One-Dimensional Collisions

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

1. Fundamentals of One-Dimensional Collisions★★☆☆☆⏱ 3 min

One-dimensional (1D) collisions are short-duration, high-force interactions where all motion before, during, and after the collision occurs along a single straight line. They are often called 'head-on collisions' on AP exams.

📘 Definition

One-dimensional collision

1D collision

A short-duration interaction between objects where all motion is constrained to a single straight line; external forces are negligible during the collision

Example:

A cart sliding straight into another stationary cart on a linear track

Because collisions are very short in duration, external forces like gravity and friction are negligible during the interaction. This means the total momentum of the closed system of colliding objects is always conserved, regardless of collision type. This topic makes up a significant portion of Unit 5 (Momentum), which accounts for 14–18% of your total AP Physics 1 score.

2. Conservation of Momentum and Collision Classification★★☆☆☆⏱ 4 min

The core principle for all 1D collisions is conservation of momentum for a closed, isolated system: total momentum before collision equals total momentum after collision, provided no net external force acts. For two objects, this gives the general equation:

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

Where $m_{1}, m_{2}$ are object masses, $v_{1 i}, v_{2 i}$ are initial velocities, and $v_{1 f}, v_{2 f}$ are final velocities. Signs are critical: velocity is negative if it points opposite your chosen positive direction. Collisions are classified by their change in total kinetic energy:

Elastic collisions: Both momentum and kinetic energy are conserved; no energy is lost to heat, sound, or deformation.
Inelastic collisions: Momentum is conserved, but some 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 objects stick together after collision, sharing the same final velocity $v_{f}$ . For this case, the momentum equation simplifies to:

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

📐 Worked Example

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?

1
Choose the positive direction to be right. Assign known values:
$v_{1i} = 3 \text{ m/s}, v_{2i} = 0 \text{ m/s}, m_1 = 2 \text{ kg}, m_2 = 1 \text{ kg}$
2
Recognize this is a perfectly inelastic collision, so both objects share the same final velocity $v_{f}$ .
3
Substitute values into the simplified momentum conservation equation:
$(2)(3) + (1)(0) = (2 + 1) v_f$
4
Simplify and solve for $v_{f}$ :
$6 = 3v_f \implies v_f = 2 \text{ m/s}$
5
The positive result confirms the combined block moves to the right.

Exam tip:

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

3. Elastic One-Dimensional Collisions★★★★☆⏱ 4 min

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 laws to get a simple linear relation between velocities. This relation gives the relative velocity rule:

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

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 cuts solving time by more than half.

📐 Worked Example

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?

1
Define positive direction as right. Assign known values:
$v_{1i} = +4 \text{ m/s}, v_{2i} = -2 \text{ m/s}, m_1 = 1 \text{ kg}, m_2 = 3 \text{ kg}$
2
Write the conservation of momentum equation:
$(1)(4) + (3)(-2) = v_{1f} + 3v_{2f} \implies -2 = v_{1f} + 3v_{2f}$
3
Write the relative velocity rule for elastic collisions:
$4 - (-2) = v_{2f} - v_{1f} \implies 6 = v_{2f} - v_{1f}$
4
Add the two equations to eliminate $v_{1 f}$ :
$4 = 4v_{2f} \implies v_{2f} = 1 \text{ m/s}$
5
Substitute back to solve for $v_{1 f}$ :
$v_{1f} = v_{2f} - 6 = -5 \text{ m/s}$
6
Final result: the 1 kg ball moves left at 5 m/s, and the 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. Memorizing it saves significant time on multi-part problems.

4. Kinetic Energy Change in Collisions★★★☆☆⏱ 3 min

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

$Δ K E = 0$ : Elastic collision, no kinetic energy lost
$Δ K E < 0$ : Inelastic collision, kinetic energy is lost to other forms (kinetic energy cannot increase in a collision between two free objects)
Maximum possible kinetic energy loss always occurs in perfectly inelastic collisions, as objects stick together and have the minimum possible final total kinetic energy consistent with momentum conservation

Lost kinetic energy is converted to heat, sound, work done to deform objects, or stored as internal potential energy. To classify a collision, you must always calculate $Δ K E$ explicitly, even if objects bounce apart.

📐 Worked Example

For the perfectly inelastic collision from the earlier example (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?

1
Calculate initial total kinetic energy:
$KE_i = \frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = 0.5(2)(3^2) + 0 = 9 \text{ J}$
2
Use the previously calculated final velocity ( $v_{f} = 2 m/s$ ) to find final total kinetic energy:
$KE_f = \frac{1}{2}(m_1 + m_2)v_f^2 = 0.5(3)(2^2) = 6 \text{ J}$
3
Calculate the change in kinetic energy:
$\Delta KE = 6 - 9 = -3 \text{ J}$
4
The negative sign indicates 3 J of kinetic energy is lost. Calculate percent loss:
$\frac{|\Delta KE|}{KE_i} \times 100 = \frac{3}{9} \times 100 = 33.3\%$

Exam tip:

Do not assume a collision is elastic just because the objects bounce off each other; most real bounces still lose some kinetic energy.

5. Concept Check★★★☆☆⏱ 2 min

✓ Quick check

Test your understanding with this AP-style multiple choice question:

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?
- $\frac{1}{3}$
- $\frac{1}{2}$
- $\frac{2}{3}$
- $\frac{3}{4}$
Reveal answer
2 —
Correct: Using conservation of momentum, final velocity is $v /3$ . Final kinetic energy is $\frac{1}{3}$ of the original, so $\frac{2}{3}$ is lost.

6. Common Pitfalls

Wrong move:

Assigning a positive velocity to an object moving opposite your chosen positive 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 to represent direction.

Correct move:

Write down your chosen positive direction explicitly 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 time-saving 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 incorrectly generalize momentum conservation to kinetic energy.

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.

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

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
Perfectly inelastic KE loss calculation
2022 · FRQ
Elastic vs inelastic momentum comparison

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.