Week 7 In-Depth — Why Momentum Is the Right Bookkeeping Variable

Read the cheatsheet first. This note explains why momentum is conserved, why elastic ≠ “stuck”, and walks through one substantial worked example per topic with every step on the page.

1 — Why momentum is the right thing to track

Newton’s Second Law is usually quoted as $\vec{F}=m\vec{a}$, but Newton actually wrote it in the form

$${\vec{F}}_{net}=\frac{d\vec{p}}{dt},\vec{p}=m\vec{v}.$$

When the mass is constant, $d\vec{p}/dt=md\vec{v}/dt=m\vec{a}$, and you recover the schoolbook version. But the momentum form is more general — it survives even when mass varies (rocket exhausts, raindrops falling into a carriage), and it slides cleanly into conservation arguments.

Why “no net external force” implies conservation

Consider a system of two objects, 1 and 2. Each one feels:

• internal forces from the other (call them ${\vec{F}}_{12}$, ${\vec{F}}_{21}$), and
• external forces (${\vec{F}}_{1,\mathrm{ext}}$, ${\vec{F}}_{2,\mathrm{ext}}$).

By Newton’s third law, ${\vec{F}}_{12}=-{\vec{F}}_{21}$. Summing the second-law equation over both objects:

$$\frac{d}{dt}({\vec{p}}_1+{\vec{p}}_2)={\vec{F}}_{12}+{\vec{F}}_{21}+{\vec{F}}_{1,\mathrm{ext}}+{\vec{F}}_{2,\mathrm{ext}}={\vec{F}}_{\mathrm{ext},\mathrm{total}}.$$

The internal pair cancels. If the net external force is zero, the total momentum ${\vec{p}}_1+{\vec{p}}_2$ has zero time derivative — i.e. it’s constant. This is conservation of momentum.

Why collisions are special

A collision is a brief, intense interaction (Slide 18). During the collision the inter-system contact force can be enormous, while ordinary external forces (gravity, friction, air drag) act over the same brief $\Delta t$ and contribute a comparatively tiny impulse. So even when external forces are technically non-zero, we treat $\sum {\vec{p}}_i=\sum {\vec{p}}_f$ across the collision to excellent approximation. The friction phase that follows the collision is handled separately, with kinematics or work-energy.

Worked example with all the steps — rain-filling carriage (Slide 15)

A 5000 kg open carriage rolls on frictionless track at $v_1=22m/s$. Over 30 minutes of rain it slows to $v_2=19m/s$. How much water did it pick up?

System: carriage + water. External horizontal force: zero (frictionless track; rain falls vertically). Momentum conserved horizontally:

$$m_1{v}_1=m_2{v}_2.$$

The water-laden carriage has mass $m_2=5000+m_{rain}$:

$$5000(22)=(5000+m_{rain})(19)⟹5000+m_{rain}=\frac{110000}{19}=5789\mathrm{kg}.$$

So

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

A more subtle point: in the carriage frame, each raindrop arrives with horizontal speed $-v_{carriage}$ and is brought to rest, transferring its momentum into the carriage. From the ground frame, the picture is “carriage + drop combine inelastically each instant” — which is exactly conservation of horizontal momentum.

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2 — Impulse: the area-under-the-curve view

For a force $\vec{F}(t)$ acting on an object,

$$\Delta \vec{p}={\int }_{t_i}^{t_f}\vec{F}(t)dt={\vec{F}}_{avg}\Delta t.$$

That integral is geometrically the area under the $F$–$t$ curve (Slide 20). The average force ${\vec{F}}_{avg}$ is the constant force that would produce the same impulse over the same $\Delta t$ — i.e. the height of a rectangle with the same area.

This recasts a hard problem (“what’s the peak force during a collision?”) into an easy one (“what’s the change in momentum, and over what time?”). The impulse–momentum theorem is how you back out forces from collisions.

Worked example — skydiver from a glider (Slide 16)

A 650 kg glider (with passengers) flies horizontally at $v_1=20m/s$. An 80 kg skydiver drops straight down out the bottom. What’s the glider’s velocity just after?

System: the original glider+skydiver. External horizontal force during the brief drop: zero.

Horizontal momentum is conserved. The skydiver leaves carrying their horizontal share, so the remaining glider mass $m_2=650-80=570\mathrm{kg}$ continues with whatever horizontal momentum is left:

$$m_1{v}_1=m_2{v}_2⟹v_2=\frac{650(20)}{570}=22.81m/s.$$

The glider speeds up. Why? Same total horizontal momentum, smaller mass to carry it. This is a useful sanity check for any “what changes when mass leaves the system” problem.

Worked example — rocket impulse (Tutorial Ex 1)

A model rocket delivers an impulse of $5.47\mathrm{N}\cdot \mathrm{s}$ with a thrust of $124\mathrm{mN}=0.124\mathrm{N}$. The burn time is

$$\Delta t=\frac{\Delta p}{F_{avg}}=\frac{5.47}{0.124}=44.11\mathrm{s}.$$

Two things worth noticing:

1. The unit $\mathrm{N}\cdot \mathrm{s}$ is numerically equal to $\mathrm{kg}\cdot m/s$ — impulse and momentum live in the same units, which is exactly what $\vec{J}=\Delta \vec{p}$ requires.
2. Read the prefix. Milli-Newtons are $10^{-3}\mathrm{N}$. Mis-reading the prefix is the easiest way to get this question wrong by a factor of 1000.

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3 — Why elastic collisions are different

In any collision (negligible external impulse during $\Delta t$):

$$\sum p_i=\sum p_f\text{always.}$$

Whether kinetic energy is also conserved is what distinguishes the three types:

• Elastic — KE conserved. No energy lost to heat, deformation, or sound. Idealised limit (billiard balls, atomic-scale collisions).
• Inelastic — some KE lost (to heat, sound, deformation, etc.).
• Totally inelastic — the maximum amount of KE the conservation law permits is lost; the objects move with a common final velocity.

Why “common final velocity” maximises KE loss

For a 1-D collision between $m_1$ moving at $v_{1,i}$ and stationary $m_2$, momentum conservation fixes the total final momentum $P=m_1{v}_{1,i}$. For a given total momentum $P$, total kinetic energy

$$KE=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}$$

is minimised (subject to $p_1+p_2=P$) when the two pieces move with the same velocity — i.e. when they’re stuck together. (You can verify this with a Lagrange multiplier or by completing the square.) Any other split of momentum between the two pieces stores more KE.

So “the totally-inelastic case is the maximum-loss case” isn’t an extra postulate — it’s a direct consequence of momentum conservation plus the algebra of kinetic energy.

Worked example with all the steps — equal-final-speed elastic (Slide 22; notes pp. 3–4)

Block $m$ at speed $u_1$ hits a stationary block $M$ in a 1-D elastic collision. After impact, both blocks travel at the same speed $v$, with $m$ rebounding. Find $M$ in terms of $m$.

Sign convention: right is positive. After the collision $m$ moves left ($-v$), $M$ moves right ($+v$).

Momentum (axis-by-axis, but 1-D so one equation):

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

Elastic condition (KE conserved):

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

Factor $u_1^2-v^2=(u_1+v)(u_1-v)$:

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

From (1): $M/m=(u_1+v)/v$. Set the two equal:

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

Cancel $u_1+v$ (non-zero) and one factor of $v$ to get

$$1=\frac{u_1-v}{v}⟹u_1=2v.$$

Substitute back:

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

The takeaway: in an elastic collision, momentum and KE conservation are two equations in two unknown final velocities — you can always solve them simultaneously.

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4 — Two-stage problems: crash reconstructions

Real-world collision problems often combine two distinct physics episodes:

1. The collision itself — short, intense, dominated by inter-object forces. Use momentum conservation.
2. The aftermath — friction slide, free-fall, swing on a rope. Use kinematics or work–energy.

You can never conserve momentum through the second phase, because friction (or gravity, or rope tension) is now the dominant external force. You handle the two episodes back-to-back, sharing the velocity at the boundary as the unknown.

Worked example — drunk-driver crash (Tutorial Ex 4)

$m_A=2300\mathrm{kg}$ (the drunk driver, moving) hits parked $m_B=1600\mathrm{kg}$ at a traffic light. The mangled wreck slides $d=28\mathrm{m}$ with ${\mu }_k=0.72$ before stopping. What was $m_A$‘s speed just before impact?

Stage 2 → 3 (the slide). Treat the wreck as a single mass with deceleration $a=-{\mu }_{k}g=-0.72(9.8)=-7.056m/{\mathrm{s}}^2$. Using $v_f^2=u^2+2ad$ with $v_f=0$:

$$0=u_{A+B}^2-2(0.72)(9.8)(28)⟹u_{A+B}=\sqrt{2(0.72)(9.8)(28)}=19.88m/s.$$

This is the speed of the combined wreck immediately after the collision — the boundary value.

Stage 1 → 2 (the coupling). The collision is totally inelastic: stationary $m_B$ is hit by $m_A$ moving at $u_A$, and they emerge moving together at $u_{A+B}=19.88m/s$. Conservation of momentum:

$$m_A{u}_A=(m_A+m_B)u_{A+B}⟹u_A=\frac{(2300+1600)(19.88)}{2300}=\frac{3900\times 19.88}{2300}=33.7m/s.$$

Convert to km/h: $33.7\times 3.6\approx 121km/h$.

That’s about three times the urban limit. The structure of the calculation — post-collision velocity from kinematics, then back-solve momentum — is the template for any reconstruction problem.

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5 — How these techniques connect

All four sub-topics use the same equation, $\sum {\vec{p}}_i=\sum {\vec{p}}_f$, but they ask different questions of it:

Technique	What’s $\vec{p}$ doing?
Conservation	$\Delta {\vec{p}}_{system}=0$ when external ${\vec{F}}_{net}=0$. The bookkeeping equation.
Impulse	$\Delta \vec{p}$ of one object equals ${\vec{F}}_{avg}\Delta t$ — backs out forces from collisions.
Elastic / inelastic class	Distinguished by whether $\sum KE$ is also conserved.
Two-stage problems	$\sum {\vec{p}}_i=\sum {\vec{p}}_f$ across the collision; kinematics across the slide.

The single mental model is: list the objects, list the forces, decide whether external impulse during $\Delta t$ is negligible, then write the conservation equation in components. Everything else is algebra.

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6 — Exam-style sample, end-to-end

Two freight cars on a level frictionless track couple together on impact. Car $A$ ($m_A=45000\mathrm{kg}$) moves at $11.6km/h$; car $B$ ($m_B=23000\mathrm{kg}$) moves at $4.18km/h$ in the same direction. Find the common final speed and the fractional kinetic-energy loss.

Setup.

Item	Value
Masses	$m_A=45000\mathrm{kg}$, $m_B=23000\mathrm{kg}$
Velocities	$v_A=11.6/3.6\approx 3.222m/s$, $v_B=4.18/3.6\approx 1.161m/s$
Collision type	Totally inelastic (couple together)
External horizontal force	Zero (frictionless)

Momentum (1-D, right positive):

$$m_A{v}_A+m_B{v}_B=(m_A+m_B)v_f.$$ $$v_f=\frac{45000(3.222)+23000(1.161)}{68000}=\frac{144990+26703}{68000}\approx 2.525m/s.$$

Initial KE:

$$K{E}_i=\frac{1}{2}(45000)(3.222{)}^2+\frac{1}{2}(23000)(1.161{)}^2\approx 2.336\times 10^5+1.550\times 10^4\approx 2.49\times 10^5\mathrm{J}.$$

Final KE:

$$K{E}_f=\frac{1}{2}(68000)(2.525{)}^2\approx 2.17\times 10^5\mathrm{J}.$$

Fractional loss:

$$\frac{K{E}_i-K{E}_f}{K{E}_i}\approx \frac{2.49-2.17}{2.49}\approx 12.98\%.$$

Answer. The wreck moves at $\approx 2.53m/s$ and loses about $13\%$ of the initial kinetic energy to deformation, sound and heat.

That layout — System, sign convention, momentum equation, energy check, answer with units — is the template for every collision question on the paper.

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Where to look next

• The cheatsheet for fast revision + the quiz.
• lecture-summary.md — the source-faithful reconstruction.
• The tutorial PDF for more practice (Wolfson Ch. 9).