Exercises — Spring-mass systems — collision problems

• Momentum conservation $m_1{u}_1+m_2{u}_2=(m_1+m_2)v$ — used during the collision (spring impulse ≈ 0 in the tiny contact time) and any time there is no external horizontal push. See Conservation of Linear Momentum.
• Energy conservation $\frac{1}{2}m{u}^2$ → spring PE $\frac{1}{2}k{x}^2$ — used after the collision while the ideal spring squishes, because nothing wastes energy. See Elastic Potential Energy of a Spring.
• Reduced mass $\mu =\frac{m_1{m}_2}{m_1+m_2}$ — the "effective mass" of the relative motion of two bodies. Only the relative-motion kinetic energy $\frac{1}{2}\mu (u_1-u_2{)}^2$ can be stored in the spring. See Reduced Mass and Two-Body Problems.

(a) Velocity at max compression → momentum conservation. What we do: set total momentum before = total momentum at max compression, giving the shared velocity $v=\frac{m_1{u}_1+m_2{u}_2}{m_1+m_2}$. Why: the two free blocks feel no external horizontal force, so their total momentum can't change; at max squish they move together (relative velocity zero), so that one common $v$ carries all the momentum. (b) Maximum compression → energy conservation. What we do: set kinetic energy before = kinetic energy of the whole system moving at $v$ plus the stored spring PE $\frac{1}{2}k{x}^2$. Why: the ideal spring wastes nothing, so total mechanical energy is conserved; solving for $x$ pulls out the compression. Answer: (a) momentum, (b) energy — the "MEME" split.

No. The wall pushes back through the spring with a real, external horizontal force, so the block's momentum is not conserved (it drops to zero at max compression). Energy is conserved — no friction, ideal spring — so we use $\frac{1}{2}m{u}^2=\frac{1}{2}k{x}^2$.

Their velocities are equal (relative velocity = 0). The gap between them shrinks only while one catches the other; the moment they match speeds the gap stops shrinking, so the spring is at its most compressed. "Max squish = same swish."

Wall = external force → energy only. All kinetic energy becomes spring PE: $\frac{1}{2}m{u}^2=\frac{1}{2}k{x}^2\Rightarrow x=u\sqrt{m/k}=3\sqrt{2/600}=3(0.05774)=\boxed{0.173\text{m}}.$

Max compression ⇔ same velocity, and momentum is conserved: $v=\frac{m_1{u}_1+m_2{u}_2}{m_1+m_2}=\frac{2(6)+4(0)}{6}=\boxed{2\text{m/s}}.$

Only the relative-motion KE is stored, with effective mass $\mu$: $\mu =\frac{2\cdot 4}{6}=\frac{4}{3}\text{kg},x=(u_1-u_2)\sqrt{\mu /k}=6\sqrt{\frac{4/3}{1200}}=6(0.03333)=\boxed{0.200\text{m}}.$ Check the picture: the spring stores $\frac{1}{2}\mu (u_1-u_2{)}^2=\frac{1}{2}(4/3)(36)=24\text{J}$; and $\frac{1}{2}k{x}^2=\frac{1}{2}(1200)(0.2{)}^2=24\text{J}$. ✓

An ideal spring returns 100% of its stored energy, so the overall encounter is a perfectly elastic collision (see Elastic and Inelastic Collisions). Use the standard 1-D elastic results with $u_2=0$: $v_1^{\prime }=\frac{m_1-m_2}{m_1+m_2}{u}_1=\frac{2-4}{6}(6)=\boxed{-2\text{m/s}}(\text{bounces back}),$ $v_2^{\prime }=\frac{2m_1}{m_1+m_2}{u}_1=\frac{4}{6}(6)=\boxed{4\text{m/s}}.$ Sanity: momentum $2(-2)+4(4)=12=2(6)$ ✓; energy $\frac{1}{2}2(4)+\frac{1}{2}4(16)=36$ ✓.

With the spring locked the blocks move together forever → perfectly inelastic. Final velocity is the same $v_{cm}$: $v=2\text{m/s},E_{\text{trapped}}=\frac{1}{2}\mu (u_1-u_2{)}^2=\frac{1}{2}\cdot \frac{4}{3}\cdot 36=\boxed{24\text{J}}.$ The point: max-compression maths is identical to L2.3; only the ending differs — released (elastic) vs trapped (inelastic).

(a) $v=\frac{3(4)+1(-2)}{4}=\frac{10}{4}=\boxed{2.5\text{m/s}}.$ The sign of $u_2$ matters — plug it in with its minus sign. (b) $\mu =\frac{3\cdot 1}{4}=0.75\text{kg}$, relative speed $u_1-u_2=4-(-2)=6\text{m/s}$: $x=(u_1-u_2)\sqrt{\mu /k}=6\sqrt{0.75/500}=6(0.03873)=\boxed{0.232\text{m}}.$ Why relative speed = 6, not 2: they approach each other, so their closing speed is the sum of the magnitudes; the formula $u_1-u_2$ captures this automatically once signs are honoured.

(i) ${\text{KE}}_0=\frac{1}{2}(3)(4^2)+\frac{1}{2}(1)(2^2)=24+2=26\text{J}$. (ii) bulk KE $=\frac{1}{2}(4)(2.5^2)=\frac{1}{2}(4)(6.25)=12.5\text{J}$. (iii) stored PE $=\frac{1}{2}k{x}^2=\frac{1}{2}(500)(0.232^2)=\frac{1}{2}(500)(0.0540)=13.5\text{J}$. Check: $26-12.5=13.5\text{J}$ ✓, and $\frac{1}{2}\mu (u_1-u_2{)}^2=\frac{1}{2}(0.75)(36)=13.5\text{J}$ ✓. All three agree — the spring holds exactly the CM-frame KE.

Wall case → energy only: $\frac{1}{2}m{u}^2=\frac{1}{2}k{x}^2\Rightarrow k=\frac{m{u}^2}{x^2}$. $k=\frac{5(4^2)}{0.10^2}=\frac{80}{0.01}=\boxed{8000\text{N/m}}.$ Why "minimum": a stiffer spring stops it in less than $0.10\text{m}$, still fine; a softer spring squishes more than $0.10\text{m}$, violating the limit. Equality gives the borderline value.

Invert the compression formula. $\mu =\frac{2\cdot 3}{5}=1.2\text{kg}$, and $x=(u_1-0)\sqrt{\mu /k}$ so $u_1=\frac{x}{\sqrt{\mu /k}}=x\sqrt{k/\mu }=0.20\sqrt{750/1.2}=0.20\sqrt{625}=0.20(25)=\boxed{5\text{m/s}}.$ Why it works: the relative approach speed sets the stored energy, and here $u_2=0$ so the relative speed is $u_1$.

As $m_2\to \infty$: $\mu =\frac{m_1{m}_2}{m_1+m_2}=\frac{m_1}{1+m_1/m_2}\to m_1$. Also the relative speed $u_1-u_2=u_1$. So $x\to u_1\sqrt{m_1/k},$ which is exactly the wall formula with $m=m_1$, $u=u_1$. Why it makes sense: an infinitely heavy $m_2$ never moves — it is a wall. And $v_{cm}=\frac{m_1{u}_1}{m_1+m_2}\to 0$, so no bulk KE survives, all of it goes into the spring, matching the wall case. See Centre of Mass Frame.

Stored energy $=\frac{1}{2}\mu (u_1-u_2{)}^2$ depends on the relative speed, which is independent of the total momentum. So two setups with the same $p$ can store wildly different energies — the split matters. The extreme is the CM frame's own frame: maximum relative speed for a given launch. For $m_1=m_2=m$, $u_1=+u$, $u_2=-u$: $\mu =\frac{m\cdot m}{2m}=\frac{m}{2}$, relative speed $u_1-u_2=2u$, so $E_{\text{stored}}=\frac{1}{2}\cdot \frac{m}{2}\cdot (2u{)}^2=\frac{1}{2}\cdot \frac{m}{2}\cdot 4u^2=\boxed{m{u}^2}.$ And $v_{cm}=0$ (total momentum zero) — so all the KE $=2\cdot \frac{1}{2}m{u}^2=m{u}^2$ is stored, consistent with links to Simple Harmonic Motion where the compressed spring is the turning point.

Let $x_1,x_2$ be positions; spring force on each is $\pm k(\text{stretch})$. Writing Newton's law for each and forming the relative coordinate $s=x_1-x_2$ gives $\mu \ddot{s}=-ks$ — the SHM equation with effective mass $\mu$. Hence $\omega =\sqrt{k/\mu },T=2\pi \sqrt{\mu /k}.$ With $\mu =\frac{2\cdot 3}{5}=1.2\text{kg}$: $T=2\pi \sqrt{1.2/750}=2\pi \sqrt{0.0016}=2\pi (0.04)=\boxed{0.251\text{s}}$. Why $\mu$ again: the two-body relative motion behaves as one body of mass $\mu$ on a spring — the same reduced-mass idea that governed $x_{max}$.