Two-Dimensional Collisions — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Conservation of momentum in perpendicular coordinate axes, component analysis for elastic and perfectly inelastic two-dimensional collisions, center of mass motion, and solution methods for AP Physics 1 exam questions.

You should already know: Conservation of one-dimensional momentum, vector component resolution, conservation of kinetic energy for elastic collisions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Two-Dimensional Collisions?

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 type of collision in the real world—from billiard balls to car crashes at intersections to subatomic particle scattering—so it is a natural extension of 1D collision concepts you already learned. According to the AP Physics 1 Course and Exam Description (CED), this topic falls under Unit 5: Momentum, which makes up 14-18% of the total AP Physics 1 exam score. Two-dimensional collision problems appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with vector skills or energy concepts to test your ability to apply momentum conservation to non-trivial scenarios. We use a standard notation convention here: 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

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. This is the core trick that turns a complex 2D problem into two separate 1D problems, which you already know how to solve. The formal equations are: $$ \sum p_{x,\text{initial}} = \sum p_{x,\text{final}} $$ $$ \sum p_{y,\text{initial}} = \sum p_{y,\text{final}} $$ This works because force is also a vector: Newton’s third law means internal forces between colliding objects cancel out in both directions, so no net change in total momentum occurs in either direction. By convention, we set the incident object (mass $m_1$) moving along the +x axis before collision, so its initial y-momentum is zero. If the target object ($m_2$) is initially at rest, both its initial x and y momentum are zero, simplifying the equations 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 ${\theta }_1$ is the angle of $m_1$ above the x-axis, and ${\theta }_2$ is the angle of $m_2$ below the x-axis, so their sine terms have opposite signs to give a total of zero.

Worked Example

Problem: 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?

Solution:

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

Exam tip: Always align your coordinate system with the initial motion of the incident object to set one initial momentum component to zero, which cuts your calculation work in half and reduces the chance of sign errors.

3. Perfectly Inelastic Two-Dimensional Collisions

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 to solve for the two unknowns. 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

Problem: 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?

Solution:

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

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, so final KE is always less than initial KE.

4. Center of Mass Motion in Two-Dimensional Collisions

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. The vector formula for center of mass velocity is: $$ \vec{v}{cm} = \frac{m_1 \vec{v}{1i} + m_2 \vec{v}{2i}}{m_1 + m_2} = \frac{m_1 \vec{v}{1f} + m_2 \vec{v}{2f}}{m_1 + m_2} $$ Because total momentum is the same before and after collision, $\vec{v}{cm}$ 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

Problem: 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.

Solution:

1. Calculate x and y components of ${\vec{v}}_{cm}$ before collision: $v_{cm,x}=\frac{m_A{v}_{A,x}+m_B{v}_{B,x}}{m_A+m_B}=\frac{(1200)(15)+(1500)(0)}{2700}\approx 6.67\text{m/s}$ $v_{cm,y}=\frac{(1200)(0)+(1500)(10)}{2700}\approx 5.56\text{m/s}$
2. Find the magnitude of $v_{cm}$ before collision: $v_{cm}=\sqrt{(6.67{)}^2+(5.56{)}^2}\approx 8.7\text{m/s}$
3. Compare to the final combined speed calculated earlier: $v_f=8.7\text{m/s}$, 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, the answer is always that it continues moving in the same straight line at constant speed, unless an external force acts on the system after collision.

5. Common Pitfalls (and how to avoid them)

• 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.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

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? A) 2.5 kg·m/s B) 6.0 kg·m/s C) 4.3 kg·m/s D) 2.0 kg·m/s

Worked Solution: By conservation of momentum, total x-momentum before collision equals total x-momentum after collision. Initial total x-momentum is $p_{x,i}=(2\text{kg})(3\text{m/s})=6\text{kgcdotpm/s}$. The x-component of the 2 kg object's final momentum is $p_{1x,f}=(2\text{kg})(2\text{m/s})\cos (30^{\circ })\approx 3.46\text{kgcdotpm/s}$. Subtract to find the x-component of the 1 kg object's momentum: $p_{2x,f}=6-3.46\approx 2.5\text{kgcdotpm/s}$. The correct answer is A.

Question 2 (Free Response)

An astronaut floating in deep space (no external forces) throws a tool in a direction perpendicular to the system's initial motion. The astronaut has a mass of 80 kg, and the tool has a mass of 5 kg. Initially, the astronaut + tool system is moving at 2 m/s in the +x direction. The astronaut throws the tool with a speed of 10 m/s in the +y direction relative to the system's center of mass. (a) What is the velocity of the center of mass of the system after the throw? Explain your reasoning. (b) What is the y-component of the astronaut's velocity after the throw? (c) Is the kinetic energy of the system after the throw greater than, less than, or equal to the kinetic energy before the throw? Justify your answer.

Worked Solution: (a) The velocity of the center of mass after the throw is still 2 m/s in the +x direction. There are no external forces acting on the astronaut-tool system, so total momentum is conserved, meaning the velocity of the center of mass cannot change from its initial constant value. (b) Total y-momentum of the system is zero before and after the throw, so $m_a{v}_{a,y}+m_t{v}_{t,y}=0$. Substituting values: $80v_{a,y}+(5)(10)=0⟹v_{a,y}=-50/80=-0.625\text{m/s}$. The y-component is -0.625 m/s (0.625 m/s in the -y direction). (c) The kinetic energy after the throw is greater than before the throw. A throw is an explosion, the reverse of a perfectly inelastic collision: internal chemical energy from the astronaut's muscles is converted to kinetic energy, so total kinetic energy increases.

Question 3 (Application / Real-World Style)

A park ranger fires a tranquilizer dart horizontally at 400 m/s, perpendicular to the direction a feral hog is running. The 0.05 kg dart hits the 50 kg hog (initially moving at 3 m/s in the +x direction) and stops inside it. What is the hog's speed immediately after being hit?

Worked Solution:

1. Assign +x as the hog's initial direction, +y as the dart's direction of motion. Calculate initial momentum components: $p_{x,i}=(50\text{kg})(3\text{m/s})=150\text{kgcdotpm/s}$, $p_{y,i}=(0.05\text{kg})(400\text{m/s})=20\text{kgcdotpm/s}$
2. Total mass after collision: $M=50+0.05=50.05\text{kg}$. Total final momentum equals initial momentum: $p_f=\sqrt{(150{)}^2+(20{)}^2}\approx 151.3\text{kgcdotpm/s}$
3. Final speed: $v_f=151.3/50.05\approx 3.02\text{m/s}$. In this context, the dart is much less massive than the hog, so it only slightly increases the hog's total speed and deflects it slightly from its original path, which matches real-world expectations.

7. Quick Reference Cheatsheet

8. 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 also reinforces the core physics principle that vector conservation laws hold independently for each component, a rule that applies across all areas of physics.

• One-Dimensional Conservation of Momentum
• Elastic and Inelastic Collisions
• Center of Mass for Systems of Objects
• Energy and Momentum in Collisions