Week 7 — Momentum, Impulse & Collisions

What this week is about

This week recasts Newton’s Second Law in momentum form and uses it to answer three related questions:

1. What stays constant when two things interact briefly? → Conservation of momentum for systems with no net external force.
2. How hard did the collision hit? → Impulse: $\vec{J}={\vec{F}}_{avg}\Delta t=\Delta \vec{p}$.
3. Did they bounce or stick? → Elastic vs. inelastic vs. totally-inelastic collisions, in 1-D and 2-D.

All three are one technique — track $\vec{p}=m\vec{v}$ before and after — applied to different questions. See the in-depth note for how impulse, conservation and collisions are the same equation in disguise.

Notes in this week

• Cheatsheet — every formula, table, and recipe in one scannable page. Includes the quiz (mixed difficulty, reshuffles every visit).
• In-depth analysis — why momentum is conserved, why elastic ≠ “stuck”, a full worked example per topic, the exam-style template.
• Lecture summary — the factual reconstruction from the slides and tutorial.

Any notes you add to this folder will appear here automatically.

What I need to know before the workshop

• Vector decomposition into $x$- and $y$-components and a chosen sign convention
• Newton’s Second Law and how kinetic energy $KE=\frac{1}{2}m{v}^2$ behaves
• Unit conversions ($km/h\to m/s$ via $÷3.6$)
• Kinematics with constant acceleration (for multi-stage crash problems)

Assessment relevance

• Exam and portfolio questions for EGD102 lean heavily on conservation of momentum and the impulse–momentum theorem.
• Multi-stage problems (collision then friction slide) are a staple of the crash-reconstruction style.

Overview

Week 7 introduces linear momentum as a vector quantity, recasts Newton’s Second Law as ${\vec{F}}_{net}=d\vec{p}/dt$, and uses it to derive conservation of momentum for systems with no net external force. The lecture then applies these tools to short-duration collisions, distinguishes elastic, inelastic and totally inelastic interactions, and introduces impulse as the area under the force-vs-time curve. The accompanying tutorial (Wolfson Ch. 9) drills impulse, 1-D elastic and completely inelastic collisions, and a multi-stage collision-plus-friction crash reconstruction.

Key concepts

• Momentum ($\vec{p}$, kg·m/s) — vector quantity describing the “amount of mass behind an object’s motion”; same direction as velocity (Slide 12).
• Momentum components ($p_x$, $p_y$) — momentum resolves into x- and y-components like any vector (Slide 12).
• Sign convention — positive: right (x), up (y); negative: left (x), down (y) (Slide 12).
• Net external force — only external forces can change the total momentum of a system; internal forces between members cancel (Slide 13).
• Conservation of momentum — when net external force is zero, $\Delta \vec{p}=0$ and $\sum {\vec{p}}_{initial}=\sum {\vec{p}}_{final}$ (Slide 14).
• Collision — brief, intense interaction; inter-system forces dominate external forces so total momentum is treated as conserved over $\Delta t$ (Slide 18).
• Impulse ($\vec{J}=\Delta \vec{p}$, N·s) — change in momentum equal to the area under the $F_x(t)$ curve (Slide 20).
• Average force (${\vec{F}}_{avg}$, N) — constant force giving the same impulse over the collision duration (Slide 20).
• Elastic collision — both momentum and total kinetic energy conserved; objects rebound (Slide 21).
• Inelastic collision — momentum conserved but kinetic energy is NOT; objects often join (Slide 21).
• Totally (completely) inelastic collision — the maximum amount of KE is lost; objects move with a common final velocity (Slide 21).
• 2-D collisions — resolve vectors into x- and y-components, conserve momentum in each plane independently, recombine (Slide 23).

Core formulas

Linear momentum (vector and component form):

$\vec{p}[\mathrm{kg}\cdot m/s]=m\vec{v}$

$p_x=m{v}_x{p}_y=m{v}_y$

Newton’s Second Law (momentum form):

${\vec{F}}_{net}=\frac{d\vec{p}}{dt}$

Conservation of momentum (no net external force):

$\Delta \vec{p}=0⟹\sum {\vec{p}}_{initial}=\sum {\vec{p}}_{final}$

For a system of objects 1, 2, 3, …:

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

Impulse–momentum theorem:

$\Delta \vec{p}={\vec{F}}_{avg}\Delta t[\mathrm{N}\cdot \mathrm{s}]$

Kinetic-energy condition for an elastic collision:

$\sum K{E}_i=\sum K{E}_f⟺\frac{1}{2}{m}_1{v}_{1,i}^2+\frac{1}{2}{m}_2{v}_{2,i}^2=\frac{1}{2}{m}_1{v}_{1,f}^2+\frac{1}{2}{m}_2{v}_{2,f}^2$

Completely inelastic collision (common final velocity):

$m_1{v}_{1,i}+m_2{v}_{2,i}=(m_1+m_2)v_f$

Worked examples

Example 1 — Rain-filling train carriage (Slide 15; notes p. 1)

A 5000 kg open carriage rolls on frictionless track at $v_1=22m/s$. After 30 min of rain it has slowed to $v_2=19m/s$. Find the mass of water collected.

Frictionless track $\Rightarrow$ no net external horizontal force on (carriage + water) system $\Rightarrow$ momentum conserved.

$\sum p_i=\sum p_f$ $m_1{v}_1=m_2{v}_2$ $5000\times 22=m_2\times 19$ $m_2=\frac{110000}{19}=5789\mathrm{kg}$

Since $m_2=5000+m_{rain}$:

$\boxed{m_{rain}=789\mathrm{kg}}$

Example 2 — Skydiver jumps from a glider (Slide 16; notes p. 2)

A 650 kg glider (including passengers) glides horizontally at $v_1=20m/s$ when an 80 kg skydiver jumps straight down out the bottom. Find the glider’s velocity just after.

Skydiver exits with the glider’s horizontal velocity (relative horizontal velocity to glider is zero at the instant of release; the notes treat the skydiver as carrying away their horizontal momentum). Using conservation of horizontal momentum with $m_2=650-80=570\mathrm{kg}$:

$m_1{v}_1=m_2{v}_2$ $650\times 20=570\times v_2$ $\boxed{v_2=\frac{13000}{570}=22.81m/s}$

The glider speeds up because shedding mass (with the original horizontal velocity) leaves the remaining mass carrying the same x-momentum.

Example 3 — 1-D elastic collision, equal final speeds (Slide 22; notes pp. 3–4)

Block $m$ hits stationary block $M$ in a 1-D elastic collision. After impact both blocks travel with the same speed $v$ (m rebounds backward, M moves forward). Express $M$ in terms of $m$.

Momentum (taking right positive; m rebounds left after collision):

$m{u}_1+M(0)=-mv+Mv$ $m(u_1+v)=Mv$ $\frac{u_1+v}{v}=\frac{M}{m}\text{...(1)}$

Elastic $\Rightarrow$ KE conserved:

$\frac{1}{2}m{u}_1^2=\frac{1}{2}m{v}^2+\frac{1}{2}M{v}^2$ $m(u_1^2-v^2)=M{v}^2$ $\frac{M}{m}=\frac{u_1^2-v^2}{v^2}=\frac{(u_1+v)(u_1-v)}{v^2}\text{...(2)}$

Equate (1) and (2):

$\frac{u_1+v}{v}=\frac{(u_1+v)(u_1-v)}{v^2}$ $1=\frac{u_1-v}{v}⟹u_1=2v$

Substitute back into (1):

$\frac{M}{m}=\frac{2v+v}{v}=3⟹\boxed{M=3m}$

Things to practise

Tutorial 7 (Wolfson Ch. 9) — solution outlines:

Exercise 1 — Impulse from a rocket thrust. Impulse $\Delta p=5.47\mathrm{N}\cdot \mathrm{s}$, thrust $F=124\mathrm{mN}=0.124\mathrm{N}$. From $\Delta p=F\Delta t$: $\Delta t=\frac{5.47}{0.124}=44.11\mathrm{s}$

Exercise 2 — 0.2 kg box hits stationary 0.2 kg box at 10 m/s. Initial $\sum K{E}_i=\frac{1}{2}(0.2)(10{)}^2=10\mathrm{J}$.

• (a) Completely inelastic: $0.2(10)=0.4v_f\Rightarrow v_f=5m/s$; $\sum K{E}_f=\frac{1}{2}(0.4)(5{)}^2=5\mathrm{J}$ (half the KE is lost).
• (b) Elastic, equal masses: solving momentum + KE simultaneously gives $v_{1,f}=0,v_{2,f}=10m/s$ (incoming block stops, stationary block leaves with the original speed). $\sum K{E}_f=10\mathrm{J}$.

Exercise 3 — Freight cars couple together. $m_A=45000\mathrm{kg}$ at $11.6km/h$, $m_B=23000\mathrm{kg}$ at $4.18km/h$ (same direction). Convert with $/3.6$, conserve momentum: $v_{A+B,f}=\frac{45000(11.6/3.6)+23000(4.18/3.6)}{68000}=2.525m/s$ $\sum K{E}_i=2.49\times 10^5\mathrm{J}$, $\sum K{E}_f=2.17\times 10^5\mathrm{J}$, $\%K{E}_{lost}=-12.98\%$.

Exercise 4 — Drunk-driver crash reconstruction (two-stage). $m_A=2300\mathrm{kg}$ (moving) hits parked $m_B=1600\mathrm{kg}$; combined wreck slides 28 m with ${\mu }_k=0.72$ before stopping.

Stage 2→3 (skid to rest), use kinematics with $a=-\mu g$: $0=u_{A+B}^2+2(-\mu g)(28)\Rightarrow u_{A+B}=\sqrt{2(0.72)(9.8)(28)}=19.88m/s$

Stage 1→2 (totally inelastic coupling): $u_A=\frac{u_{A+B}{m}_{A+B}}{m_A}=\frac{19.88\times 3900}{2300}=33.7m/s\approx 121.34km/h$

Common pitfalls

• Forgetting that momentum is a vector. Always set a sign convention (positive right, positive up) and treat rebounds as negative. In Example 3 the rebounding block carries $-mv$, not $+mv$.
• Confusing momentum conservation with KE conservation. Momentum is conserved in every collision (provided external forces are negligible). KE is conserved only in elastic collisions.
• Unit slips. $\vec{p}$ is in kg·m/s; impulse is in N·s; convert km/h to m/s by dividing by 3.6 before using in $p=mv$ (Exercise 3).
• Wrong “combined” mass. In a totally inelastic collision the final mass is $m_1+m_2$. In Example 1 (rain) the final mass is $5000+m_{rain}$, not just 5000.
• Treating “elastic” as “objects stick together”. That is the totally inelastic case. Elastic means the objects rebound and KE is conserved.
• Ignoring the assumption “external forces negligible during $\Delta t$”. Friction, gravity etc. do exist but act over much shorter impulses than the collision force during the brief collision time — that is why we can use $\sum p_i=\sum p_f$ across the collision itself.
• Multi-stage problems. When a collision is followed by sliding to rest (Exercise 4), you must use momentum across the collision and kinematics/work-energy across the slide; do not try to conserve momentum through the friction phase.
• 2-D problems. Resolve into x and y, conserve each independently, then recombine — do not add magnitudes.

Source citations

• Lecture slides: EGD102-Physics/Lecture7_CTP1.pdf (Slides 1, 10–23, 25)
• Worked-example notes: EGD102-Physics/EGD102 - Lecture7 - Notes.pdf (pp. 1–4, Examples 1–3)
• Tutorial sheet: EGD102-Physics/Tutorial 7.pdf (Exercises 1–4)
• Tutorial solutions: EGD102-Physics/Tutorial 7_Solutions.pdf (Exercises 1–4)
• Textbook reference: Wolfson, R. (2020). Essential University Physics, Vol. 1, 4th Ed. SI, Ch. 9 (Slide 25; Tutorial cover slide).