Worked examples — Newton's third law — action-reaction, common misconceptions

This is a worked-example dojo for Newton's Third Law. The parent note built the idea; here we drill it against every kind of situation the law can throw at you — so that when an exam or a real problem appears, you have already seen its shape.

The scenario matrix

Think of every third-law problem as living in one cell of this table. Our goal: cover them all.

Every numeric answer below ends with its own Verify line, and each is machine-checked.

Ex 1 — Static book on a table (Cell C1: equilibrium, fake pair)

Step 1 — List forces on the book. Weight $W=mg=2\times 9.8=19.6\text{N}$ downward, and normal $N$ upward. Why this step? Before naming pairs we must know which forces even touch the book. In the figure's left tower, both arrows sit on the same dot (the book).

Step 2 — Use equilibrium to get $N$. The book does not accelerate, so by Newton's Second Law the net force is zero: $N-W=0⟹N=19.6\text{N}.$ Why this step? $N=W$ is what tempts people into calling them a pair. But this equality comes from equilibrium, not the third law — remove the equilibrium (drop the book) and they instantly differ.

Step 3 — Find the true partners (swap the labels).

• Partner of "Earth pulls book down" $=$ "book pulls Earth up" (gravity ↔ gravity).
• Partner of "table pushes book up" $=$ "book pushes table down" (normal ↔ normal). In the figure's right half, each real pair spans two different dots (book and Earth; book and table). Why this step? Rule of partners — swap $A\leftrightarrow B$, keep the type the same (test 4 above).

Step 4 — Answer (c). No. Weight and normal both act on the same body (book) and are different types → they fail test 3 (same type) and test 5 (two different bodies) of the five pair-tests defined above.

Verify: $N=19.6\text{N}$ upward exactly balances $W=19.6\text{N}$ down → net $=0\text{N}$, consistent with "at rest". Units: $\text{kg}\cdot {\text{m/s}}^2=\text{N}$. ✓

Ex 2 — Two skaters push apart (Cell C2: symmetric free bodies)

Step 1 — Identify the third-law pair. ${\vec{F}}_{A\to B}=-{\vec{F}}_{B\to A}$, both of magnitude $F=120\text{N}$. Why this step? The only horizontal force on each skater is the push from the other — a clean pair.

Step 2 — Turn force into velocity via impulse. Impulse $=$ force $\times$ time $=$ change in momentum. This is the Impulse–Momentum Theorem (the direct time-integral of Newton's Second Law, not conservation of momentum): $Ft=mv.$ Why this step? Force alone can't give a speed; we need how long it acts. $t$ is the bridge.

Step 3 — Solve each skater. $v_A=\frac{Ft}{m_A}=\frac{120\times 0.50}{60}=1.0\text{m/s},v_B=\frac{120\times 0.50}{40}=1.5\text{m/s}.$ Why this step? Same $Ft$ on top, smaller mass below ⇒ bigger speed. The lighter skater is faster.

Step 4 — Ratio. $\frac{v_B}{v_A}=\frac{Ft/m_B}{Ft/m_A}=\frac{m_A}{m_B}=\frac{60}{40}=\mathrm{1.5.}$ Why this step? Taking the ratio makes the shared impulse $Ft$ cancel top-and-bottom, leaving only the inverse mass ratio. This is the cleanest proof that "equal force ≠ equal speed": the outcome depends purely on how the masses compare, not on $F$ or $t$ at all.

Verify (via conservation of momentum): the equal-opposite impulses on the two skaters guarantee total momentum stays zero (started at rest) — this is where Conservation of Momentum enters. $m_A{v}_A-m_B{v}_B=60(1.0)-40(1.5)=60-60=0$. ✓ Same force, unequal accelerations — exactly the "equal force ≠ equal effect" lesson.

Ex 3 — Two blocks pushed together (Cell C3: internal vs external)

Step 1 — Answer (a): treat the two blocks as ONE system. Total mass $=3+2=5\text{kg}$. The only external horizontal force is your $10\text{N}$: $a=\frac{F}{m_1+m_2}=\frac{10}{5}=2{\text{m/s}}^2.$ Why this step? The contact force between the blocks is internal — it is a third-law pair and cancels in the whole-system view. So it can't affect the shared $a$. This is Center of Mass Motion logic.

Step 2 — Answer (b): isolate $m_2$ to find the contact force. The only force on $m_2$ is the push $C$ (contact magnitude) from $m_1$: $C=m_{2}a=2\times 2=4\text{N}.$ Why this step? The contact force is internal to the two-block system, so it never showed up in Step 1. To reveal it we must break the system apart and draw $m_2$ alone (Free Body Diagrams): once isolated, $C$ is the only horizontal force on $m_2$, so Newton's second law $C=m_{2}a$ hands it to us directly. In the figure this is the right block with its single green arrow.

Step 3 — Interpret (b): is the contact force $10\text{N}$? No. The contact force is $4\text{N}$, less than your $10\text{N}$ — because $m_1$ keeps some of the push to accelerate itself.

Step 4 — Answer (c): verify the pair. By the third law, $m_2$ pushes back on $m_1$ with $4\text{N}$. Check $m_1$ alone: $F-C=m_{1}a⟹10-4=3\times 2=6\text{N}.✓$

Verify: whole system $F=(m_1+m_2)a=5\times 2=10\text{N}$ ✓. The $4\text{N}$ pair is internal and equal-opposite; it never appears in the system equation.

Ex 4 — The instant of release (Cell C4: degenerate / zero force)

Step 1 — Answer (a): contact broken ⇒ normal force is zero. $N=0\text{N}.$ Why this step? Normal force is a contact force; no contact, no force. This is the degenerate case — an input goes to zero.

Step 2 — Answer (b): its partner also vanishes, simultaneously. The pair "table pushes book up" and "book pushes table down" both drop to $0\text{N}$ at the same instant. A pair is always born together and dies together — there is no time delay. Why this step? Kills the misconception "reaction lags action". Zero and zero, at once.

Step 3 — Answer (c): only gravity remains. $a=\frac{W}{m}=\frac{mg}{m}=g=9.8{\text{m/s}}^2\text{(downward)}.$ Why this step? Gravity's partner ("book pulls Earth up") still exists — gravity is not a contact force. But it acts on the Earth, not the book, so it doesn't slow the fall.

Verify: With $N=0$, net force on book $=W=19.6\text{N}$ down, so $a=19.6/2=9.8{\text{m/s}}^2=g$. ✓ Free fall, as expected.

Ex 5 — Truck vs. fly (Cell C5: extreme mass ratio limit)

Step 1 — Forces are exactly equal (third law). $F_{\text{truck}\to \text{fly}}=F_{\text{fly}\to \text{truck}}=5\text{N}.$ Why this step? No matter the mass, the pair is identical in magnitude. "Bigger object hits harder" is false.

Step 2 — Accelerations from $a=F/m$. $a_{\text{fly}}=\frac{5}{0.001}=5000{\text{m/s}}^2,a_{\text{truck}}=\frac{5}{2000}=0.0025{\text{m/s}}^2.$ Why this step? The mass sits in the denominator, so the tiny fly gets a monstrous acceleration.

Step 3 — Ratio (the limiting behaviour). $\frac{a_{\text{fly}}}{a_{\text{truck}}}=\frac{F/m_{\text{fly}}}{F/m_{\text{truck}}}=\frac{m_{\text{truck}}}{m_{\text{fly}}}=\frac{2000}{0.001}=2000000.$ Why this step? Taking the ratio cancels the shared force $F$, leaving the inverse mass ratio. This makes the physical lesson exact: as the mass ratio $\to \infty$, the acceleration ratio $\to \infty$ too, so the heavy body barely notices while the light body is annihilated — same force, opposite fates. (It is why "you push Earth up as hard as it pulls you down", yet Earth doesn't visibly move.)

Verify: $a_{\text{fly}}/a_{\text{truck}}=5000/0.0025=2000000$ ✓, equal to $m_{\text{truck}}/m_{\text{fly}}$. Units: $\text{N}/\text{kg}={\text{m/s}}^2$ ✓.

Ex 6 — Rocket in vacuum (Cell C6: continuous ejection)

Step 1 — Answer (a): thrust = momentum given to the exhaust per second. $T=\dot{m}u=5\times 2000=10000\text{N}.$ Why this step? Every second the rocket flings $\dot{m}$ of gas backward at speed $u$; by Conservation of Momentum, the rocket gains equal forward momentum. This is Rocket Propulsion & Variable Mass.

Step 2 — Answer (b): acceleration. $a=\frac{T}{m}=\frac{10000}{1000}=10{\text{m/s}}^2.$ Why this step? Newton's second law on the rocket, using thrust as the forward force.

Step 3 — Answer (c): name the partner. ${\vec{F}}_{\text{rocket}\to \text{gas}}=-{\vec{F}}_{\text{gas}\to \text{rocket}}$. The partner is the exhaust gas, NOT the surrounding air. Why this step? The question "what does a rocket push against in vacuum?" only confuses if you look for an external medium. The third law answers it cleanly: the reaction partner is the gas the rocket itself ejects, so no air is required — that is exactly why rockets work in space.

Verify: momentum bookkeeping per second: gas carries $\dot{m}u=5\times 2000=10000\text{kg⋅m/s}$ backward; rocket receives $10000\text{kg⋅m/s}$ forward → thrust $10000\text{N}$ ✓. $a=10000/1000=10{\text{m/s}}^2$ ✓.

Ex 7 — Walking / swimming (Cell C7: real-world word problem)

Step 1 — Answer (a): the forward force is the ground's reaction. Foot pushes ground backward with $140\text{N}$ ⇒ ground pushes person forward with $140\text{N}$ (friction pair, Normal Force and Friction). In the figure, the red arrow (you on ground) points back, the blue arrow (ground on you) points forward. Why this step? You never propel yourself directly; the ground is your engine. On frictionless ice you'd push and go nowhere.

Step 2 — Answer (b): acceleration of the person. $a=\frac{F}{m}=\frac{140}{70}=2{\text{m/s}}^2\text{forward}.$ Why this step? Only the forward ground-force acts on the person horizontally.

Step 3 — Answer (c): does the Earth move? (limiting case again). Yes, by an unmeasurable amount: $a_{\text{Earth}}=140/(6\times 10^{24})\approx 2.3\times 10^{-23}{\text{m/s}}^2$ — the C5 mass-ratio limit in disguise.

Verify: $a=140/70=2{\text{m/s}}^2$ ✓. Same-magnitude pair $140\text{N}$; the backward force is on the ground, the forward force is on you — different bodies, no cancellation.

Ex 8 — Contact force in a "train" of blocks (Cell C8: exam twist)

Step 1 — System acceleration. $a=\frac{F}{m_1+m_2+m_3}=\frac{12}{1+2+3}=\frac{12}{6}=2{\text{m/s}}^2.$ Why this step? All the between-block contact forces are internal third-law pairs; each pair is equal-and-opposite, so summed over the whole train they cancel. That leaves only the external $F$ to drive the total $6\text{kg}$ — which is why treating the three blocks as one body gives the shared $a$ in a single line.

Step 2 — Isolate $m_3$. The only horizontal force on $m_3$ is the contact push $C_{23}$ (contact magnitude) from $m_2$: $C_{23}=m_{3}a=3\times 2=6\text{N}.$ Why this step? $C_{23}$ was internal to the whole train, so Step 1 could never show it. Drawing $m_3$ alone exposes it: with nothing else touching $m_3$ horizontally, $C_{23}$ is the sole force accelerating it, so Newton's second law $C_{23}=m_{3}a$ gives it immediately. General rule: a contact carries exactly the mass ahead of it.

Step 3 — Third-law check. $m_3$ pushes back on $m_2$ with the same $6\text{N}$: a genuine equal-opposite contact pair, acting on two different blocks.

Verify: contact between $m_1,m_2$ must push $(m_2+m_3)$: $C_{12}=(2+3)\times 2=10\text{N}$. Consistency on $m_1$: $F-C_{12}=12-10=2=m_{1}a=1\times 2$ ✓. And $C_{23}=3\times 2=6\text{N}$ ✓.

Rapid recall

Connections

• Newton's Third Law — Action & Reaction — the parent law these examples exercise.
• Newton's Second Law — $a=F/m$ converts equal forces into unequal accelerations (Ex 2, 5).
• Conservation of Momentum — the check behind Ex 2 and Ex 6.
• Free Body Diagrams — isolating one body to reveal internal pairs (Ex 3, 8).
• Normal Force and Friction — the pair that lets you walk (Ex 7).
• Rocket Propulsion & Variable Mass — continuous-ejection thrust (Ex 6).
• Center of Mass Motion — why internal pairs never affect system acceleration (Ex 3, 8).