Spring-mass systems — collision problems: Visual walkthrough

This page rebuilds the two-block spring collision result from absolute zero, one picture at a time. We will earn every symbol before we use it, and by the end you will see why the maximum squish of the spring is exactly

$x_{ma x} = (u_{1} - u_{2}) \frac{μ}{k}, μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} .$

If any symbol here looks scary — good. We will meet each one slowly. Parent note: 1.3.13 Spring-mass systems — collision problems (Hinglish).

Step 1 — Draw the two carts and name everything

WHAT. Two blocks slide on frictionless ice. The left one, mass $m_{1}$ , moves right at speed $u_{1}$ . The right one, mass $m_{2}$ , moves right at speed $u_{2}$ . A spring of stiffness $k$ is glued to one of them.

WHY. Before any physics, we must fix names and a direction, otherwise a "minus sign" later has no meaning. We choose rightward = positive. Every velocity is a signed number: a block drifting left just carries a negative $u$ .

PICTURE. Look at the figure. The two arrows are the velocities; their lengths are the speeds. Because $u_{1}$ is drawn longer than $u_{2}$ , the left block is catching up — that is the only reason the spring will ever touch.

Figure — Spring-mass systems — collision problems

For the spring to compress at all we need the left block to gain on the right block, i.e. the approach speed $u_{1} - u_{2}$ must be positive. Hold that quantity in mind — it is the hero of the whole story.

Step 2 — The collision instant: why momentum, not energy

WHAT. At the very first touch, we ask: what quantity is safe to carry across the instant of contact?

WHY. The spring pushes with force $k x$ — where $x$ is how much it is squished. At the first touch $x = 0$ , so the force is essentially zero, and even a moment later it is only a finite force acting for a tiny time. Force $\times$ time is called impulse, and it is the only thing that can change momentum. A finite force over a near-zero time gives near-zero impulse.

PICTURE. The figure shows the spring force $k x$ starting at zero and rising gently — nothing sudden. Compare that to a hammer-hard rigid collision (dashed spike). The spring never spikes, so momentum simply flows through untouched.

Step 3 — What "maximum compression" actually means

WHAT. The spring squishes by a length $x$ that changes with time. We want the biggest value $x$ ever reaches, called $x_{ma x}$ .

WHY. The squish $x$ grows while the two blocks get closer and shrinks while they separate. The gap between them shrinks only while the left block is still faster than the right: while $v_{1} > v_{2}$ . The gap stops shrinking — so $x$ stops growing — at the exact instant the two velocities become equal.

PICTURE. Two velocity curves are drawn against time: the fast block slowing, the slow block speeding up. Where the curves cross (same height = same velocity) the squish is deepest. Before the cross, they approach; after it, they separate.

Prove-it-to-yourself: reveal the reason below.

Why must the velocities be equal at max compression?

If they differed, one block would still be closing on (or opening from) the other, so

x

would still be changing — hence not yet at a maximum/turning point.

Step 4 — Solve for the shared velocity $v$

WHAT. At the max-compression instant both blocks move at the same $v$ . Plug $v_{1} = v_{2} = v$ into the momentum equation of Step 2.

WHY. Momentum was conserved (Step 2), and now we know both final speeds are the same number $v$ . That collapses two unknowns into one, which we can solve.

PICTURE. The figure shows the pair briefly locked at one velocity $v$ — the same arrow on both blocks — with the spring at its deepest squish in between.

$m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v$

Left side: total incoming momentum (unchanged since Step 2).
$(m_{1} + m_{2})$ : the two blocks now move as one lump, so their masses add.
$v$ : the one velocity they share.

Step 5 — The energy books: where did the kinetic energy go?

WHAT. Now switch tools. Once the blocks are in contact and moving, the only force is the conservative, frictionless spring, so mechanical energy is conserved from first touch to max compression.

WHY. Kinetic energy (energy of motion, $\frac{1}{2} m v^{2}$ ) does not vanish — it gets stored in the squished spring as elastic potential energy (see Elastic Potential Energy of a Spring). Write "KE before = KE at max squish + energy locked in spring."

PICTURE. A bar chart: the tall "before" KE bar splits into a shorter "still-moving" KE bar (the whole lump gliding at $v$ ) plus an orange "stored in spring" bar. The stored bar is the piece we want.

$KE before touching \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = KE of the moving lump \frac{1}{2} (m_{1} + m_{2}) v^{2} + energy stored in the spring \frac{1}{2} k x_{ma x}^{2}$

$\frac{1}{2} m u^{2}$ — kinetic energy of a mass $m$ at speed $u$ .
$\frac{1}{2} (m_{1} + m_{2}) v^{2}$ — the whole system is still gliding at $v$ ; that motion is not available to squish the spring.
$\frac{1}{2} k x_{ma x}^{2}$ — energy a spring holds when squished by $x_{ma x}$ .

Step 6 — Algebra: the reduced mass $μ$ appears

WHAT. Rearrange Step 5 to isolate the stored energy, then substitute $v$ from Step 4.

WHY. We want $x_{ma x}$ in terms of the given numbers ( $m_{1}, m_{2}, u_{1}, u_{2}, k$ ), not the derived $v$ . When we substitute and simplify, a beautiful cancellation packs everything into one tidy grouping.

PICTURE. The figure shows the stored-energy expression shrinking, term by term, into the single compact block $\frac{1}{2} μ (u_{1} - u_{2})^{2}$ — the messy sum folding into one neat parcel.

Isolate stored energy:

$\frac{1}{2} k x_{ma x}^{2} = \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - \frac{1}{2} (m_{1} + m_{2}) v^{2} .$

Substituting $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ and grinding through the algebra, everything collapses to

$\frac{1}{2} k x_{ma x}^{2} = \frac{1}{2} μ (u_{1} - u_{2})^{2}, μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} .$

$(u_{1} - u_{2})$ — the approach speed from Step 1. Only relative motion can squish the spring.
$μ$ — the reduced mass: the "effective single mass" for the relative motion of two bodies (see Reduced Mass and Two-Body Problems).

Solving the boxed line for $x_{ma x}$ :

$x_{ma x} = (u_{1} - u_{2}) \frac{μ}{k}$

Step 7 — Sanity: every edge and degenerate case

WHAT. Test the formula against extreme inputs. A good formula must never break.

WHY. If a limiting case gives nonsense, the derivation is wrong. Each case below is a "free experiment."

PICTURE. Four mini-panels, one per case, each showing the carts and the resulting squish.

Equal speeds, $u_{1} = u_{2}$ . Approach speed $= 0$ → $x_{ma x} = 0$ . They never close in, spring never touches. ✔
Target at rest, $u_{2} = 0$ . $x_{ma x} = u_{1} μ / k$ — the standard textbook case. ✔
Very heavy target, $m_{2} \to \infty$ . Then $μ \to m_{1}$ , and $x_{ma x} \to u_{1} m_{1} / k$ — exactly the wall result, because an infinite mass acts like an immovable wall. ✔ (See Elastic and Inelastic Collisions.)
Blocks moving apart, $u_{1} < u_{2}$ . Approach speed is negative; $(u_{1} - u_{2})^{2}$ is still positive but the spring is never engaged in the first place, so physically $x_{ma x} = 0$ — the formula only applies once contact begins. ✔

The one-picture summary

Everything on one canvas: the two carts (Step 1) → momentum flows through the gentle spring force (Step 2) → curves cross at equal velocity $v$ = deepest squish (Steps 3–4) → the energy bar splits, and the stored slice equals $\frac{1}{2} μ (u_{1} - u_{2})^{2}$ (Steps 5–6) → giving $x_{ma x} = (u_{1} - u_{2}) μ / k$ .

Recall Feynman retelling — the whole walkthrough in plain words

Two carts on ice; the fast one has a springy bumper and catches the slow one. When they touch, the push starts from nothing and builds up gently, so during the quick bump we just say the total momentum is unchanged — no sudden hammer-blow to change it. Now the spring squishes. It keeps squishing only while the fast cart is still gaining ground; the deepest squish happens at the one instant both carts roll at the same speed $v$ — after that the spring shoves them apart. To find how deep, we count energy: some motion energy stays as the whole pair glides at $v$ , and the leftover — which is exactly the energy of them squeezing toward each other — is what the spring stores. That leftover depends only on how fast they were closing in, $u_{1} - u_{2}$ , and on a special combined mass $μ$ . Set that stored energy equal to the spring's $\frac{1}{2} k x^{2}$ and out pops $x_{ma x} = (u_{1} - u_{2}) μ / k$ . Every extreme case checks out: same speed → no squish; infinite target → the wall answer.

Connections

1.3.13 Spring-mass systems — collision problems (Hinglish)
Conservation of Linear Momentum
Elastic and Inelastic Collisions
Reduced Mass and Two-Body Problems
Elastic Potential Energy of a Spring
Centre of Mass Frame
Simple Harmonic Motion