Worked examples — Spring-mass systems — collision problems

This page is the drill floor. The parent note built the two-phase idea (momentum during the moment, energy for the compression). Here we hunt down every kind of situation the topic can hand you — every sign of velocity, both a wall and free blocks, the equal-mass surprise, the huge-mass limit, a latched (locked) spring, and a real-world word problem — and solve each one from line one.

Before any formula appears, here is the vocabulary we will lean on, in plain words:

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The scenario matrix

Every problem in this topic is one cell of this grid. The worked examples below are labelled by cell.

Cell	What changes	Which law(s)	Example
A	One block hits a wall-fixed spring	Energy only (wall = external force)	Ex 1
B	Two free blocks, target at rest ($u_2=0$)	Momentum + Energy	Ex 2
C	Two free blocks, head-on ($u_1>0$, $u_2<0$)	Momentum + Energy, watch signs	Ex 3
D	Two free blocks, same direction chase ($u_1,u_2$ both $+$)	Momentum + Energy	Ex 4
E	Equal masses $m_1=m_2$ (degenerate: velocities swap)	Elastic-collision limit	Ex 5
F	Huge target $m_2\gg m_1$ (limit → wall case)	Limit check	Ex 6
G	Spring latched at max compression (inelastic outcome)	Momentum, energy trapped	Ex 7
H	Real-world word problem (bumper cars)	Full toolkit	Ex 8

We cover: both signs of every velocity, zero input ($u_2=0$), the degenerate equal-mass case, the $m_2\to \infty$ limit, the elastic vs inelastic outcome branch, and a word problem.

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Cell A — Block into a wall-fixed spring

Forecast: Will the block stop and bounce straight back at the same speed, or slower? Guess before reading.

1. Which law? Why this step? The wall is fixed — it can push on the spring with any force it likes. That is an external horizontal force on the block, so momentum is not conserved. But nothing wastes energy (frictionless, ideal spring), so mechanical energy is conserved. This is why Cell A is the odd one out: no momentum, energy only.

2. Set energy before = energy at max compression. At max compression the block is momentarily at rest (the wall doesn't move, so "common velocity" is zero). Why this step? All the kinetic energy has nowhere to go except into the spring: $\frac{1}{2}m{u}^2=\frac{1}{2}k{x}_{max}^2\Rightarrow x_{max}=u\sqrt{\frac{m}{k}}$

3. Plug in. $x_{max}=3\sqrt{\frac{2}{600}}=3\times 0.05774=0.173\text{m}$

4. Release velocity. Why this step? An ideal spring returns 100% of the stored energy, so the block leaves with the same speed but reversed direction: $v^{\prime }=-3\text{m/s}$.

Verify: Units of $u\sqrt{m/k}$: $\frac{\text{m}}{\text{s}}\sqrt{\frac{\text{kg}}{\text{N/m}}}=\frac{\text{m}}{\text{s}}\sqrt{\frac{\text{kg}\cdot \text{m}}{\text{kg}\cdot {\text{m/s}}^2}}=\frac{\text{m}}{\text{s}}\cdot \text{s}=\text{m}$. ✓ Length, good. Energy check: $\frac{1}{2}(2)(3^2)=9\text{J}$ in, $\frac{1}{2}(600)(0.173^2)=9.0\text{J}$ stored. ✓

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Cell B — Two free blocks, target at rest

Forecast: Does the light block bounce back, or keep going forward slower?

1. (a) Common velocity. Why this step? Max compression happens when both move together — momentum is conserved in Phase 2 (no wall now): $v=\frac{2(6)+4(0)}{2+4}=\frac{12}{6}=2\text{m/s}$

2. (b) Max compression. Why reduced mass? Only the relative-motion kinetic energy gets stored; the whole system still glides at $v$. That relative KE is $\frac{1}{2}\mu (u_1-u_2{)}^2$: $\mu =\frac{2\times 4}{6}=\frac{4}{3},x_{max}=|6-0|\sqrt{\frac{4/3}{1200}}=6\times 0.03333=0.200\text{m}$

3. (c) Final velocities. Why elastic? The ideal spring gives every joule back → the overall encounter is a perfectly elastic collision (see Elastic and Inelastic Collisions): $v_1^{\prime }=\frac{m_1-m_2}{m_1+m_2}{u}_1=\frac{-2}{6}(6)=-2\text{m/s},v_2^{\prime }=\frac{2m_1}{m_1+m_2}{u}_1=\frac{4}{6}(6)=4\text{m/s}$

Verify: Momentum: before $2(6)=12$; after $2(-2)+4(4)=-4+16=12$. ✓ Energy: before $\frac{1}{2}(2)(36)=36\text{J}$; after $\frac{1}{2}(2)(4)+\frac{1}{2}(4)(16)=4+32=36\text{J}$. ✓ Light block bounces back — Forecast test!

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Cell C — Head-on collision (opposite signs)

Forecast: Which way is the combined system drifting at max squeeze — left or right?

1. Fix signs first. Why this step? This is where the $\pm$ discipline matters. Rightward $=+$, so $u_1=+4$, $u_2=-6$. The approach speed is the gap $u_1-u_2=4-(-6)=10\text{m/s}$ — they close in at $10\text{m/s}$, faster than either alone. This is exactly why we wrote $|u_1-u_2|$: subtracting a negative adds.

2. Common velocity. $v=\frac{3(4)+1(-6)}{3+1}=\frac{12-6}{4}=\frac{6}{4}=1.5\text{m/s}$ Positive → the pair drifts right at max compression (the heavier, rightward block wins the tug).

3. Max compression. $\mu =\frac{3\times 1}{4}=0.75,x_{max}=|4-(-6)|\sqrt{\frac{0.75}{500}}=10\times 0.03873=0.387\text{m}$

Verify: Momentum: before $3(4)+1(-6)=6$; at max compression $(3+1)(1.5)=6$. ✓ Stored energy $\frac{1}{2}\mu (u_1-u_2{)}^2=\frac{1}{2}(0.75)(100)=37.5\text{J}$; spring PE $\frac{1}{2}k{x}^2=\frac{1}{2}(500)(0.387^2)=37.5\text{J}$. ✓ Drift is rightward — Forecast!

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Cell D — Same-direction chase (both positive)

Forecast: The gap they close at is only $8-3=5$, not $8$. Will the spring squish more or less than if $m_2$ were at rest?

1. Common velocity. $v=\frac{2(8)+3(3)}{2+3}=\frac{16+9}{5}=\frac{25}{5}=5\text{m/s}$

2. Approach speed is the difference, not the sum. Why this step? Both move right, so the only motion the spring "feels" is the relative closing speed $|8-3|=5\text{m/s}$. A chasing collision squishes less than a head-on one at the same absolute speeds — this is the key contrast with Cell C. $\mu =\frac{2\times 3}{5}=1.2,x_{max}=5\sqrt{\frac{1.2}{400}}=5\times 0.05477=0.274\text{m}$

Verify: Momentum: before $2(8)+3(3)=25$; after $(5)(5)=25$. ✓ Stored energy $\frac{1}{2}(1.2)(5^2)=15\text{J}$; spring PE $\frac{1}{2}(400)(0.274^2)=15.0\text{J}$. ✓

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Cell E — Equal masses (degenerate case)

Forecast: Something famous happens when equal masses collide elastically. Guess the two final speeds.

1. Final velocities — watch the formula collapse. Why this step? Plug $m_1=m_2=m$ into the elastic results: $v_1^{\prime }=\frac{m-m}{2m}{u}_1=0,v_2^{\prime }=\frac{2m}{2m}{u}_1=u_1=5\text{m/s}$ The moving block stops dead; the target leaves at the incoming speed. They swap velocities — the signature of an equal-mass elastic collision.

2. Max compression. $\mu =\frac{2\times 2}{4}=1,x_{max}=5\sqrt{\frac{1}{800}}=5\times 0.03536=0.177\text{m}$

Verify: Momentum: before $2(5)=10$; after $2(0)+2(5)=10$. ✓ Energy: before $\frac{1}{2}(2)(25)=25\text{J}$; after $\frac{1}{2}(2)(25)=25\text{J}$. ✓ Velocity swap confirmed — Forecast!

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Cell F — Huge target (the limit back to a wall)

Forecast: If $m_2$ is essentially immovable, what should the light block do — and how close should this be to hitting a wall?

1. Common velocity ≈ 0. Why this step? A tiny mass sharing momentum with a mountain barely moves the mountain: $v=\frac{1(4)+1000(0)}{1001}=\frac{4}{1001}=0.003996\text{m/s}\approx 0$ So at max compression both are essentially at rest — exactly the wall picture.

2. Reduced mass ≈ $m_1$. Why this step? $\mu =\frac{1\times 1000}{1001}=0.999\text{kg}\approx m_1$. When $m_2\to \infty$, $\mu \to m_1$, so $x_{max}\to u_1\sqrt{m_1/k}$ — the Cell A wall formula. $x_{max}=4\sqrt{\frac{0.999}{250}}=4\times 0.06322=0.2529\text{m}$

3. Rebound velocity ≈ $-u_1$. $v_1^{\prime }=\frac{1-1000}{1001}(4)=\frac{-999}{1001}(4)=-3.992\text{m/s}\approx -4\text{m/s}$ It bounces back at (almost) its incoming speed — just like off a wall.

Verify: Wall-formula prediction $u_1\sqrt{m_1/k}=4\sqrt{1/250}=0.2530\text{m}$; our $0.2529\text{m}$ matches to $<0.1\%$. ✓ $v_1^{\prime }\approx -4$ ✓.

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Cell G — Latched spring (inelastic outcome)

Forecast: The max compression maths is identical to Cell B — but does the light block still bounce back?

1. Same $x_{max}$. Why this step? The squeezing phase is unchanged up to the instant of max compression, so $x_{max}=0.200\text{m}$ exactly as in Cell B. What changes is only what happens next.

2. They stay together. Why this step? With the spring locked, the blocks are now a single rigid object moving at the common velocity — that is the definition of a perfectly inelastic collision: $v_{\text{final}}=v=2\text{m/s}\text{(both blocks, forever)}$

3. Trapped energy. Why this step? The relative-motion KE that Cell B returned is now permanently held in the compressed spring: $E_{\text{trapped}}=\frac{1}{2}\mu (u_1-u_2{)}^2=\frac{1}{2}\left(\frac{4}{3}\right)(6^2)=24\text{J}$

Verify: KE before $=\frac{1}{2}(2)(36)=36\text{J}$. KE after (moving together) $=\frac{1}{2}(6)(2^2)=12\text{J}$. Lost $=36-12=24\text{J}=E_{\text{trapped}}$. ✓ Momentum still conserved: $12=(6)(2)$. ✓ No bounce-back — Forecast contrast confirmed.

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Cell H — Real-world word problem (bumper cars)

Forecast: Will car A stop, keep rolling forward, or reverse? Its mass is bigger than B's.

1. Common velocity at deepest squeeze. $v=\frac{150(2.0)+100(0)}{250}=\frac{300}{250}=1.2\text{m/s}$

2. Reduced mass and compression. $\mu =\frac{150\times 100}{250}=60\text{kg},x_{max}=|2.0-0|\sqrt{\frac{60}{30000}}=2.0\times 0.04472=0.0894\text{m}$ So the bumper squishes about 8.9 cm — a believable, visible dent depth.

3. Speeds after separation (elastic bumper). $v_1^{\prime }=\frac{150-100}{250}(2.0)=\frac{50}{250}(2.0)=0.4\text{m/s},v_2^{\prime }=\frac{2(150)}{250}(2.0)=\frac{300}{250}(2.0)=2.4\text{m/s}$ Car A keeps drifting forward slowly ($+0.4$); car B shoots ahead at $2.4\text{m/s}$.

Verify: Momentum: before $150(2.0)=300$; after $150(0.4)+100(2.4)=60+240=300$. ✓ Energy: before $\frac{1}{2}(150)(4)=300\text{J}$; after $\frac{1}{2}(150)(0.16)+\frac{1}{2}(100)(5.76)=12+288=300\text{J}$. ✓ Car A did not reverse (heavier than B) — Forecast!

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Recall Which cell am I in? (quick decision list)

Is a wall involved? ::: Cell A — energy only, no momentum, $x=u\sqrt{m/k}$. Two free blocks — which law during the squeeze? ::: Momentum for $v$, energy for $x_{max}$; always both. Blocks moving toward each other (opposite signs)? ::: Cell C — approach speed = $|u_1-u_2|$ adds the magnitudes. Blocks chasing (same sign)? ::: Cell D — approach speed = difference; squishes less. Equal masses, elastic? ::: Cell E — velocities swap; incoming block stops. One mass enormous? ::: Cell F — reduces to the wall case; $\mu \to m_1$, block rebounds at $-u_1$. Spring gets latched? ::: Cell G — same $x_{max}$, but outcome is perfectly inelastic; energy $\frac{1}{2}\mu (u_1-u_2{)}^2$ trapped.

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Connections

• 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