Week 7 Cheatsheet — Momentum, Impulse & Collisions

How this week breaks down

Three views of the same equation, $\Delta \vec{p}=0$. Skim this once, then revise from the in-depth note.

Topic	What you do
Momentum	Identify the system; resolve $\vec{p}=m\vec{v}$ into components with a sign convention.
Conservation	Set $\sum {\vec{p}}_i=\sum {\vec{p}}_f$ (axis-by-axis if 2-D). Solve for the unknown.
Impulse	Use $\vec{J}=\Delta \vec{p}={\vec{F}}_{avg}\Delta t$ (or area under $F$–$t$).
Collision class	Check whether KE is conserved (elastic), partially lost (inelastic), or maximally lost with stuck-together pieces (totally inelastic).

---

1 — Core quantities at a glance

Quantity	Symbol	Definition	SI units
Momentum	$\vec{p}$	$m\vec{v}$	$\mathrm{kg}\cdot m/s$
Components	$p_x,p_y$	$m{v}_x,m{v}_y$	$\mathrm{kg}\cdot m/s$
Impulse	$\vec{J}$	$\Delta \vec{p}={\vec{F}}_{avg}\Delta t$	$\mathrm{N}\cdot \mathrm{s}$
Average force	${\vec{F}}_{avg}$	$\vec{J}/\Delta t$	$\mathrm{N}$
Kinetic energy	$KE$	$\frac{1}{2}m{v}^2$	$\mathrm{J}$

Sign convention (Slide 12): positive $x$ = right, positive $y$ = up. Velocities to the left or downward are negative.

---

2 — Conservation of momentum: the recipe

1. Define the system — what’s inside? What pushes it? Are external forces negligible during $\Delta t$?
2. Pick a sign convention and resolve every velocity into $x,y$.
3. Write $\sum {\vec{p}}_i=\sum {\vec{p}}_f$ (per axis if needed).
4. Solve for the unknown.

Generic system (objects 1, 2, 3, …):

$$(p_1+p_2+p_3{)}_{initial}=(p_1+p_2+p_3{)}_{final}.$$

---

3 — Collision class table

Type	Momentum conserved?	KE conserved?	Final state
Elastic	Yes	Yes	Objects rebound
Inelastic	Yes	No (some lost)	Objects rebound but slower / hotter
Totally (completely) inelastic	Yes	No (maximum lost)	Objects move with a common $v_f$

Two key special equations:

• Totally inelastic, 1-D: $m_1{v}_{1,i}+m_2{v}_{2,i}=(m_1+m_2)v_f.$
• Elastic, 1-D, equal masses, target at rest: velocities swap (incoming stops; target leaves with original speed).

---

4 — Impulse at a glance

Form	When to use
$\vec{J}=\Delta \vec{p}=m{\vec{v}}_f-m{\vec{v}}_i$	You know the velocity change.
$\vec{J}={\vec{F}}_{avg}\Delta t$	You know average force and duration.
$\vec{J}=\int F(t)dt$ (area under $F$–$t$)	You have an $F$–$t$ graph.

Equivalent unit identity: $1\mathrm{N}\cdot \mathrm{s}=1\mathrm{kg}\cdot m/s$.

---

5 — Worked snippets

Problem	Setup	Result
Rain in a frictionless carriage	$5000(22)=m_2(19)$	$m_{rain}=789\mathrm{kg}$
Glider drops a skydiver	$650(20)=(650-80)v_2$	$v_2=22.81m/s$
Elastic, equal-final-speed: $m$ hits $M$, both move at $v$ ($m$ rebounds)	momentum + KE	$u_1=2v,M=3m$
Rocket impulse $5.47\mathrm{N}\cdot \mathrm{s}$, thrust $0.124\mathrm{N}$	$\Delta t=\Delta p/F$	$\Delta t=44.11\mathrm{s}$
0.2 kg + 0.2 kg, totally inelastic at $10m/s$	$0.2(10)=0.4v_f$	$v_f=5m/s$; half KE lost
0.2 kg + 0.2 kg, elastic	velocities swap	$v_{1,f}=0,v_{2,f}=10m/s$
Freight cars couple	$45000\frac{11.6}{3.6}+23000\frac{4.18}{3.6}=68000v_f$	$v_f=2.525m/s$; $\approx 13\%$ KE lost
Drunk-driver crash: $2300\mathrm{kg}$ hits parked $1600\mathrm{kg}$, skids $28\mathrm{m}$ at ${\mu }_k=0.72$	$u_{A+B}=\sqrt{2\mu gd}$, then momentum	$u_A\approx 33.7m/s\approx 121km/h$

---

Common mistakes

1. Forgetting that $\vec{p}$ is a vector. Always set a sign convention. Rebounds carry negative momentum.
2. Mixing up elastic with totally inelastic. Elastic = bounces off, KE conserved. Totally inelastic = sticks, KE maximally lost.
3. Forgetting the “combined mass”. In a totally inelastic collision the final mass is $m_1+m_2$. In the rain example the final mass is $5000+m_{rain}$.
4. Unit slips. $km/h\to m/s$ via $÷3.6$. Impulse is in $\mathrm{N}\cdot \mathrm{s}$, momentum in $\mathrm{kg}\cdot m/s$ (numerically equal).
5. Conserving momentum through a friction slide. You can’t. Use momentum across the collision and kinematics/work-energy across the slide.
6. 2-D mistakes. Add components, never magnitudes. Each axis is conserved independently.
7. Substituting before differentiating is the related-rates error; here, the analogous trap is substituting before picking a sign.

---

Key formulas

$$\vec{p}=m\vec{v}{p}_x=m{v}_x{p}_y=m{v}_y$$ $${\vec{F}}_{net}=\frac{d\vec{p}}{dt}\Delta \vec{p}=0⟺\sum {\vec{p}}_i=\sum {\vec{p}}_f$$ $$\vec{J}=\Delta \vec{p}={\vec{F}}_{avg}\Delta t[\mathrm{N}\cdot \mathrm{s}]$$ $$\text{Elastic:}\frac{1}{2}{m}_1{v}_{1,i}^2+\frac{1}{2}{m}_2{v}_{2,i}^2=\frac{1}{2}{m}_1{v}_{1,f}^2+\frac{1}{2}{m}_2{v}_{2,f}^2$$ $$\text{Totally inelastic (1-D):}{m}_1{v}_{1,i}+m_2{v}_{2,i}=(m_1+m_2)v_f$$

For the why and the full derivations, see the in-depth note.