Conservation of linear momentum and collisions

AP Physics C: Mechanics· AP Physics C: Mechanics CED — Systems of Particles and Linear Momentum· 14 min read

1. Core Principle: Conservation of Linear Momentum★★☆☆☆⏱ 4 min

Conservation of linear momentum is a fundamental principle governing interactions between objects in isolated systems, and is the primary tool for analyzing collision events. This topic accounts for 14-18% of the total AP Physics C Mechanics exam weight, appearing in both multiple choice and free response, often combined with energy or kinematics concepts.

📘 Definition

Conservation of Linear Momentum

$P_{ini t ia l} = P_{f ina l}$ , $p_{i} = m_{i} v_{i}$

The total linear momentum of an isolated system (with zero net external force) remains constant over time, even when internal forces change the momentum of individual system components.

Example:

Momentum is conserved during a collision between two billiard balls on a frictionless table.

The formal condition for momentum conservation comes from Newton's second law for systems:

\frac{d\vec{P}_{total}}{dt} = \vec{F}_{net,ext}

If $F_{n e t, e x t} = 0$ , then total momentum is constant, because internal forces cancel out in equal and opposite pairs per Newton's third law. A key result is that the velocity of the center of mass of an isolated system is always constant.

📐 Worked Example

A 1000 kg rocket moving horizontally at 50 m/s explodes into two fragments. A 400 kg fragment stops immediately after the explosion. What is the velocity of the second fragment, and what is the velocity of the system’s center of mass after the explosion?

1
Define the system as both rocket fragments. No net external force acts in the horizontal direction during the short explosion, so momentum is conserved horizontally.
2
Calculate initial total momentum:
$P_i = M v_i = (1000\ \text{kg})(50\ \text{m/s}) = 50000\ \text{kg·m/s}$
3
Set equal to final momentum, solve for $v_{2}$ :
$P_f = m_1 v_1 + m_2 v_2 = (400\ \text{kg})(0) + (600\ \text{kg})v_2 = 50000 \\ v_2 = \frac{50000}{600} \approx 83.3\ \text{m/s}$
4
Calculate center of mass velocity from total final momentum:
$v_{cm} = \frac{P_f}{M_{total}} = \frac{50000}{1000} = 50\ \text{m/s}$

Exam tip:

Always explicitly define your system before writing a momentum conservation equation—this helps you catch unaccounted net external forces and earns you reasoning points on FRQs.

2. One-Dimensional Collisions★★★☆☆⏱ 4 min

Collisions in one dimension (all motion along a single line) are classified by whether kinetic energy is conserved during the interaction:

Perfectly inelastic: Objects stick together after collision and move with a common final velocity. Kinetic energy is lost to heat, deformation, or sound.
Inelastic: Kinetic energy is lost, but objects remain separate after collision.
Elastic: Both momentum and kinetic energy are conserved (an idealization, common for billiard balls or atomic collisions).

For perfectly inelastic collisions, the momentum conservation equation is:

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

For 1D elastic collisions, two conservation equations can be rearranged to a convenient relative velocity relation that avoids solving a quadratic equation, saving time on the exam:

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

📐 Worked Example

A 2 kg block moving right at 6 m/s collides elastically head-on with a 1 kg block moving left at 3 m/s. Find the final velocities of both blocks.

1
Set right as positive, so $v_{1 i} = + 6 m/s$ , $v_{2 i} = - 3 m/s$ , $m_{1} = 2 kg$ , $m_{2} = 1 kg$ .
2
Write momentum conservation:
$2(6) + 1(-3) = 2v_{1f} + 1v_{2f} \implies 9 = 2v_{1f} + v_{2f}$
3
Apply the elastic collision relative velocity relation:
$v_{1i} - v_{2i} = v_{2f} - v_{1f} \implies 6 - (-3) = v_{2f} - v_{1f} \implies 9 = v_{2f} - v_{1f}$
4
Solve the system of equations to get final velocities:
$0 = 3v_{1f} \implies v_{1f} = 0\ \text{m/s}, \quad v_{2f} = 9\ \text{m/s}$

Exam tip:

Memorize the relative velocity relation for 1D elastic collisions—it saves 2-3 minutes of algebra and avoids quadratic solution errors.

3. Two-Dimensional Collisions★★★★☆⏱ 4 min

For collisions in two dimensions, momentum is a vector, so it is conserved separately in the $x$ and $y$ directions. For elastic 2D collisions, we add a third equation for conservation of kinetic energy to solve for unknowns. The most common AP exam problem is a glancing collision between a moving mass and a stationary target.

\sum p_{x,i} = \sum p_{x,f}, \quad \sum p_{y,i} = \sum p_{y,f}

📐 Worked Example

A 0.1 kg billiard ball moving at 2 m/s along the $x$ -axis collides elastically with an identical stationary billiard ball. After collision, the first ball moves at $3 0^{\circ}$ above the $x$ -axis. Find the speed of each ball after collision.

1
Both masses are equal ( $m_{1} = m_{2} = m$ ), so we can cancel $m$ from all equations. Initial momentum: $p_{x, i} = 2 m$ , $p_{y, i} = 0$ .
2
Write momentum conservation for each component:
$x: 2 = v_{1f}\cos 30^\circ + v_{2f}\cos\theta_2 \\ y: 0 = v_{1f}\sin 30^\circ - v_{2f}\sin\theta_2$
3
Apply kinetic energy conservation for the elastic collision:
$4 = v_{1f}^2 + v_{2f}^2$
4
Square and add the momentum equations, use $cos^{2} θ + sin^{2} θ = 1$ to simplify:
$4 - 2\sqrt{3}v_{1f} + v_{1f}^2 = v_{2f}^2$
5
Substitute $v_{2 f}^{2} = 4 - v_{1 f}^{2}$ from KE conservation and solve:
$v_{1f} = \sqrt{3} \approx 1.73\ \text{m/s}, \quad v_{2f} = 1\ \text{m/s}$

Exam tip:

Always split momentum into $x$ and $y$ components for 2D collisions—never add momentum magnitudes directly, as momentum is a vector quantity.

4. AP Style Worked Practice Problems★★★☆☆⏱ 6 min

✓ Quick check

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

A 5 kg bowling ball moving right at 4 m/s hits a 1 kg bowling pin moving left at 1 m/s. After the collision, the bowling ball continues moving right at 2 m/s. What is the speed of the pin immediately after the collision, assuming momentum is conserved?
- 4 m/s
- 9 m/s
- 11 m/s
- 21 m/s
Reveal answer
9 m/s —
Correct. Setting right as positive, momentum conservation gives $5 (4) + 1 (- 1) = 5 (2) + 1 v_{f}$ , solving gives $v_{f} = 9$ m/s.

📐 Worked Example

A 2 kg block is attached to a 0.5 kg stationary ballistic pendulum. A 0.05 kg bullet is fired horizontally into the block and embeds itself. The entire system swings upward to a maximum height of 0.2 m. Find (a) the speed immediately after collision, (b) the initial bullet speed, (c) the fraction of kinetic energy lost during collision.

1
(a) After collision, mechanical energy is conserved as the system swings. Let $M = 2.55$ kg be total mass:
$\frac{1}{2}M v_f^2 = M g h \\ v_f = \sqrt{2(9.8)(0.2)} \approx 1.98\ \text{m/s}$
2
(b) Momentum is conserved during the collision, so solve for bullet speed:
$m_b v_b = M v_f \\ v_b = \frac{(2.55)(1.98)}{0.05} \approx 101\ \text{m/s}$
3
(c) Calculate the fraction of kinetic energy lost:
$KE_i \approx 255\ \text{J}, \quad KE_f \approx 5\ \text{J} \\ \text{Fraction lost} = \frac{KE_i - KE_f}{KE_i} \approx 0.98 = 98\%$

📐 Worked Example

A 1000 kg cannon on frictionless ice fires a 20 kg shell horizontally at 250 m/s relative to the ground. A 100 N constant friction force stops the cannon after firing. How far does the cannon slide before stopping?

1
Total initial momentum is zero, so conservation of momentum gives recoil velocity:
$0 = m_c v_c + m_s v_s \\ v_c = -\frac{(20)(250)}{1000} = -5\ \text{m/s}$
2
Use the work-kinetic energy theorem to find stopping distance:
$-f d = 0 - \frac{1}{2}m_c v_c^2 \\ d = \frac{(1000)(25)}{2(100)} = 125\ \text{m}$

5. Common Pitfalls

Wrong move:

Forgetting to use negative signs for velocity when objects move in opposite directions in 1D collisions.

Why:

Students confuse velocity (vector) with speed (scalar), so they add all velocities as positive, leading to wrong total momentum.

Correct move:

Always set a coordinate system, assign negative signs to velocities opposite your positive direction before writing the momentum equation.

Wrong move:

Applying kinetic energy conservation to perfectly inelastic collisions.

Why:

Students incorrectly assume all collisions conserve KE after learning elastic collision rules.

Correct move:

Only use KE conservation if the problem explicitly states the collision is elastic; for all other collisions, solve for unknowns with momentum only.

Wrong move:

Claiming momentum is conserved for a system with large net external impulse during the interaction.

Why:

Students overapply the 'short interaction time' approximation to all problems.

Correct move:

Only approximate momentum as conserved if the external impulse ( $F_{e x t} Δ t$ ) is much smaller than the total momentum change of the system.

Wrong move:

Combining the masses of colliding objects for elastic collisions when calculating final momentum.

Why:

Students mix up perfectly inelastic (sticking) and elastic (separate) collisions.

Correct move:

Only use $(m_{1} + m_{2}) v_{f}$ when objects stick together after a perfectly inelastic collision.

Wrong move:

Calculating center of mass velocity by averaging fragment velocities equally, not weighting by mass.

Why:

Students confuse average velocity with center of mass velocity.

Correct move:

Always use $v_{c m} = (m_{1} v_{1} + m_{2} v_{2}) / M_{t o t a l}$ to find center of mass velocity.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Momentum conservation condition	$\frac{d P _{t o t a l}}{d t} = F_{n e t, e x t} = 0 ⟹ P_{ini t ia l} = P_{f ina l}$	Applies to any direction with zero net external force, even if other directions have non-zero net force
CM velocity (isolated systems)	$v_{c m} = \frac{P _{t o t a l}}{M _{t o t a l}} = constant$	True before, during, and after collisions/explosions
Perfectly inelastic 1D collision	$m_{1} v_{1 i} + m_{2} v_{2 i} = (m_{1} + m_{2}) v_{f}$	Objects stick after collision; kinetic energy is not conserved
1D elastic collision	$m_{1} v_{1 i} + m_{2} v_{2 i} = m_{1} v_{1 f} + m_{2} v_{2 f} v_{1 i} - v_{2 i} = v_{2 f} - v_{1 f}$	Both momentum and KE conserved; relative velocity avoids quadratics
2D collision momentum conservation	$\sum p_{x, i} = \sum p_{x, f}, \sum p_{y, i} = \sum p_{y, f}$	Momentum conserved per component; never add magnitudes directly
Recoil problem (initial rest)	$0 = m_{1} v_{1} + m_{2} v_{2}$	Applies to systems initially at rest before firing/explosion
KE change in collision	$Δ K E = K E_{f ina l} - K E_{ini t ia l}$	Negative $Δ K E$ means energy lost to heat or deformation

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
1D elastic collision velocity calculation
2022 · FRQ
Ballistic pendulum momentum and energy
2021 · MCQ
2D glancing collision speed calculation

What's Next

Conservation of linear momentum is a prerequisite for nearly all advanced topics in AP Physics C Mechanics, starting with rotational motion and angular momentum, which share the same core conservation logic. You will next apply system momentum principles to analyze rolling motion, collisions involving rotation, and variable mass systems like rocket propulsion, which extends the conservation framework you learned here to systems that change mass over time. Without mastering component-wise momentum conservation and collision classification from this guide, solving rotational collision problems or variable mass motion will be difficult, as all these topics rely on the same logic of isolating a system and distinguishing between internal and external forces. This topic also builds the foundation for understanding energy transfer in interacting systems, which is tested across nearly all FRQ problems on the exam.