta_1 + m_2 v_{2f} \sin\theta_2	Sines have opposite signs for objects on opposite sides of x-axis
Perfectly inelastic final speed	v_f = \frac{\sqrt{(\sum p_{x,i})^2 + (\sum p_{y,i})^2}}{m_1 + m_2}	Only for collisions where objects stick together
Center of mass velocity	\vec{v}_{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}	Constant for all collisions if no external forces
Kinetic energy conservation	\sum KE_{\text{initial}} = \sum KE_{\text{final}}	Only applies to elastic 2D collisions, never inelastic
Perfectly inelastic KE	KE_{\text{final}} < KE_{\text{initial}}	Energy lost to deformation, heat, and sound
Explosion KE	KE_{\text{final}} > KE_{\text{initial}}	Internal energy converted to kinetic energy

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Solve for speed after 2D collision
2022 · FRQ
Center of mass after 2D collision

What's Next

Two-dimensional collisions are the capstone of the momentum unit, applying vector skills and momentum conservation you learned throughout Unit 5 to the most realistic collision scenarios. Mastering component conservation for 2D collisions is a prerequisite for understanding exploding projectiles, a common multi-concept topic in AP Physics 1 FRQs, and also builds vector problem-solving skills you will need for circular motion, gravitation, and torque. After this topic, you will connect momentum and energy conservation to solve multi-concept system problems, and extend momentum principles to extended objects and rotational motion. This topic reinforces the core principle that vector conservation laws hold independently for each component, a rule that applies across all areas of physics.