Conservation of linear momentum and collisions — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Conditions for linear momentum conservation, center of mass motion during collisions, 1D elastic/inelastic/perfectly inelastic collision analysis, 2D collision techniques, and recoil problem-solving methods.

You should already know: Newton's third law of motion, vector component addition, center of mass definition for systems of particles.

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 C: Mechanics 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 Conservation of linear momentum and collisions?

Conservation of linear momentum is a fundamental principle governing interactions between objects in isolated systems, and is the primary tool for analyzing collision events. Per the AP Physics C CED, this topic accounts for 14-18% of the total exam weight, and appears in both multiple choice (MCQ) and free response (FRQ) sections, often combined with energy, kinematics, or circular motion concepts. By definition, the principle states that the total linear momentum of an isolated system (a system with zero net external force acting on it) remains constant over time, even when internal forces act between the system’s components to change the momentum of individual parts. Standard notation for this topic uses ${\vec{P}}_{total}$ for total system momentum, ${\vec{p}}_i=m_i{\vec{v}}_i$ for the momentum of the $i$-th particle, so conservation is written as ${\vec{P}}_{initial}={\vec{P}}_{final}$. Collisions are short-duration interactions between two or more objects where internal forces are much larger than any external forces, allowing us to approximate momentum as conserved even if small external forces (like friction or gravity) are present.

2. Conditions for Momentum Conservation and Center of Mass Motion

The first and most critical step in any momentum problem is confirming when momentum is conserved. The formal condition comes directly from Newton’s second law for systems of particles: the rate of change of total system momentum equals the net external force acting on the system: $$\frac{d{\vec{P}}_{total}}{dt}={\vec{F}}_{net,ext}$$ If ${\vec{F}}_{net,ext}=0$, then $\frac{d{\vec{P}}_{total}}{dt}=0$, so ${\vec{P}}_{total}$ is a constant vector. This works because internal forces come in equal and opposite pairs per Newton’s third law, so they cancel out when summed over the entire system—only forces from objects outside the system can change total momentum.

Common cases where momentum is conserved (or can be approximated as conserved) include: (1) collisions/explosions (interaction time $\Delta t$ is so small that impulse from external forces $\vec{J}={\vec{F}}_{ext}\Delta t\approx 0$), (2) motion along one direction with zero net external force in that direction (even if net force is non-zero in the perpendicular direction, e.g., a projectile exploding mid-air has conserved horizontal momentum), and (3) systems on frictionless surfaces with no other external forces. A key related result is that the velocity of the center of mass of an isolated system is always constant, since $v_{cm}=P_{total}/M_{total}$.

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: $P_i=M{v}_i=(1000\text{kg})(50\text{m/s})=50000\text{kgcdotpm/s}$.
3. Set equal to final momentum: $P_f=m_1{v}_1+m_2{v}_2=(400\text{kg})(0)+(600\text{kg})v_2=50000$. Solve for $v_2=50000/600\approx 83.3\text{m/s}$ in the original direction of motion.
4. Calculate center of mass velocity: $v_{cm}=P_f/M_{total}=50000/1000=50\text{m/s}$, matching the initial velocity.

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.

3. One-Dimensional Collisions: Elastic, Inelastic, and Perfectly Inelastic

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 (this is an idealization, but common for billiard balls or atomic collisions).

For perfectly inelastic collisions, we only need one momentum conservation equation: $$m_1{v}_{1i}+m_2{v}_{2i}=(m_1+m_2)v_f$$ which can be solved directly for $v_f$. For elastic collisions, we have two conservation equations, which can be rearranged to give the convenient relative velocity relation: $$v_{1i}-v_{2i}=v_{2f}-v_{1f}$$ This relation avoids solving the quadratic equation that comes from kinetic energy conservation, saving time on the exam.

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 $v_{1i}=+6\text{m/s}$, $v_{2i}=-3\text{m/s}$, $m_1=2\text{kg}$, $m_2=1\text{kg}$.
2. Write momentum conservation: $2(6)+1(-3)=2v_{1f}+1v_{2f}⟹9=2v_{1f}+v_{2f}$.
3. Apply the elastic collision relative velocity relation: $v_{1i}-v_{2i}=v_{2f}-v_{1f}⟹6-(-3)=v_{2f}-v_{1f}⟹9=v_{2f}-v_{1f}$.
4. Solve the system of equations: Subtract the second equation from the first to get $0=3v_{1f}⟹v_{1f}=0\text{m/s}$, then $v_{2f}=9\text{m/s}$. Checking confirms momentum and kinetic energy are both conserved.

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

4. Two-Dimensional Collisions

For collisions in two dimensions, momentum is a vector, so it is conserved separately in the $x$ and $y$ directions: $$\sum p_{x,i}=\sum p_{x,f},\sum p_{y,i}=\sum p_{y,f}$$ 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, where one angle is given and we solve for speeds or the second angle.

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 $30^{\circ }$ 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: $p_{x,i}=2m$, $p_{y,i}=0$.
2. Write momentum components: $x:2=v_{1f}\cos 30^{\circ }+v_{2f}\cos {\theta }_2$ $y:0=v_{1f}\sin 30^{\circ }-v_{2f}\sin {\theta }_2$
3. Kinetic energy conservation (elastic): $4=v_{1f}^2+v_{2f}^2$.
4. Square and add the two momentum equations, use the Pythagorean identity ${\cos }^2\theta +{\sin }^2\theta =1$ to simplify to $4-2\sqrt{3}{v}_{1f}+v_{1f}^2=v_{2f}^2$.
5. Substitute $v_{2f}^2=4-v_{1f}^2$ from KE conservation: $v_{1f}=\sqrt{3}\approx 1.73\text{m/s}$, $v_{2f}=1\text{m/s}$.

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

5. Common Pitfalls (and how to avoid them)

• 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 ($F_{ext}\Delta t$) 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 $v_{cm}=(m_1{v}_1+m_2{v}_2)/M_{total}$ to find center of mass velocity.

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

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? A) 4 m/s B) 9 m/s C) 11 m/s D) 21 m/s

Worked Solution: Set right as the positive direction, so the initial velocity of the pin is $-1$ m/s. Conservation of momentum gives $m_b{v}_{b,i}+m_p{v}_{p,i}=m_b{v}_{b,f}+m_p{v}_{p,f}$. Substitute values: $(5)(4)+(1)(-1)=(5)(2)+(1)v_{p,f}⟹19=10+v_{p,f}⟹v_{p,f}=9$ m/s. Speed is the magnitude of velocity, so the correct answer is B.

Question 2 (Free Response)

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 in the block. The entire system swings upward, reaching a maximum height of 0.2 m above the initial position. (a) What is the speed of the system immediately after the bullet embeds itself? (b) What is the initial speed of the bullet before impact? (c) What fraction of the bullet's initial kinetic energy is lost during the collision?

Worked Solution: (a) After the collision, mechanical energy is conserved as the system swings upward. Let $M=2+0.5+0.05=2.55$ kg be the total mass. Kinetic energy converts to gravitational potential energy: $\frac{1}{2}M{v}_f^2=Mgh$. Mass cancels, so $v_f=\sqrt{2gh}=\sqrt{2(9.8)(0.2)}\approx 1.98$ m/s. (b) Momentum is conserved during the short collision, so $m_b{v}_b=M{v}_f$. Solve for bullet speed: $v_b=\frac{M{v}_f}{m_b}=\frac{(2.55)(1.98)}{0.05}\approx 101$ m/s. (c) Initial kinetic energy: $K{E}_i=\frac{1}{2}{m}_b{v}_b^2\approx 255$ J. Final kinetic energy after collision: $K{E}_f=\frac{1}{2}M{v}_f^2\approx 5$ J. Fraction of energy lost: $\frac{K{E}_i-K{E}_f}{K{E}_i}\approx 0.98$, or 98%.

Question 3 (Application / Real-World Style)

A 1000 kg cannon is mounted on a frictionless frozen lake, and fires a 20 kg shell horizontally at 250 m/s relative to the ground. After firing, a constant friction force of 100 N acts on the cannon to stop it. How far does the cannon slide on the ice before stopping?

Worked Solution: The system of cannon + shell has zero initial momentum, so conservation of momentum gives $0=m_c{v}_c+m_s{v}_s$. Solve for cannon recoil velocity: $v_c=-\frac{m_s{v}_s}{m_c}=-\frac{(20)(250)}{1000}=-5$ m/s (negative indicates direction opposite the shell, speed is 5 m/s). Use the work-kinetic energy theorem to find stopping distance: work done by friction equals change in kinetic energy: $-fd=0-\frac{1}{2}{m}_c{v}_c^2$. Solve for $d=\frac{m_c{v}_c^2}{2f}=\frac{(1000)(25)}{2(100)}=125$ m. In context, the cannon slides 125 meters backward on the ice before friction brings it to a complete stop.

7. Quick Reference Cheatsheet

8. 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 chapter, solving rotational collision problems or variable mass motion will be impossible, 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.

Angular momentum and rotational conservation Variable mass and rocket propulsion Rotational collision problems