One-Dimensional Collisions — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Conservation of momentum for closed collision systems, classification of one-dimensional collisions (elastic, inelastic, perfectly inelastic), kinetic energy change calculations, solving for unknown final velocities, and relative speed rules for collisions.

You should already know: Conservation of linear momentum for closed systems, definition of kinetic energy, coordinate system sign conventions for velocity in 1D.

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 One-Dimensional Collisions?

One-dimensional collisions are interactions between two or more objects where all motion before, during, and after the collision occurs along a single straight line. This simplifies problem-solving to scalar calculations with sign conventions (positive for one direction, negative for the opposite) instead of splitting vectors into multiple components. In the AP Physics 1 Course and Exam Description (CED), this topic is part of Unit 5: Momentum, which accounts for 14–18% of the total exam score. One-dimensional collisions regularly appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with energy concepts for 5–10 point parts on FRQs. A collision is defined as a short-duration interaction with a large internal force, so external forces like friction or gravity can be neglected during the collision, making momentum conserved for the system of colliding objects. You may also see the synonym "head-on collision" used on the exam to describe 1D collisions. This topic is the foundation for all collision problem-solving in AP Physics 1.

2. Conservation of Momentum and Collision Classification

The core principle governing all 1D collisions is conservation of momentum for a closed, isolated system: the total momentum of the system before the collision equals the total momentum after the collision, as long as no net external force acts during the collision. For two objects in 1D, this gives the general equation: $m_{1} v_{1 i} + m_{2} v_{2 i} = m_{1} v_{1 f} + m_{2} v_{2 f}$ Where $m_{1}, m_{2}$ are the masses of the two objects, $v_{1 i}, v_{2 i}$ are initial velocities, and $v_{1 f}, v_{2 f}$ are final velocities. Signs are critical: if an object moves opposite your chosen positive direction, its velocity and momentum are negative. Collisions are classified by their kinetic energy change:

Elastic collisions: Both momentum and kinetic energy are conserved, with no energy converted to heat, sound, or deformation.
Inelastic collisions: Momentum is conserved, but kinetic energy is lost to other forms. Most real-world collisions fall into this category.
Perfectly inelastic collisions: A special case of inelastic collision where the two objects stick together after the collision, moving with the same final velocity $v_{f}$ . This simplifies the conservation equation to: $m_{1} v_{1 i} + m_{2} v_{2 i} = (m_{1} + m_{2}) v_{f}$

Worked Example

Problem: A 2 kg block sliding right at 3 m/s on a frictionless track hits a stationary 1 kg block. After the collision, the two blocks stick together. What is their final speed? Steps:

Choose the positive direction to be right, so $v_{1 i} = 3 m/s$ , $v_{2 i} = 0 m/s$ , $m_{1} = 2 kg$ , $m_{2} = 1 kg$ .
Recognize this is a perfectly inelastic collision, so the objects share the same final velocity $v_{f}$ .
Substitute into the simplified conservation of momentum equation: $(2) (3) + (1) (0) = (2 + 1) v_{f}$ .
Simplify to solve: $6 = 3 v_{f} ⟹ v_{f} = 2 m/s$ . The positive value confirms the combined block moves right.

Exam tip: Always define your positive direction explicitly at the start of any 1D collision problem, even if the question does not ask you to. This prevents sign errors that AP graders will mark harshly.

3. Elastic One-Dimensional Collisions

For elastic 1D collisions, we have two independent conservation equations: one for momentum, one for kinetic energy. Instead of solving a system of one linear and one quadratic equation (which is slow and error-prone on the exam), we can combine the two conservation laws to derive a simple linear relation between initial and final velocities. Starting from: $m_{1} v_{1 i} + m_{2} v_{2 i} = m_{1} v_{1 f} + m_{2} v_{2 f}$ $\frac{1}{2} m_{1} v_{1 i}^{2} + \frac{1}{2} m_{2} v_{2 i}^{2} = \frac{1}{2} m_{1} v_{1 f}^{2} + \frac{1}{2} m_{2} v_{2 f}^{2}$ Rearranging and factoring both equations gives the relative velocity rule: $v_{1 i} - v_{2 i} = v_{2 f} - v_{1 f}$ This means the relative speed of approach before the collision equals the relative speed of separation after the collision. This rule is only valid for elastic 1D collisions, and it cuts solving time by more than half.

Worked Example

Problem: A 1 kg ball moving right at 4 m/s collides elastically head-on with a 3 kg ball moving left at 2 m/s. What are the final velocities of the two balls? Steps:

Define positive direction as right, so $v_{1 i} = + 4 m/s$ , $v_{2 i} = - 2 m/s$ , $m_{1} = 1 kg$ , $m_{2} = 3 kg$ .
Write the two equations for elastic collision:
1. Conservation of momentum: $(1) (4) + (3) (- 2) = v_{1 f} + 3 v_{2 f} ⟹ - 2 = v_{1 f} + 3 v_{2 f}$
2. Relative velocity rule: $4 - (- 2) = v_{2 f} - v_{1 f} ⟹ 6 = v_{2 f} - v_{1 f}$
Add the two equations to eliminate $v_{1 f}$ : $4 = 4 v_{2 f} ⟹ v_{2 f} = 1 m/s$ .
Substitute back to find $v_{1 f} = v_{2 f} - 6 = - 5 m/s$ . Final result: 1 kg ball moves left at 5 m/s, 3 kg ball moves right at 1 m/s.

Exam tip: If you forget the relative velocity rule, you can always derive it from the two conservation laws during the exam. But memorizing it saves significant time for elastic collision problems.

4. Kinetic Energy Change in Collisions

AP Physics 1 regularly asks you to calculate or interpret the change in kinetic energy during a 1D collision. For any collision, momentum is always conserved for an isolated system, but kinetic energy change $Δ K E = K E_{total, f} - K E_{total, i}$ defines the collision type:

$Δ K E = 0$ : Elastic collision, no kinetic energy lost
$Δ K E < 0$ : Inelastic collision, kinetic energy lost (all real non-elastic collisions have negative $Δ K E$ ; you cannot gain kinetic energy in a collision between two free objects)
The maximum possible kinetic energy loss always occurs in perfectly inelastic collisions, because the objects stick together, so they have the minimum possible total kinetic energy after the collision consistent with momentum conservation. Lost kinetic energy is converted to heat, sound, work done to deform objects, or stored as internal potential energy.

Worked Example

Problem: For the perfectly inelastic collision from Section 2 (2 kg block at 3 m/s hitting a stationary 1 kg block, sticks together), how much kinetic energy is lost, and what percent of the initial kinetic energy is lost? Steps:

Calculate initial total kinetic energy: $K E_{i} = \frac{1}{2} m_{1} v_{1 i}^{2} + \frac{1}{2} m_{2} v_{2 i}^{2} = 0.5 (2) (3^{2}) + 0 = 9 J$ .
Use the final velocity we found earlier ( $v_{f} = 2 m/s$ ) to calculate final total kinetic energy: $K E_{f} = \frac{1}{2} (m_{1} + m_{2}) v_{f}^{2} = 0.5 (3) (2^{2}) = 6 J$ .
Calculate kinetic energy change: $Δ K E = 6 - 9 = - 3 J$ . The negative sign indicates 3 J of kinetic energy is lost.
Calculate percent loss: $\frac{∣Δ K E ∣}{K E _{i}} \times 100 = \frac{3}{9} \times 100 = 33.3%$ .

Exam tip: If you are asked to classify a collision, always calculate $Δ K E$ explicitly. Do not assume a collision is elastic just because the objects bounce off each other — most real bounces still lose some kinetic energy.

5. Common Pitfalls (and how to avoid them)

Wrong move: Assigning a positive velocity to an object moving opposite your chosen direction, e.g. a 2 kg mass moving left gets $v = + 2 m/s$ instead of $- 2 m/s$ . Why: Students confuse speed (a scalar) with velocity (a vector) and forget 1D motion still requires signs for direction. Correct move: Write down your positive direction at the top of the problem, and check every velocity's sign before plugging into the momentum equation.
Wrong move: Using the relative velocity relation $v_{1 i} - v_{2 i} = v_{2 f} - v_{1 f}$ for inelastic or perfectly inelastic collisions. Why: Students memorize the useful relation and forget it only applies when kinetic energy is conserved. Correct move: Only use the relative velocity relation after you confirm the problem explicitly states the collision is elastic.
Wrong move: Accidentally writing $m_{1} K E_{1 i} + m_{2} K E_{2 i} = m_{1} K E_{1 f} + m_{2} K E_{2 f}$ for elastic collisions, adding an extra mass factor. Why: Students mix the form of momentum and kinetic energy equations, since both are additive for the system. Correct move: Remember that mass is already inside the kinetic energy term ( $K E = \frac{1}{2} m v^{2}$ ), so do not add it again when summing total kinetic energy.
Wrong move: Claiming kinetic energy is conserved in all collisions, just like momentum. Why: Students confuse the two conservation laws and generalize momentum conservation to kinetic energy incorrectly. Correct move: Recite "momentum always conserved, KE only conserved for elastic" to yourself before starting any collision problem.
Wrong move: For perfectly inelastic collisions, treating $v_{1 f}$ and $v_{2 f}$ as separate unknowns, leading to an unsolvable system. Why: Students forget that sticking together means the two objects move at the same velocity. Correct move: Factor out the common final velocity immediately to get $(m_{1} + m_{2}) v_{f}$ on the right-hand side of the momentum equation.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A cart of mass $m$ moving at speed $v$ collides head-on with a stationary cart of mass $2 m$ on a frictionless track. The collision is perfectly inelastic. What fraction of the original kinetic energy of the moving cart is lost during the collision? A) $\frac{1}{3}$ B) $\frac{1}{2}$ C) $\frac{2}{3}$ D) $\frac{3}{4}$

Worked Solution: First apply conservation of momentum for the perfectly inelastic collision. Initial total momentum is $m v + (2 m) (0) = m v$ , and final total momentum is $(m + 2 m) v_{f} = 3 m v_{f}$ . Equating gives $v_{f} = v /3$ . Initial kinetic energy is $K E_{i} = \frac{1}{2} m v^{2}$ , and final kinetic energy is $\frac{1}{2} (3 m) (v /3)^{2} = \frac{1}{6} m v^{2} = \frac{1}{3} K E_{i}$ . The kinetic energy lost is $K E_{i} - K E_{f} = \frac{2}{3} K E_{i}$ , so the fraction lost is $\frac{2}{3}$ . Correct answer: C.

Question 2 (Free Response)

A 0.05 kg dart is fired horizontally at a 1.0 kg wooden block that is initially at rest on a horizontal frictionless surface. The dart sticks into the block, and the combined system moves at 0.6 m/s after the collision. (a) Calculate the initial speed of the dart. (b) Is this collision elastic or inelastic? Justify your answer with a calculation of kinetic energy before and after the collision. (c) Suppose a second dart of the same mass and initial speed collides elastically with the same stationary block. Do the magnitude of the final momentum of the block increase, decrease, or stay the same compared to the inelastic case? Justify your answer without full calculation.

Worked Solution: (a) Use conservation of momentum for perfectly inelastic collision: $m_{d} v_{d i} + m_{b} v_{bi} = (m_{d} + m_{b}) v_{f}$ Substitute values: $0.05 v_{d i} + 0 = (1.05) (0.6) ⟹ v_{d i} = 0.63/0.05 = 12.6 m/s$ .

(b) Calculate initial and final kinetic energy: $K E_{i} = \frac{1}{2} (0.05) (12.6)^{2} \approx 3.97 J$ $K E_{f} = \frac{1}{2} (1.05) (0.6)^{2} \approx 0.19 J$ $Δ K E = 0.19 - 3.97 = - 3.78 J < 0$ , so kinetic energy is not conserved. The collision is inelastic.

(c) The magnitude of the block's final momentum will be larger for the elastic collision. In the elastic collision, the dart bounces backward after colliding, so its final momentum is negative (opposite its initial direction). To conserve total momentum, the block must carry all of the initial momentum of the dart plus additional momentum to cancel the dart's negative final momentum. In the inelastic case, the dart moves with the block, so the block only carries a fraction of the initial momentum. Thus, the block's final momentum is larger for the elastic collision.

Question 3 (Application / Real-World Style)

In a car crash test, a 1200 kg car moving forward at 15 m/s collides head-on with a stationary 1800 kg truck. The two vehicles lock together after the collision. What is the velocity of the combined wreckage immediately after the collision, and how much kinetic energy is dissipated (lost) in the crash?

Worked Solution:

Define forward as positive, and apply conservation of momentum for perfectly inelastic collision: $1200 (15) + 1800 (0) = (1200 + 1800) v_{f} ⟹ 18000 = 3000 v_{f} ⟹ v_{f} = 6 m/s forward$ .
Calculate initial and final kinetic energy: $K E_{i} = 0.5 (1200) (1 5^{2}) = 135000 J = 135 kJ$ $K E_{f} = 0.5 (3000) (6^{2}) = 54000 J = 54 kJ$
Kinetic energy lost = $135 - 54 = 81 kJ$ . In context, this 81 kJ of energy is dissipated as heat from friction, sound, and work done to deform the metal of the car and truck during the crash.

7. Quick Reference Cheatsheet

Category	Formula	Notes
General conservation of momentum	$m_{1} v_{1 i} + m_{2} v_{2 i} = m_{1} v_{1 f} + m_{2} v_{2 f}$	Applies to all 1D collisions in isolated systems
Perfectly inelastic collision	$m_{1} v_{1 i} + m_{2} v_{2 i} = (m_{1} + m_{2}) v_{f}$	Objects stick together, share the same final velocity
Elastic collision relative velocity	$v_{1 i} - v_{2 i} = v_{2 f} - v_{1 f}$	Only for elastic 1D collisions; approach speed = separation speed
Elastic collision kinetic energy	$\frac{1}{2} m_{1} v_{1 i}^{2} + \frac{1}{2} m_{2} v_{2 i}^{2} = \frac{1}{2} m_{1} v_{1 f}^{2} + \frac{1}{2} m_{2} v_{2 f}^{2}$	Only holds for elastic collisions
Kinetic energy change	$Δ K E = K E_{f} - K E_{i}$	$Δ K E = 0$ = elastic; $Δ K E < 0$ = inelastic
Maximum KE loss	Occurs for perfectly inelastic collisions	Always true for 1D collisions between two free objects

8. What's Next

After mastering one-dimensional collisions, you will extend the same conservation of momentum principles to two-dimensional collisions, where you split momentum into x and y components and apply conservation to each axis separately. Without mastering sign conventions and core conservation rules for 1D collisions, solving 2D collision problems will be significantly harder, as you will repeat the same 1D logic for each axis. This topic also forms the foundation for understanding center of mass motion, impulse, and even rotational collision problems later in AP Physics 1, as the same core conservation laws apply to all interactions. One-dimensional collisions are the simplest case of momentum conservation for interacting systems, so building fluency here makes all more advanced momentum problems easier to solve.

One-Dimensional Collisions — AP Physics 1 Study Guide

1. What Is One-Dimensional Collisions?

2. Conservation of Momentum and Collision Classification

Worked Example

3. Elastic One-Dimensional Collisions

Worked Example

4. Kinetic Energy Change in Collisions

Worked Example

5. Common Pitfalls (and how to avoid them)

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

Question 2 (Free Response)

Question 3 (Application / Real-World Style)

7. Quick Reference Cheatsheet

8. What's Next

More study guides