ARCHETYPE: Two Blocks & a Spring

Every elastic collision you have ever solved is secretly a spring problem played too fast to watch. When two steel balls "bounce," they really compress like stiff springs for a few microseconds, store the collision's energy as deformation, and then pay every joule back as they un-squash. This archetype slows that movie down until you can see every frame: put an actual spring on the front of one block, slide another block into it, and the whole hidden anatomy of a collision — the compression, the moment of maximum squeeze, the rebound — stretches out over visible fractions of a second. The setup earns its place in JEE papers because it forces the one question students dodge: what exactly is conserved, and when? Momentum: always — the spring's pushes are internal to the two-block system. Kinetic energy: only at the ends of the story, when the spring is relaxed; in between, some of it hides in the spring as elastic potential energy. And the star moment — maximum compression — happens at the exact instant the blocks move with a common velocity, because the spring stops compressing only when the blocks stop approaching each other. Master this and you get three archetypes for one: run it forward, it is an elastic collision in slow motion; latch the spring at full squeeze, it becomes a perfectly inelastic collision with the "lost" energy visibly stored; run it backward from a compressed start, it is an explosion. One spring, the whole momentum chapter.

Key takeaways

Block $m_1$ (speed $u$ ) slides into a light spring (constant $k$ ) mounted on block $m_2$ at rest, on a smooth floor. Momentum is conserved throughout — the spring is internal.
Maximum compression happens when both blocks move with the common velocity $v_c = \dfrac{m_1 u}{m_1+m_2}$ — the instant the approach stops.
Energy hidden in the spring at that moment: $\tfrac12 k x_{\max}^2 = \tfrac12\,\dfrac{m_1 m_2}{m_1+m_2}\,u^2$ , so $x_{\max} = u\sqrt{\dfrac{\mu}{k}}$ with $\mu = \dfrac{m_1m_2}{m_1+m_2}$ .
The spring then pays everything back: the blocks separate with exactly the elastic-collision velocities.
Latch the spring at full squeeze → perfectly inelastic outcome, with the "lost" KE stored, not destroyed.
Released compressed spring between blocks → explosion: speeds inverse to masses.
Key idea: the spring makes a collision's interior visible — momentum holds at every instant, kinetic energy dips into the spring and returns, and maximum compression coincides with common velocity.

The core insight

Momentum conservation holds at every instant ( $m_1u = m_1v_1 + m_2v_2$ ); maximum compression occurs when relative velocity vanishes (both blocks at $v_c = \frac{m_1u}{m_1+m_2}$ ), storing $\frac12 kx_{\max}^2 = \frac12\mu u^2$ ( $\mu$ = reduced mass) — and if the spring relaxes again, the encounter finishes as a perfectly elastic collision.

Where this lives today

Railway shunting yards run Example 1 all day: buffer springs between wagons are sized by exactly $\tfrac12\mu u^2$ — the relative-motion energy they must swallow so that couplings survive; the COM of the wagon pair sails on regardless, just as equation (2) says. Spacecraft docking systems are the gentle version: soft-capture rings on the ISS ports are sprung so that an approaching craft compresses, reaches common velocity with the station (frame 2 of the hero figure), and is latched — Example 4's "latch" is literal flight hardware. Vehicle safety engineering splits the same budget the other way: bumpers act as springs (payback, elastic, no damage) for parking-speed hits, while crumple zones deliberately refuse to pay back — converting the relative-motion energy into deformation, an inelastic outcome by design. And at the smallest scale, chemists model molecular collisions as atoms on spring-like potentials: "elastic" bounces and "reactive capture" (a latch made of chemistry) are frames 3 and 2 of this article, run at picosecond speed.

The setup

On a frictionless floor, block $m_2$ (spring of stiffness $k$ fixed to its face) rests; block $m_1$ arrives at speed $u$ . Find: the common velocity, the maximum compression, and the final velocities after separation. Variants: spring latched at full squeeze; compressed spring released between two blocks; spring backed by a wall.

Figure

Three frames of the encounter — approach, maximum squeeze, payback

The heart of the archetype is one sentence about frame 2: the spring keeps compressing as long as the blocks keep approaching. Compression $x$ grows whenever $m_1$ moves faster than $m_2$ ; it shrinks whenever $m_2$ outruns $m_1$ . Maximum squeeze is therefore the turning point where the relative velocity is exactly zero — both blocks momentarily gliding at the same speed. Not " $m_1$ stops" (it never does here), not " $m_2$ hasn't started" (it started the instant the spring touched it): common velocity. From that one insight, momentum hands you $v_c$ and energy hands you $x_{\max}$ , and the problem is over.

The master solution pattern

Momentum, valid at every instant. The spring's forces are internal to the pair, the floor is smooth, so at any moment:

m_1 u = m_1 v_1 + m_2 v_2 \qquad (1)

where: $v_1, v_2$ — the instantaneous velocities. This is a running conservation law, not just a before/after statement — you may deploy it at maximum compression, at half compression, anywhere.

The common velocity at maximum squeeze. Setting $v_1 = v_2 = v_c$ in (1):

v_c = \frac{m_1\,u}{m_1 + m_2} \qquad (2)

where: $v_c$ is also the velocity of the centre of mass — it must be, since at that instant the system moves as one piece. The blocks never both stop (momentum forbids it); they sail together at $v_c$ while the spring holds its maximum charge.

Energy at maximum squeeze. Kinetic energy is conserved plus spring energy: comparing the start with the moment of max compression,

\tfrac12 m_1 u^2 = \tfrac12 (m_1+m_2)v_c^2 + \tfrac12 k x_{\max}^2 \quad\Longrightarrow\quad \tfrac12 k x_{\max}^2 = \tfrac12\,\frac{m_1 m_2}{m_1+m_2}\,u^2 \qquad (3)

where: the combination $\mu = \dfrac{m_1m_2}{m_1+m_2}$ (the reduced mass) does the bookkeeping in one stroke: the storable energy is exactly the energy of relative motion, $\tfrac12\mu u^2$ — the centre-of-mass share $\tfrac12(m_1+m_2)v_c^2$ is untouchable, protected by momentum conservation. Hence

x_{\max} = u\,\sqrt{\frac{\mu}{k}} = u\,\sqrt{\frac{m_1 m_2}{k\,(m_1+m_2)}} \qquad (4)

The payback. The compressed spring now accelerates $m_2$ and decelerates $m_1$ until it returns to natural length — at which point it has repaid every joule (light spring, no friction). Momentum (1) plus full kinetic-energy recovery is precisely the definition of an elastic collision, so the exit velocities are the standard elastic results:

v_1' = \frac{m_1 - m_2}{m_1+m_2}\,u, \qquad v_2' = \frac{2m_1}{m_1+m_2}\,u \qquad (5)

where: for $m_1 < m_2$ , $v_1'$ is negative — the light block bounces back. Equal masses swap: $v_1' = 0$ , $v_2' = u$ , the collision chapter's exchange rule, now visible in slow motion.

variant	what changes	result
standard (spring on $m_2$ 's face)	—	eqs. (2)–(5)
spring latched at max squeeze	energy stays stored	inelastic outcome: both at $v_c$ ; spring holds $\tfrac12\mu u^2$
latch released later	stored energy repaid	final velocities still (5) — timing irrelevant
compressed spring released from rest	explosion	$m_1v_1 = m_2v_2$ ; energies inverse to masses
spring backed by a wall	momentum NOT conserved	energy alone: $\tfrac12 m u^2 = \tfrac12 kx_{\max}^2$
spring attached to BOTH blocks	they never separate	oscillation about the COM; velocities trade periodically

Derivation ACommon velocity at maximum compression, and the reduced-mass energy split

We prove the two pillars — that maximum compression coincides with common velocity, and that the storable energy is $\tfrac12\mu u^2$ — because both are asserted-by-habit in solutions and JEE tests whether you know why. Strategy: track the compression through the relative velocity, then split the kinetic energy into a centre-of-mass part and a relative part and watch momentum conservation freeze the first. Keep an eye on what the spring can and cannot touch.

Step 1. Compression grows with approach. The spring's length shortens at the rate at which the blocks close in: the compression $x$ increases while $v_1 > v_2$ , decreases while $v_1 < v_2$ . At the maximum, the rate of change is zero:

v_1 = v_2 \equiv v_c

Why this is airtight: $x$ is a running total of (approach speed × time); a running total peaks exactly when the thing being totalled crosses zero. No forces, no calculus formalities — just the definition of compression.

Step 2. Momentum fixes $v_c$ . Equation (1) with both velocities equal: $v_c = \dfrac{m_1u}{m_1+m_2}$ — which is the centre-of-mass velocity, constant from the first touch to the last. The blocks momentarily ride along with their own COM.

Step 3. Split the kinetic energy. Quite generally, the pair's kinetic energy separates into a COM part plus a relative part:

KE = \tfrac12(m_1+m_2)v_{\text{cm}}^2 + \tfrac12\,\mu\,v_{\text{rel}}^2, \qquad \mu = \frac{m_1m_2}{m_1+m_2}

(Substituting $v_{\text{cm}}$ and $v_{\text{rel}} = v_1 - v_2$ and expanding reproduces $\tfrac12m_1v_1^2 + \tfrac12m_2v_2^2$ — a two-line algebra check worth doing once in your life.)

Step 4. Watch what the spring can touch. Momentum conservation keeps $v_{\text{cm}}$ constant, so the COM share $\tfrac12(m_1+m_2)v_{\text{cm}}^2$ is locked — no internal force can raise or spend it. The spring trades only with the relative share, initially $\tfrac12\mu u^2$ (since $v_{\text{rel}} = u$ at the start). At maximum compression $v_{\text{rel}} = 0$ : the relative share sits entirely in the spring:

\tfrac12 k x_{\max}^2 = \tfrac12\,\mu\,u^2 \quad\Longrightarrow\quad x_{\max} = u\sqrt{\mu/k}

Step 5. The payback and the elastic ending. As the spring re-extends it returns the stored energy to the relative motion; at natural length $v_{\text{rel}}$ has recovered its full magnitude $u$ but reversed in direction (the blocks now separate as fast as they once approached — the $e = 1$ signature). Momentum plus reversed relative velocity is exactly the elastic-collision system, giving (5). Verified: numbers — $m_1 = 2$ , $m_2 = 4$ kg, $u = 6$ m/s, $k = 800$ N/m: $v_c = 2$ m/s; COM share $= \tfrac12(6)(4) = 12$ J; relative share $= 36 - 12 = 24$ J $= \tfrac12\mu u^2 = \tfrac12(4/3)(36)$ ✓; $x_{\max} = \sqrt{2(24)/800} = 0.245$ m ✓. Limits: $m_2 \to \infty$ : $v_c \to 0$ , $\mu \to m_1$ , wall formula recovered; $k \to \infty$ : $x_{\max} \to 0$ , the invisible-spring (instant collision) limit.

Limiting cases

case	result	meaning
$m_2 \to \infty$ (wall)	$v_c \to 0$ , $\tfrac12kx_{\max}^2 = \tfrac12 m_1u^2$	all KE storable — the wall version's licence
$m_1 = m_2$	$v_c = u/2$ ; half the KE stored at max squeeze	then full exchange: $v_1' = 0$ , $v_2' = u$
$k$ very stiff	tiny $x_{\max}$ , brief contact	the hidden spring of "instant" elastic collisions
$k$ very soft	large, slow squeeze	same energies, gentler forces, longer contact
latched at $x_{\max}$	both blocks at $v_c$ for good	perfectly inelastic, with the loss visibly banked
spring attached to both	no separation, SHM about COM	velocities oscillate between $u,0$ and $v_1',v_2'$

Common trap“At maximum compression the front block has stopped — or the whole system has”

Neither. If $m_1$ had stopped while $m_2$ moved, or both had stopped, momentum would not add up — equation (1) forbids it (the total is $m_1u \ne 0$ throughout). At maximum compression the blocks move together at $v_c = \frac{m_1u}{m_1+m_2}$ : the approach has paused, not the motion. The related energy blunder writes $\tfrac12 m_1u^2 = \tfrac12 kx_{\max}^2$ — dumping all the kinetic energy into the spring. That is legal only when the spring is backed by a wall (or $m_2$ is effectively infinite); for a free $m_2$ , the centre-of-mass share $\tfrac12(m_1+m_2)v_c^2$ can never be touched by any internal force, and only the relative-motion energy $\tfrac12\mu u^2$ is available for storage. Using the wall formula on a free block inflates $x_{\max}$ by the factor $\sqrt{(m_1+m_2)/m_2}$ — for equal masses, a 41% overshoot, and yes, that wrong answer is among the options.

Worked examples

1 · Board level — common velocity and maximum compression (full method)

A $2.0\ \mathrm{kg}$ block moving at $6.0\ \mathrm{m/s}$ on a smooth floor runs into a light spring ( $k = 800\ \mathrm{N/m}$ ) mounted on a stationary $4.0\ \mathrm{kg}$ block. Find (a) the velocity of each block at maximum compression, (b) the maximum compression.

Step 1. At maximum compression the blocks share a common velocity (approach has paused). Momentum: $v_c = \dfrac{2.0 \times 6.0}{6.0} = \boxed{2.0\ \mathrm{m/s}}$ — both blocks.

Step 2. Energy audit at that instant. Initial KE: $\tfrac12(2.0)(36) = 36$ J. KE at common velocity: $\tfrac12(6.0)(4.0) = 12$ J. Stored in the spring: $36 - 12 = 24$ J.

Step 3. Compression: $\tfrac12 k x_{\max}^2 = 24$ , so $x_{\max} = \sqrt{\dfrac{48}{800}} = \sqrt{0.06} = \boxed{0.245\ \mathrm{m}}$ (about 24.5 cm).

Step 4. Shortcut check via reduced mass: $\mu = \dfrac{2\times4}{6} = \dfrac43$ kg; $\tfrac12\mu u^2 = \tfrac12 \times \tfrac43 \times 36 = 24$ J ✓ — one line instead of three.

2 · Board level — the spring as an explosion (release from rest)

A spring compressed by $0.10\ \mathrm{m}$ ( $k = 600\ \mathrm{N/m}$ ) is held between blocks of $1.0\ \mathrm{kg}$ and $2.0\ \mathrm{kg}$ on a smooth floor, everything at rest. The system is released. Find each block's final speed.

Step 1. Stored energy: $\tfrac12(600)(0.10)^2 = 3.0$ J. Momentum from rest: $1.0\,v_1 = 2.0\,v_2$ (opposite directions), so $v_1 = 2v_2$ .

Step 2. Energy: $\tfrac12(1.0)v_1^2 + \tfrac12(2.0)v_2^2 = 3.0$ . Substitute $v_1 = 2v_2$ : $2v_2^2 + v_2^2 = 3.0$ , so $v_2^2 = 1.0$ .

Step 3. $v_2 = \boxed{1.0\ \mathrm{m/s}}$ , $v_1 = \boxed{2.0\ \mathrm{m/s}}$ , opposite ways. Audit: energies $2.0$ J and $1.0$ J — inverse to the masses, the explosion module's split, with the spring as the "charge."

3 · Competitive level — the final velocities (JEE-Main pattern, ~3 minutes)

In the setup of Example 1 ( $2.0$ kg at $6.0$ m/s, $4.0$ kg at rest), the spring re-extends and the blocks separate. Find the final velocity of each block.

Step 1. Recognize the ending: light spring returns all its energy ⇒ momentum conserved and kinetic energy fully recovered ⇒ elastic collision in slow motion. Use the standard results.

Step 2. $v_1' = \dfrac{m_1 - m_2}{m_1+m_2}u = \dfrac{-2}{6}\times6.0 = \boxed{-2.0\ \mathrm{m/s}}$ — the light block rebounds.

Step 3. $v_2' = \dfrac{2m_1}{m_1+m_2}u = \dfrac{4}{6}\times6.0 = \boxed{+4.0\ \mathrm{m/s}}$ .

Step 4. Audits: momentum $2(-2) + 4(4) = 12 = 2(6)$ ✓; energy $\tfrac12(2)(4) + \tfrac12(4)(16) = 4 + 32 = 36$ J ✓; separation speed $4-(-2) = 6$ = approach speed ✓ ( $e = 1$ ). Timing note: ~2 min if you spot "elastic in disguise" and quote (5); ~6 min if you re-derive from scratch. The tell is always "light spring" + "smooth floor" + "after the blocks separate." Don't re-solve what the collisions module already solved.

4 · Competitive level — JEE-Advanced stretch: the latch (~4 minutes)

Same collision again, but at the instant of maximum compression a latch locks the spring. (a) Describe the motion now. (b) The latch is released much later. Find the final velocities and comment on the timing.

Step 1. Locked at max squeeze: the pair moves as one body at $v_c = 2.0$ m/s, carrying KE $= 12$ J, with $24$ J banked in the latched spring. This is a perfectly inelastic collision whose "lost" energy is stored instead of burned — the latch is what $e = 0$ looks like when nothing is allowed to become heat.

Step 2. On release (wherever, whenever): the spring pushes the blocks apart, spending its 24 J on their relative motion while momentum stays $12\ \mathrm{kg\,m/s}$ . These are the same two conservation statements as the unlatched story.

Step 3. Same equations, same answers: $v_1' = -2.0\ \mathrm{m/s}$ , $v_2' = +4.0\ \mathrm{m/s}$ .

Step 4. The moral, which is the actual JEE-Advanced content: the final state depends only on the conserved totals (momentum $12\ \mathrm{kg\,m/s}$ , energy $36$ J), not on the schedule of storage and release. Latch it for a second or a year — physics does not bill by the hour. Timing note: ~3 min once you see that release restores the elastic budget. The distractor answer treats the latched phase as "energy lost" and applies inelastic formulas to the ending — but the latch stores, it does not burn.

Exam traps

Wall formula on a free block. $\tfrac12 m_1u^2 = \tfrac12 kx^2$ is legal only when the spring is backed by a wall/infinite mass. For a free $m_2$ , subtract the untouchable COM energy first — or jump straight to $\tfrac12\mu u^2 = \tfrac12 kx_{\max}^2$ .

"Maximum compression = block 1 stops." It never stops mid-encounter (unless $m_1 < m_2$ after separation, which is a different moment). Max squeeze = common velocity = COM velocity. Write $v_1 = v_2 = v_c$ as the working line.

Forgetting the encounter ends elastic. After the spring relaxes, don't invent new physics: quote the elastic-collision results (5). Equal masses exchange; light block off heavy block rebounds; heavy block barely notices — all inherited intact.

Reduced-mass slips. $\mu = \dfrac{m_1m_2}{m_1+m_2}$ — the product over the sum, always smaller than either mass. Writing the sum over the product (or using $m_1 + m_2$ ) inflates the stored energy. Sanity: for equal masses $\mu = m/2$ .

Signs after separation. $v_1'$ is negative when $m_1 < m_2$ . Carry the sign; "speed" answers then need the magnitude, and momentum audits need the sign. The rebound direction itself is a favourite one-mark MCQ.

Attached versus resting. A spring merely resting against $m_2$ (or the incoming block) pushes but cannot pull: the story ends at natural length with the elastic outcome. A spring attached to both blocks pulls back past natural length — the blocks never separate and oscillate about their COM forever. Read the mounting before predicting the ending.

Impulsive-force reflex. There is nothing "sudden" here — forces are finite ( $kx$ ), contact lasts visible time. Do not import collision-time approximations; conversely, this is why the floor must be stated smooth: over a long contact, friction would have time to act (unlike in the bullet-block impact).

Origins · 1 minRobert Hooke and how this was discovered

Robert Hooke

1635–1703 · London

Hooke published his spring law in 1676 as an anagram — ceiiinosssttuv — staking priority while he kept experimenting; unscrambled two years later, it read ut tensio, sic vis: as the stretch, so the force. That linear law, $F = kx$ , is why this archetype is exactly solvable: linear force means quadratic energy $\tfrac12kx^2$ , and the whole encounter reduces to two conservation lines. Hooke was the great mechanic of the early Royal Society — city surveyor after the Great Fire, microscopist who coined the word "cell," Newton's most bitter rival — and his law turned out to be the small-deformation truth of all solids: the steel balls, railway buffers and colliding atoms that "bounce elastically" are all running Hooke's law in miniature, storing and returning energy exactly as the visible spring does here.

History

1660

— Hooke discovers the proportionality of spring force and extension; publishes it as an anagram in 1676, solving it in 1678.

1669

— Wallis, Wren and Huygens deliver collision laws to the Royal Society — momentum conserved always, "vis viva" only in the elastic case.

1687

— Newton's Principia frames both as consequences of the laws of motion; the scholium measures restitution ball by ball.

1880

s — Hertz solves the contact of elastic spheres: real collisions ARE spring problems, with

k

from elasticity theory.

1900

s — Railway buffer-and-coupler design industrializes the archetype: springs sized by

\tfrac12\mu u^2

to swallow shunting collisions.

Going further

Texts. HC Verma, Concepts of Physics Vol. 1, Ch. 9 (the two-blocks-and-spring family in worked examples and exercises) · NCERT Physics Class 11, Ch. 6 · Kleppner & Kolenkow, Ch. 4 (collisions and the centre-of-mass frame). For the frame-based view of this archetype: Centre-of-Mass Frame Collisions (UG deep dive).

Provenance. All results are standard. A verifier should recompute: Example 1's chain ( $v_c = 2.0$ m/s, stored 24 J, $x_{\max} = 0.245$ m) and the $\mu = 4/3$ kg shortcut; Example 2's explosion split ( $2.0$ and $1.0$ m/s, energies 2:1); Example 3's elastic exit ( $-2.0$ , $+4.0$ m/s) with all three audits; the KE decomposition identity of Derivation A Step 3 by direct expansion.

Prerequisites. Elastic Collisions in 1D · Spring Force & Hooke's Law · Conservation of Linear Momentum. Leads to. Which Conservation Law? (technique) · Centre-of-Mass Frame Collisions (UG deep dive).

Why exactly does maximum compression happen at common velocity?

Because compression is accumulated approach. The spring shortens at the rate $v_1 - v_2$ (how fast the blocks close in); its length is the running total of that rate. A running total peaks when the rate crosses zero — that is, when $v_1 = v_2$ . After that instant the rear block is slower than the front block, the gap grows, and the spring re-extends. No dynamics needed: it is the same logic as "your bank balance peaks when deposits stop exceeding withdrawals."

Is momentum really conserved even while the spring is pushing?

Yes — at every instant. The spring pushes $m_1$ backward and $m_2$ forward with equal magnitudes (Newton's third law through a massless spring), so the pair's total momentum never budges: $m_1v_1 + m_2v_2 = m_1u$ from first touch to final separation. This is stronger than the usual "before and after" statement, and exam questions exploit it: "find $v_2$ when the spring holds 10 J" is solvable precisely because momentum holds mid-encounter too.

What is the reduced mass really doing here?

It converts a two-body problem into a one-body summary. The pair's kinetic energy always splits as (COM part) + (relative part), and the relative part is $\tfrac12\mu v_{\text{rel}}^2$ with $\mu = \frac{m_1m_2}{m_1+m_2}$ — as if the relative motion belonged to a single fictitious particle of mass $\mu$ . Since internal forces can only touch relative motion, $\mu$ tells you instantly how much energy is up for grabs: storable in springs, losable in inelastic collisions ( $\tfrac12\mu u^2(1-e^2)$ ), transferable in encounters. It is the momentum chapter's best shortcut and the rotation chapter will hand you its twin.

What changes if the spring is attached to both blocks?

The ending. Everything up to natural length is identical — but an attached spring can pull, so as the blocks try to separate the spring stretches and hauls them back. Result: no separation, ever; the blocks oscillate about their common centre of mass (which drifts at $v_c$ ), trading velocities periodically between $(u, 0)$ and $(v_1', v_2')$ . In the COM frame it is pure SHM with effective mass $\mu$ , angular frequency $\omega = \sqrt{k/\mu}$ — a preview of the oscillations chapter riding on this module's algebra.

Does the stiffness k affect the final velocities?

No — and that is worth internalizing. $k$ sets the choreography: stiff spring, small $x_{\max}$ , quick encounter; soft spring, deep slow squeeze. But the final velocities come from momentum + full energy return, and both are $k$ -blind. This is exactly why "instantaneous" elastic collisions can be solved without knowing anything about the contact forces: the invisible atomic springs have some enormous $k$ , and it cancels out of the answers just as it does here.

Real springs have mass. How bad is the "light spring" idealization?

For exam setups, excellent. A spring of mass $m_s$ carries some kinetic energy of its own (roughly a third of $m_s$ moves with the contact end), so a real encounter returns slightly less than 100% to the blocks and the outcome is very slightly inelastic. With $m_s$ a few percent of the block masses, corrections are below the precision of any answer you'd give. The idealization to actually watch is the floor: any friction acts throughout the long contact and genuinely breaks the momentum bookkeeping — which is why the problem always says "smooth."