Spring-mass systems — collision problems

WHY split the problem into phases?

Phase 2 setup: the "common velocity" insight

Deriving the common velocity (from scratch)

Block of mass $m_1$ moving at $u_1$ hits a block $m_2$ (initially $u_2$ ) with a spring between/attached. Take rightward positive.

At maximum compression, both move at $v$ . Momentum is conserved (Phase 2 has no external horizontal force):

$m_1 u_1 + m_2 u_2 = (m_1+m_2)\,v$

Deriving maximum compression $x_{max}$

Energy before compression = energy at max compression (KE turns partly into spring PE):

$\tfrac12 m_1 u_1^2 + \tfrac12 m_2 u_2^2 = \tfrac12 (m_1+m_2)v^2 + \tfrac12 k x_{max}^2$

Why this step? Total KE isn't fully stored — the system is still moving as a whole at $v$ . Only the KE in the centre-of-mass frame can be stored in the spring.

Solve for $x_{max}$ . Substituting $v$ and simplifying gives the clean reduced-mass form:

$\tfrac12 k x_{max}^2 = \tfrac12\,\mu\,(u_1-u_2)^2,\qquad \mu=\frac{m_1 m_2}{m_1+m_2}$

Figure — Spring-mass systems — collision problems

Worked Example 1 — block hits a spring fixed to a wall

A $2\,\text{kg}$ block slides at $3\,\text{m/s}$ into a spring ( $k=600\,\text{N/m}$ ) attached to a wall. Find max compression.

Why no momentum here? The wall exerts an external force, so momentum is not conserved. But energy is (frictionless, ideal spring).

$\tfrac12 m u^2 = \tfrac12 k x^2 \;\Rightarrow\; x = u\sqrt{m/k}$ Why this step? All KE converts to spring PE because the wall brings the block fully to rest.

$x = 3\sqrt{2/600}=3\times0.0577 = 0.173\,\text{m}$

Worked Example 2 — two free blocks, spring on one

$m_1=2\,\text{kg}$ at $u_1=6\,\text{m/s}$ hits $m_2=4\,\text{kg}$ (at rest) carrying a spring $k=1200\,\text{N/m}$ . Find (a) velocity at max compression, (b) max compression, (c) final velocities.

(a) Common velocity — Why? Max compression ⇔ same velocity. $v=\frac{2(6)+4(0)}{6}=2\,\text{m/s}$

(b) Max compression — Why reduced mass? Only relative-motion KE is stored. $\mu=\frac{2\times4}{6}=\frac43,\quad x=(6-0)\sqrt{\frac{4/3}{1200}}=6\times0.0333=0.200\,\text{m}$

(c) Final velocities — The spring eventually springs back fully → the encounter is effectively a perfectly elastic collision (spring restores all energy). Use elastic-collision results: $v_1'=\frac{m_1-m_2}{m_1+m_2}u_1=\frac{-2}{6}(6)=-2\,\text{m/s},\quad v_2'=\frac{2m_1}{m_1+m_2}u_1=\frac{4}{6}(6)=4\,\text{m/s}$ Why elastic? An ideal spring stores energy then returns 100% of it — no loss → elastic.

Worked Example 3 — spring stays partly compressed?

If the blocks are stuck/latched at max compression (a clip locks the spring), the collision is now perfectly inelastic: they move together at $v=v_{cm}$ forever, and the energy $\tfrac12\mu(u_1-u_2)^2$ is stored (not returned). Same $x_{max}$ , but final state differs.

Why this matters: "max compression" maths is identical; what changes is whether the stored energy is released (elastic outcome) or trapped (inelastic outcome).

Common mistake Steel-manning the classic errors

Wrong idea 1: "Use energy conservation during the collision too." Why it feels right: spring is ideal, so surely energy is always conserved. The fix: During the instantaneous contact, finite spring impulse ≈ 0, so use momentum. Energy is only safe to use once you correctly track stored PE — and only if no latch/friction removes it.

Wrong idea 2: "All KE goes into the spring, so $\tfrac12 m_1 u_1^2 = \tfrac12 k x^2$ for two free blocks." Why it feels right: worked for the wall case. The fix: with two free blocks the system still moves at $v_{cm}$ , so only the CM-frame KE $=\tfrac12\mu(u_1-u_2)^2$ is stored. Subtract the $\tfrac12(m_1+m_2)v^2$ .

Wrong idea 3: "Max compression is when the front block stops." Why it feels right: compression is biggest when target moves most. The fix: compression grows while they approach. It's maximal when their velocities are equal, not when either is zero.

Recall Feynman: explain to a 12-year-old

Imagine two carts on ice, one with a springy bumper. When they bump, the spring squishes. It keeps squishing as long as one cart is catching up to the other. The squish is biggest at the exact moment they're rolling at the same speed — then the spring pushes them apart again. During the quick bump, pushing is so fast we just say "total push-power (momentum) stays the same." A perfect spring is a perfect trampoline: it gives back every bit of energy, so the carts bounce off without losing any pep.

Flashcards

Which conservation law applies during the instantaneous collision, and why?

Momentum — the spring's finite force gives negligible impulse in the tiny collision time.

At what instant is spring compression maximum?

When both blocks have the same velocity (relative velocity = 0).

What is the common velocity at max compression for

m_1u_1+m_2u_2

v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}

(the centre-of-mass velocity).

Formula for max compression of two free blocks?

x_{max}=(u_1-u_2)\sqrt{\mu/k}

with

\mu=\frac{m_1m_2}{m_1+m_2}

Why does reduced mass

\mu

appear in

x_{max}

Only the relative-motion (CM-frame) KE

\tfrac12\mu(u_1-u_2)^2

can be stored; relative motion has effective mass

\mu

Block hits a wall-fixed spring: which law and result?

Energy only (wall = external force);

x=u\sqrt{m/k}

For two free blocks with an ideal spring, what kind of collision is the overall outcome?

Perfectly elastic — the spring returns 100% of stored energy.

If a latch locks the spring at max compression, what type of collision is it?

Perfectly inelastic — energy

\tfrac12\mu(u_1-u_2)^2

stays trapped; blocks move together.

How much KE is stored in the spring at max compression?

\tfrac12\mu(u_1-u_2)^2

(the CM-frame kinetic energy).

Connections

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, spring-mass collision problem ka asli funda yeh hai ki ise do phases mein todo. Pehla phase: collision ka exact moment. Yahan spring ka force chhota hota hai aur time bahut hi kam, isliye uska impulse almost zero — matlab is instant pe momentum conserve hota hai, energy ko abhi haath mat lagao. Doosra phase: ab spring dhire-dhire compress hone lagta hai, koi friction ya loss nahi (ideal spring), isliye ab mechanical energy conserve karo.

Sabse important baat: spring tab tak compress hota rahega jab tak dono blocks ek doosre ke paas aa rahe hain. Jis instant dono ki velocity same ho jaati hai (relative velocity zero), wahi maximum compression hai. Us common velocity ko nikalo: $v = (m_1u_1+m_2u_2)/(m_1+m_2)$ — yeh bas centre of mass ki velocity hai.

Maximum compression ke liye energy equation lagao, lekin yaad rakho — saari KE spring mein nahi jaati! System abhi bhi $v_{cm}$ se chal raha hai, isliye sirf relative motion wali KE store hoti hai: $\tfrac12\mu(u_1-u_2)^2$ , jahan $\mu$ reduced mass hai. Isse $x_{max}=(u_1-u_2)\sqrt{\mu/k}$ aata hai. Agar wall se spring juda ho to momentum mat lagao (wall external force deta hai), seedha $\tfrac12 mu^2=\tfrac12 kx^2$ .

Aur ek bonus: ideal spring 100% energy wapas de deta hai, to do free blocks ka overall result perfectly elastic collision jaisa hota hai. Agar spring ko latch laga ke lock kar do max compression pe, to wahi perfectly inelastic ban jaata hai — energy phans jaati hai. Bas yeh logic clear rakho aur har problem solve ho jayegi.

Spring-mass systems — collision problems

WHY split the problem into phases?

Phase 2 setup: the "common velocity" insight

Deriving the common velocity (from scratch)

Deriving maximum compression $x_{max}$

Worked Example 1 — block hits a spring fixed to a wall

Worked Example 2 — two free blocks, spring on one

Worked Example 3 — spring stays partly compressed?

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Work, Energy & Power

Connections