Linear Momentum (Calculus-based) — AP Physics C: Mechanics Study Guide

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

Covers: Definition of momentum, impulse-momentum theorem with variable forces, conservation of momentum in collisions and explosions, elastic vs inelastic collisions, center of mass motion — AP Physics C Mechanics Unit 4 (calculus-based).

You should already know: Newton's laws (Unit 2), work-energy theorem (Unit 3), single-variable calculus (derivatives, integrals).

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 papers.

1. Why Linear Momentum Matters

Unit 4 is small (about 12-18% of AP Physics C: Mechanics) but indispensable — every collision and every explosion problem rests on momentum conservation. Calculus-based treatment lets you handle variable mass (rockets), variable force (cushions), and the bridge from impulse to time-averaged force.

Two foundational concepts:

1. Momentum $\vec{p}=m\vec{v}$ — a vector quantity, conserved in any closed system.
2. Impulse $\vec{J}=\int \vec{F}dt=\Delta \vec{p}$ — the time integral of force, equals change in momentum.

2. Impulse-momentum theorem

For a single particle: $${\vec{F}}_{\text{net}}=\frac{d\vec{p}}{dt}$$

Integrating over time gives the impulse-momentum theorem: $$\vec{J}={\int }_{t_1}^{t_2}{\vec{F}}_{\text{net}}dt={\vec{p}}_2-{\vec{p}}_1=\Delta \vec{p}$$

For constant mass: $\Delta \vec{p}=m\Delta \vec{v}$.

For varying mass (e.g. a rocket): ${\vec{F}}_{\text{ext}}=m\frac{d\vec{v}}{dt}+\vec{v}\frac{dm}{dt}$. Be careful — this is only equivalent to "$F=ma$" when $dm/dt=0$.

Average force: $\bar{F}=J/\Delta t$. Useful for collision problems where peak force is hard to measure but impulse is.

3. Conservation of momentum

If the net external force on a system is zero, total momentum of the system is conserved: $$\sum {\vec{p}}_i^{\text{before}}=\sum {\vec{p}}_i^{\text{after}}$$

Internal forces (within the system) cannot change total momentum — they cancel in pairs by Newton's third law. So during a collision, the impulsive contact force between two cars is internal to the (car-car) system; total momentum is conserved.

Components: in 2D problems, conserve $p_x$ and $p_y$ separately.

4. Collisions

Two-body collisions are classified by what happens to kinetic energy:

For 1D elastic collision with $m_1,v_1$ hitting stationary $m_2$: $$v_1^{\prime }=\frac{m_1-m_2}{m_1+m_2}{v}_1,v_2^{\prime }=\frac{2m_1}{m_1+m_2}{v}_1$$

Special cases:

• $m_1=m_2$: $v_1^{\prime }=0$, $v_2^{\prime }=v_1$ (the moving ball stops, the stationary ball takes off).
• $m_1\gg m_2$: $v_1^{\prime }\approx v_1$, $v_2^{\prime }\approx 2v_1$.
• $m_1\ll m_2$: $v_1^{\prime }\approx -v_1$, $v_2^{\prime }\approx 0$ (ball bounces back).

5. Center of mass

For a system of $N$ particles: $${\vec{R}}_{\text{cm}}=\frac{1}{M_{\text{total}}}\sum _i{m}_i{\vec{r}}_i,{\vec{v}}_{\text{cm}}=\frac{1}{M_{\text{total}}}\sum _i{m}_i{\vec{v}}_i$$

For a continuous body: ${\vec{R}}_{\text{cm}}=\frac{1}{M}\int \vec{r}dm$.

The CM moves as if all mass were concentrated there and the net external force acted on it: $$M{\vec{a}}_{\text{cm}}={\vec{F}}_{\text{ext}}$$

So an isolated system's CM moves at constant velocity regardless of internal motions. A spinning, exploding firework still has its CM tracing a smooth parabolic arc until each fragment hits the ground.

6. Worked Example

A 0.50 kg ball moving rightward at 8.0 m/s collides head-on with a 1.5 kg block at rest. After collision, the ball rebounds leftward at 2.0 m/s. (a) Calculate the block's velocity after collision. (b) Calculate the impulse on the ball. (c) If the collision lasted 0.020 s, calculate the average force on the ball. (d) Determine whether the collision was elastic, inelastic, or perfectly inelastic.

Solution.

(a) Conservation of momentum (taking rightward as positive): $(0.50)(8.0)+(1.5)(0)=(0.50)(-2.0)+(1.5)v_2^{\prime }$ $4.0=-1.0+1.5v_2^{\prime }$ $v_2^{\prime }=5.0/1.5=3.33$ m/s rightward.

(b) Impulse on ball: $J=\Delta p_{\text{ball}}=m_1(v_1^{\prime }-v_1)=0.50(-2.0-8.0)=-5.0$ kg·m/s. Magnitude: 5.0 kg·m/s, direction leftward.

(c) Average force: $\bar{F}=J/\Delta t=-5.0/0.020=-250$ N. Magnitude: 250 N leftward.

(d) Initial KE: $\frac{1}{2}(0.50)(8.0{)}^2+0=16.0$ J. Final KE: $\frac{1}{2}(0.50)(2.0{)}^2+\frac{1}{2}(1.5)(3.33{)}^2=1.0+8.32=9.32$ J. KE decreased — inelastic (not perfectly: the bodies didn't stick).

7. Common Pitfalls

• Confusing momentum with force: momentum is a vector at one instant; impulse is the time-integral of force = change in momentum.
• Closed system not closed: external forces (e.g. gravity, friction) over the time interval add impulse and break conservation. For brief collisions, gravity/friction impulse is often negligible compared to internal forces — verify.
• Elastic ≠ collision where things bounce: many bouncing collisions are partially inelastic. Elastic specifically means KE is conserved — rare in macroscopic collisions.
• Sign errors in 1D: pick a positive direction and stick to it. A leftward-moving object has negative velocity in your sign convention.

8. Practice Questions (CED Style)

1. A 0.20 kg ball is dropped from 5.0 m onto the floor and rebounds to 3.0 m. Calculate (a) impulse on ball during contact; (b) if contact lasts 0.010 s, the average force.
2. A 50 kg cart moving at 4 m/s collides with a stationary 30 kg cart and the two stick together. Find their common velocity and the fraction of initial KE lost.
3. Two skaters (60 kg, 80 kg) push off each other, the 60-kg skater moving right at 2.0 m/s. Find the 80-kg skater's velocity. Assume initially at rest.

9. Quick Reference Cheatsheet

• $\vec{p}=m\vec{v}$ (vector).
• ${\vec{F}}_{\text{net}}=d\vec{p}/dt$.
• Impulse: $\vec{J}=\int \vec{F}dt=\Delta \vec{p}$.
• Average force: $\bar{F}=J/\Delta t$.
• Conservation: $\sum {\vec{p}}_{\text{before}}=\sum {\vec{p}}_{\text{after}}$ if no net external force.
• Elastic: $p$ and KE conserved.
• Perfectly inelastic: bodies stick, $p$ conserved, max KE lost.
• 1D elastic onto stationary target: $v_1^{\prime }=(m_1-m_2)/(m_1+m_2)\cdot v_1$, $v_2^{\prime }=2m_1/(m_1+m_2)\cdot v_1$.
• CM: ${\vec{v}}_{\text{cm}}$ unchanged by internal forces.

10. What's Next

Linear momentum bridges to Unit 5 (Rotation) via angular momentum (cross-product analogue), and connects backward to energy in elastic vs inelastic distinction. Use Ollie to step through 2D collision problems (which require component-wise conservation) or rocket-equation derivations for variable-mass systems.