TECHNIQUE: Which Conservation Law?

Particle physicists run this module's audit millions of times per second. Every recorded collision at the LHC is triaged: momentum conservation is imposed per axis (transverse components must close, exactly like the fragment problems), while the energy books are allowed spectacular entries — mass created, mass destroyed — provided the total balances. "Missing transverse momentum" is Question 5 of the audit, automated: an axis whose account doesn't close signals an invisible particle. Crash investigators run the two-ledger split on wrecked cars: momentum across the impact reconstructs velocities (crumpling can't hide momentum), then work–energy along the skid marks rewinds each vehicle's approach — Drill 5 with insurance consequences. And every engineering simulation of impacts, from phone drops to bird-strike tests on jet engines, is built on solvers that enforce momentum exactly at contacts while tracking energy loss through material models — the triage, compiled.

The two ledgers keep books on different things, so they are audited by different inspectors. Momentum's inspector only checks the boundary: internal forces — however enormous, however dissipative — always cancel in third-law pairs, so a bomb can shred itself without moving the momentum total by a hair. The only entries that count are pushes from outside the system, and over a millisecond event, finite outside forces (gravity, friction) haven't the time to matter. Momentum across an event fails in exactly one way: something rigid — a wall, a hinge, the ground — supplies an impulsive external force.

Energy's inspector doesn't care about the boundary at all — energy has no direction and can't be "carried off" by the system's motion. It checks the books inside: did any kinetic energy leave the mechanical accounts, into heat, sound, deformation? Sticking, embedding, tearing, rubbing — each is a leak. Smooth swings, frictionless slides, spring compressions — no leaks, energy conserved. Energy across a stretch fails in exactly one way: dissipation (or an un-tracked energy source, like an explosion's chemistry — then energy books still balance, but only with the $Q$ entry added).

One more picture, because it decides the hard cases: momentum is a vector with a separate account per direction. A ball bouncing obliquely off a smooth floor takes a vertical impulse from the ground (vertical account raided) while nothing horizontal touches it (horizontal account intact): conserve $p_x$, apply restitution to $p_y$. A wall clamps one direction, not all three. Whenever "is momentum conserved?" feels ambiguous, ask it one axis at a time.

where: the signals and stamps are catalogued below — this table is the technique; the drills install it.

where: the last three rows are the vector rows — the ones that separate JEE-Main from JEE-Advanced scores. Note what is never a signal: the size of the forces, the drama of the event, or the roughness of a surface during a brief impact (impulse, not work, decides events — the bullet-block module's central lesson).

The five-question audit ("is momentum conserved here?")

Question 1. What exactly is my system? Name the bodies. (You may redraw: putting the wall or the Earth inside the system makes its push internal — sometimes that is the winning move, usually it just hides the question.)

Question 2. What is my time window? Conservation claims are before/after statements across a named window. "Across the collision" and "across the swing" are different claims with different verdicts.

Question 3. List the external forces in that window. Gravity? Normal forces? Friction? Tension anchored outside?

Question 4. Impulsive or finite? Finite forces (gravity, ordinary friction) over a millisecond event: negligible impulse — momentum survives the event. The same forces over a half-second process: fatal. Impulsive external forces (wall, hinge, ground, taut string anchored outside): fatal even in a millisecond.

Question 5. Axis by axis. A verdict per direction: an oblique floor bounce conserves $p_x$ and not $p_y$; a shell in flight conserves nothing over the whole flight but everything across the burst.

The reflexes this technique replaces are themselves dangerous when inverted. Three failure modes deserve names. Ledger-swapping: energy across a sticky event (the ballistic-pendulum sin — wrong by $\sqrt{(m+M)/m}$) or momentum across a swing (gravity and tension drain it). Window-stretching: "momentum before launch = momentum at landing" across seconds of gravity — conservation windows must be short or force-free, and a whole projectile flight is neither. Boundary-blindness: applying momentum conservation to a block resting against a wall, or a ball hitting the ground vertically — the rigid constraint supplies whatever impulse it likes, and the missing momentum went into the planet. When you catch yourself writing a conservation line, say the window and the system out loud; if you cannot, the line is not yet physics. And one honest limitation: triage tells you which laws to write, not how many unknowns you have. Partial-information collisions (given $e$, or given a KE-loss percentage) need the momentum line plus the given statement as a second equation. "Elastic" is the only word that hands you two conservation equations for free.

Drill 1 · Board level — stamp a two-act story (full method)

A $2.0\ \mathrm{kg}$ block slides from rest down a smooth ramp of height $0.80\ \mathrm{m}$, then hits an identical block at the bottom and sticks to it. Take $g = 10\ \mathrm{m/s^2}$. Find the final speed and the kinetic energy lost.

Step 1. Triage. Act 1: "smooth ramp" → energy. Act 2: "sticks" → momentum only. Two stamps, one boundary at the moment of impact.

Step 2. Act 1 (energy): $v = \sqrt{2gh} = \sqrt{2\times10\times0.80} = 4.0\ \mathrm{m/s}$ at the bottom.

Step 3. Act 2 (momentum): $2.0\times4.0 = 4.0\,V$, so $V = \boxed{2.0\ \mathrm{m/s}}$.

Step 4. KE lost in the sticky hit: $\tfrac12(2)(16) - \tfrac12(4)(4) = 16 - 8 = \boxed{8.0\ \mathrm{J}}$ — exactly half, as equal-mass sticking always destroys half. Note what was never written: energy across the impact, momentum along the ramp.

Drill 2 · Board level — one word changes the physics

Same setup, but the blocks collide "perfectly elastically" instead of sticking. Find both final velocities.

Step 1. Re-triage: "elastic" → both laws across the hit. For equal masses, both-laws has a memorized verdict: complete exchange.

Step 2. The moving block stops; the struck block leaves at $\boxed{4.0\ \mathrm{m/s}}$. KE lost: zero.

Step 3. The lesson, stated plainly: identical numbers, one changed word, different law set, different answer ($2.0$ m/s shared vs $4.0$ m/s transferred). The word is the physics. Train your eye to hit "sticks/elastic/$e = \dots$" before anything else in a collision sentence.

Drill 3 · Board level — partial information: "loses 36% of its kinetic energy"

A $1.0\ \mathrm{kg}$ ball moving at $10\ \mathrm{m/s}$ hits a stationary $1.0\ \mathrm{kg}$ ball head-on. The collision destroys 36% of the initial kinetic energy. Find the final velocities.

Step 1. Triage: a collision, neither "elastic" nor "sticks" — momentum line plus the given energy statement (not energy conservation).

Step 2. Momentum: $v_1 + v_2 = 10$. Energy statement: $\tfrac12v_1^2 + \tfrac12v_2^2 = 0.64\times50 = 32$, i.e. $v_1^2 + v_2^2 = 64$.

Step 3. Solve: $(v_1+v_2)^2 = 100 \Rightarrow 2v_1v_2 = 36 \Rightarrow v_1v_2 = 18$. Then $v_1, v_2$ are roots of $t^2 - 10t + 18 = 0$: $t = 5 \pm \sqrt{7} \approx 7.65$ and $2.35$ m/s. Physical assignment (the struck ball must end faster): $v_2 = \boxed{7.6\ \mathrm{m/s}}$, $v_1 = \boxed{2.4\ \mathrm{m/s}}$.

Step 4. Consistency: $e = (v_2-v_1)/10 = 2\sqrt7/10 \approx 0.53$ — between 0 and 1 ✓. Partial-information collisions always resolve into "momentum + one more equation."

Drill 4 · Competitive level — the oblique bounce: one axis lives, one dies (JEE-Main pattern, ~3 minutes)

A ball hits a smooth floor at $45^\circ$ to the vertical with speed $14.1\ \mathrm{m/s}$ ($\approx 10\sqrt2$). The coefficient of restitution is $e = 0.50$. Find the rebound speed and angle.

Step 1. Triage per axis. The floor is smooth: no horizontal force ⇒ $p_x$ conserved: $v_x = 10\ \mathrm{m/s}$ unchanged. Vertical: the ground's impulsive normal acts ⇒ momentum dead; restitution rules: $v_y' = e\,v_y = 0.5\times10 = 5.0\ \mathrm{m/s}$ upward.

Step 2. Rebound speed: $\sqrt{10^2 + 5^2} = \sqrt{125} = \boxed{11.2\ \mathrm{m/s}}$.

Step 3. Rebound angle to the vertical: $\tan\phi = v_x/v_y' = 10/5 = 2$, so $\phi = \boxed{63^\circ}$ — the ball rebounds flatter than it arrived (the vertical component was taxed, the horizontal was not). Timing note: ~2.5 min. The whole question is the per-axis verdict; students who treat $e$ as acting on the speed (giving 7.05 m/s at 45°) have applied a scalar tax to a vector event — the flagship distractor.

Drill 5 · Competitive level — three stamps in a row (JEE-Main/NEET pattern, ~4 minutes)

A pendulum bob of mass $1.0\ \mathrm{kg}$ on a string of length $0.80\ \mathrm{m}$ is released from the horizontal. At the lowest point it strikes a $1.0\ \mathrm{kg}$ block elastically. The block then slides onto a rough patch ($\mu = 0.40$). Take $g = 10\ \mathrm{m/s^2}$. How far into the patch does the block go?

Step 1. Triage the three acts: swing (smooth, string) → energy; strike (elastic, equal masses) → both laws = exchange; rough slide → work–energy.

Step 2. Act 1: $v = \sqrt{2gL} = \sqrt{16} = 4.0\ \mathrm{m/s}$ at the bottom.

Step 3. Act 2: equal-mass elastic hit — the bob stops, the block takes $4.0\ \mathrm{m/s}$.

Step 4. Act 3: $d = \dfrac{v^2}{2\mu g} = \dfrac{16}{2\times0.4\times10} = \boxed{2.0\ \mathrm{m}}$. Timing note: ~3 min, almost all of it bookkeeping. Every act used a different law, and no law crossed an act boundary — that is the entire technique performing. A wrong stamp anywhere (momentum on the swing, energy on a sticky version of the hit) poisons every number downstream, which is why examiners love the chain format.

Energy across a sticky event. The classic. "Embeds/sticks" destroys $\tfrac12\mu u_{\text{rel}}^2$ of kinetic energy, non-negotiably. If your solution equates KE before and after a capture, stop.

Momentum across a smooth process. Gravity and tension have all the time they need over a swing or slide. Momentum lines are for events (or genuinely force-free stretches), never for "from the top of the swing to the bottom."

The whole-flight window. No conservation of momentum from launch to landing of a projectile — gravity's impulse over seconds is the whole story of the flight. Bursts during flight get a momentum line only across the burst itself.

Wall, hinge, ground, clamp. Rigid external constraints deliver impulses on demand: momentum along the constrained direction dies. Ball hitting the ground vertically, block against a wall, rod hinged at one end — each has a "missing momentum" that went into the Earth.

Scalar-taxing a vector. Restitution acts on the normal component only; smooth surfaces spare the parallel component entirely. Applying $e$ to the whole speed of an oblique bounce is the single most-harvested wrong answer in this topic.

"Both laws" without licence. Two conservation equations require the word "elastic." Given $e$ or a percentage loss, you get momentum plus that statement — not energy conservation. Given nothing, you get momentum alone (and possibly an unknown too many — check whether the question is really asking for a range or an inequality).

System-boundary theatrics. "Include the Earth and momentum is always conserved" is true and almost always useless — the Earth's velocity change is unmeasurable, and the exam wants the block's. Redraw boundaries to make forces internal only when the newly included body's motion is actually solvable or negligible-by-choice.

Texts. HC Verma, Concepts of Physics Vol. 1, Ch. 9 (the chapter this technique compresses) · NCERT Physics Class 11, Ch. 6 · For the symmetry story behind the ledgers: any modern-physics account of Noether's theorem.

Provenance. All results are standard consequences of the momentum and work–energy theorems. A verifier should recompute: Drill 1 (4.0 → 2.0 m/s, 8 J lost); Drill 3's quadratic ($v = 5\pm\sqrt7$, $e = 2\sqrt7/10 \approx 0.53$); Drill 4's per-axis bounce ($11.2$ m/s at $63^\circ$ to the vertical); Drill 5's chain (4.0 m/s → exchange → 2.0 m).

Prerequisites. Conservation of Linear Momentum · Conservation of Mechanical Energy & Efficiency · Inelastic Collisions & Coefficient of Restitution. Leads to. Force Method or Energy Method? (technique) · Option Elimination & Answer Checking (track capstone).