NBPhO 2025

Overview

The dumbbell is a “two heavy point masses on a light spring” once one notices how the masses compare. Each ball weighs

$$m \;=\; \tfrac{4}{3}\pi r^{3}\rho \;\approx\; 3.27\times 10^{-2}\,\text{kg},$$

while the rod weighs only

$$m_{\mathrm{rod}} \;=\; \pi (d/2)^{2}\,l\,\rho \;\approx\; 6.1\times 10^{-4}\,\text{kg}.$$

The ratio $m_{\mathrm{rod}}/m \approx 1.9\times 10^{-2}$ is small enough that, to first approximation, the rod is a massless elastic spring connecting two heavy balls. Two finite-mass corrections — Hertzian deformation of the balls themselves, and longitudinal standing waves within the rod — are subleading and the problem instructs us to ignore them.

The relevant scales are

Symbol	Meaning	Order
$c = \sqrt{Y/\rho}$	speed of sound in steel	$5\times 10^{3}\,\text{m s}^{-1}$
$\tau \sim 2r/c$	ball–wall contact time	$\sim 4\,\mu\text{s}$
$l/c$	wave transit through rod	$\sim 20\,\mu\text{s}$
$T = 2\pi\sqrt{ml/(2YA)}$	rod-spring breathing period	$\sim 6\times 10^{-4}\,\text{s}$

These scales separate as

$$\tau \;\ll\; l/c \;\ll\; T,$$

and almost everything in the problem follows from this hierarchy. During a wall contact, the rod cannot transmit any signal between balls (so the wall talks only to the front ball). On the breathing timescale $T$, the contact time is invisible — wall hits look impulsive. And in part (iv), where the rod additionally rotates at rate $\Omega$, the further small parameter

$$\frac{\Omega}{\omega} \;\sim\; \frac{2v\sin\alpha}{\omega l}$$

lets us linearise the rod’s slow tumble against its fast breathing. The hint $\min(\sin x/x)\approx -0.217$ is precisely the threshold of that linearised equation.

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Part (i) — Free longitudinal oscillation period

Why the rod is a massless spring

A thin rod of length $L$, cross-section $A$, Young’s modulus $Y$ deforms by $\Delta L = FL/(YA)$ under axial load $F$. So as a spring it has stiffness

$$k_{\mathrm{rod}} \;=\; \frac{YA}{L}.$$

We want the period at which the balls oscillate, with the rod acting purely as a restoring element. Two assumptions are baked in:

• Rod mass negligible. From the numbers above, $m_{\mathrm{rod}}/m \approx 0.02$. The lightest standing-wave mode within a rod with free ends would have frequency $c/(2l)\approx 2.5\times 10^{4}\,\text{s}^{-1}$, far above the breathing frequency we are about to compute — and the problem explicitly excludes those internal-rod modes.
• Hertz deformation of the balls is fast. A ball under a force $F$ from the rod compresses by an amount that adjusts on timescale $\sim r/c$, much shorter than $T$. So the balls behave as rigid bodies of mass $m$ for the breathing motion.

Why it is the half-rod stiffness, not the full rod

By symmetry of the dumbbell about its midpoint, in the centre-of-mass frame the centre of the rod is stationary during the breathing oscillation. Each ball is therefore connected, effectively, to a fixed wall by half a rod of length $l/2$. The relevant spring constant per ball is

$$k \;=\; \frac{YA}{l/2} \;=\; \frac{2YA}{l},$$

twice that of the full rod. Forgetting this factor of two is the most common pitfall in this part — using $k = YA/l$ would give a period too long by $\sqrt{2}$.

Period

The equation of motion of one ball (about its rest position, in the CoM frame) is $m\ddot x = -kx$, so

$$\omega \;=\; \sqrt{\frac{k}{m}} \;=\; \sqrt{\frac{2YA}{ml}}, \qquad \boxed{\; T \;=\; 2\pi\sqrt{\frac{ml}{2YA}} \;}$$

Substituting $A = \pi(d/2)^{2}$ and $m = \tfrac{4}{3}\pi r^{3}\rho$,

$$T \;=\; 2\pi\sqrt{\frac{\tfrac{4}{3}\pi r^{3}\rho\, l}{2\,Y\,\pi(d/2)^{2}}} \;=\; 2\pi\sqrt{\frac{8\,r^{3}\,l\,\rho}{3\,Y\,d^{2}}}.$$

Numerically with $r = 10^{-2}\,\text{m}$, $d = 10^{-3}\,\text{m}$, $l = 10^{-1}\,\text{m}$, $\rho = 7800\,\text{kg m}^{-3}$, $Y = 2\times 10^{11}\,\text{Pa}$:

$$T \;\approx\; 6.4\times 10^{-4}\,\text{s} \;\approx\; 0.64\,\text{ms}.$$

The corresponding angular frequency is $\omega \approx 9.8\times 10^{3}\,\text{s}^{-1}$.

Consistency checks

• Dimensions. $[ml/(YA)] = \text{kg}\cdot\text{m}/(\text{Pa}\cdot\text{m}^{2}) = \text{kg}/(\text{N}/\text{m}) = \text{s}^{2}$. ✓
• Limits. $Y\to\infty$ (rigid rod) gives $T\to 0$, as expected — a rigid dumbbell has no internal oscillation. $l\to\infty$ at fixed cross-section gives $T\to\infty$ (a long thin rod is a soft spring). $r\to 0$ (massless balls) gives $T\to 0$ (light masses, fast oscillation). All sensible.
• Cross-check via the speed of sound. $\omega = \sqrt{2YA/(ml)} = c\sqrt{2A/(Vl)\cdot(\rho V/m)}\cdot(\text{algebra})$ — one can show $\omega l/c = \sqrt{2 m_{\mathrm{rod}}/m}\approx 0.20$. So $\omega l \approx 0.20\, c$, well below $c/(2l)\cdot 2\pi =$ lowest standing wave in the rod, validating the massless-rod approximation a posteriori.

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Part (ii) — Ball–wall contact time

Order-of-magnitude argument

When a steel ball strikes a steel wall at speed $v$, a compression wave is launched into the ball at the speed of sound in steel,

$$c \;=\; \sqrt{\frac{Y}{\rho}} \;=\; \sqrt{\frac{2\times 10^{11}}{7800}} \;\approx\; 5.06\times 10^{3}\,\text{m s}^{-1}.$$

The wave traverses the ball, reflects from the far surface, and returns — bringing back the information that the ball is finite. After the round trip the contact ends and the ball departs. The natural estimate is

$$\tau \;\sim\; \frac{2r}{c} \;=\; 2r\sqrt{\frac{\rho}{Y}}.$$

With $r = 10^{-2}\,\text{m}$,

$$\boxed{\;\tau \;\sim\; \frac{2r}{c} \;\approx\; 4\,\mu\text{s}\;}$$

This is an order-of-magnitude estimate: the precise contact time depends on a Hertz-contact analysis (and is weakly dependent on $v$, scaling like $v^{-1/5}$), but the qualitative scale $\tau\sim r/c$ is what matters here.

Why this number is the protagonist of every later part

Compare the three timescales:

$$\tau \;\approx\; 4\,\mu\text{s} \;\ll\; l/c \;\approx\; 20\,\mu\text{s} \;\ll\; T \;\approx\; 640\,\mu\text{s}.$$

So during one wall contact:

• The rod’s breathing barely advances ($\omega\tau \approx 0.04$ radians).
• The compression wave inside the rod cannot reach the rear ball ($\tau < l/c$).

The wall therefore acts only on the front ball, instantaneously on the breathing timescale — exactly the impulse approximation. This decoupling is the engine of parts (iii)–(v).

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Part (iii) — $v_{x}(t)$ for the front ball, normal incidence

Setup

Place the wall at $x=0$, dumbbell at $x>0$, axis along $\hat x$. Both balls have velocity $-v\hat x$. Let $t=0$ be the moment the front ball first contacts the wall.

Phase 1: the first contact ($0\le t\lesssim\tau$)

During contact, the wall delivers an impulse purely along $+\hat x$ on the front ball. By the timescale argument above, the rear ball is still uninformed about the collision (the wave needs $l/c\approx 5\tau$ to reach it). So the wall sees a single elastic ball of mass $m$ bouncing off a rigid surface, and reverses its $x$-velocity:

$$v_{F,x}: \;\; -v \;\longrightarrow\; +v, \qquad v_{R,x} = -v\;\;\text{(unchanged)}.$$

Phase 2: free breathing ($\tau \lesssim t < T/2$)

Just after contact the system has

$$\vec v_{F} = (+v,0), \qquad \vec v_{R} = (-v,0), \qquad \vec v_{\mathrm{CoM}} = 0.$$

In the CoM frame the two balls converge toward the centre, exciting the breathing mode of part (i) with maximum amplitude. Each ball oscillates as

$$v_{F,x}(t) \;=\; v\cos(\omega t), \qquad v_{R,x}(t) \;=\; -v\cos(\omega t).$$

At $t = T/4$ the rod is maximally compressed (both balls instantaneously at rest in the lab frame). At $t = T/2$ the rod is back at its rest length and the velocities have reversed:

$$v_{F,x}(T/2) \;=\; -v, \qquad v_{R,x}(T/2) \;=\; +v.$$

The front ball is back at its rest position — which, since it left the wall at this position, is the wall — and is moving toward the wall at speed $v$.

Phase 3: the second contact (around $t=T/2$)

Same impulsive analysis as before. The wall reverses $v_{F,x}$ from $-v$ to $+v$. After this contact:

$$\vec v_{F} = (+v,0), \qquad \vec v_{R} = (+v,0).$$

Both balls move at $+v\hat x$, the rod is unstrained, the dumbbell flies away as a rigid body.

The sketch

v_F,x ↑
      |
   +v |── ─ ─ ─●━━●─.                                .─━━●━━━━━━━━━ ...
      |             `.                            ,'
      |               `.                        ,'
      |                 `.                    ,'
    0 +───────────────────●──────────────────●────────────────────► t
      |                   |  `.            ,'  |
      |                  T/4   `.        ,'   T/2
      |                          `.    ,'
   −v |── ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ●━━●─── (no, this jumps up at T/2)
      |
      0    τ                                       T/2

A cleaner verbal description:

• $t<0$ : flat at $-v$.
• $t\approx 0$ : near-vertical jump from $-v$ to $+v$ over time $\tau\ll T$.
• $0<t<T/2$ : smooth cosine, $v\cos(\omega t)$, descending through $0$ at $t=T/4$ and reaching $-v$ at $t=T/2$.
• $t\approx T/2$ : near-vertical jump from $-v$ back to $+v$.
• $t>T/2$ : flat at $+v$.

So the trace looks like two near-vertical risers separated by a half-cosine “swing” of the front ball.

Educational remark — why two bounces, not one

A rigid dumbbell would bounce once: the wall would absorb the entire $4mv$ of momentum in a single contact and the dumbbell would depart at $+v$. The elastic rod splits this into two bounces, each transferring $2mv$:

• bounce 1 reverses only the front ball, leaving energy stored in the rod;
• the rod stores that energy as breathing for half a period;
• bounce 2 transfers the rebuilt momentum back to the wall via the front ball, and the dumbbell departs at $+v$.

The total contact-with-wall lasts $\sim 2\tau\approx 8\,\mu$s — the interval between contacts $T/2\approx 320\,\mu$s is the rod’s signature. If we replaced steel with a stiffer, “more rigid” material ($Y\to\infty$) we would recover the single-bounce picture as $T\to 0$.

Energy and momentum check

	front $v_x$	rear $v_x$	$p_x$	KE
before	$-v$	$-v$	$-2mv$	$mv^{2}$
after 1st bounce	$+v$	$-v$	$0$	$mv^{2}$
after 2nd bounce	$+v$	$+v$	$+2mv$	$mv^{2}$

Total impulse from wall: $4mv$ (split evenly), kinetic energy preserved at every step. ✓

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Part (iv) — Critical angle $\alpha_{0}$

Geometry

Tilt the dumbbell so that, at the moment of first contact, the front ball is at the origin and the rod axis points (from front to rear) along

$$\hat n_{0} \;=\; (\cos\alpha, \sin\alpha).$$

The rear ball is at $\hat n_{0}\,l$. Both balls have velocity $-v\hat x$. We restrict to $0<\alpha<\pi/2$.

First contact: only the front ball reverses $v_{x}$

Same hierarchy as before — during $\tau$ the wall talks only to the front ball, and the wall’s impulse is purely normal (along $+\hat x$). So the front ball’s $y$-velocity is unaffected and its $x$-velocity reverses:

$$\vec v_{F} = (-v,0)\;\longrightarrow\;(+v,0), \qquad \vec v_{R} = (-v,0)\;\text{(unchanged)}.$$

The total impulse from the wall is $2mv\hat x$, so the CoM goes from velocity $-v\hat x$ to rest in the lab frame. (Note: same conclusion as in part (iii), and independent of $\alpha$ — the wall always reverses just the front ball’s $v_{x}$.)

Decomposition: rotation + breathing

In the CoM frame, just after the bounce, the front ball is at $-\tfrac{l}{2}\hat n_{0}$ moving at $(+v,0)$, and the rear at $+\tfrac{l}{2}\hat n_{0}$ moving at $(-v,0)$.

The relative velocity $\vec v_{F}-\vec v_{R} = (2v,0)$ decomposes onto the rod axis $\hat n_{0}$ and the perpendicular $\hat n_{\perp,0} = (-\sin\alpha,\cos\alpha) = \hat z\times\hat n_{0}$:

$$(2v,0)\cdot\hat n_{0} \;=\; 2v\cos\alpha \quad\text{(along the rod — *breathing*)},$$ $$(2v,0)\cdot\hat n_{\perp,0} \;=\; -2v\sin\alpha \quad\text{(perpendicular to the rod — *rotation*)}.$$

The rotation rate is fixed by $\vec v_{F}-\vec v_{R} = \vec\Omega\times(\vec r_{F}-\vec r_{R}) = -\Omega l\,\hat n_{\perp,0}$, giving

$$\boxed{\;\Omega \;=\; \frac{2v\sin\alpha}{l}\;}$$

(positive $\Omega$ = counterclockwise spin viewed from $+\hat z$). The breathing oscillation has each ball moving toward the CoM along $\hat n_{0}$ at speed $v\cos\alpha$, so the front-ball displacement along $+\hat n_{0}$ is

$$u_{F}(t) \;=\; \frac{v\cos\alpha}{\omega}\sin(\omega t), \qquad \omega = \sqrt{\frac{2YA}{ml}}.$$

Two small parameters of the same order

For “ordinary” launch speeds (say $v\sim 1$–$10\,\text{m s}^{-1}$),

$$\frac{\Omega}{\omega} \;=\; \frac{2v\sin\alpha}{\omega l} \;\sim\; \frac{v}{500\,\text{m s}^{-1}}\,\sin\alpha \;\ll\; 1,$$

so the rod tumbles only a little during one breathing period. Likewise the breathing amplitude $v\cos\alpha/\omega$ is much smaller than $l/2$. We will linearise in both, keeping only the leading-in-$v$ terms.

Position of the front ball

In the lab frame, the CoM stays put at $(\tfrac l2\cos\alpha,\,\tfrac l2\sin\alpha)$. The rod axis at time $t$ is

$$\hat n(t) \;=\; \bigl(\cos(\alpha+\Omega t),\;\sin(\alpha+\Omega t)\bigr),$$

and the front ball sits at $\vec r_{F}(t) = \vec r_{\mathrm{CoM}} - \bigl(\tfrac l2 - u_{F}(t)\bigr)\hat n(t)$. The $x$-component is

$$x_{F}(t) \;=\; \tfrac{l}{2}\cos\alpha \;-\; \bigl(\tfrac{l}{2}-u_{F}(t)\bigr)\cos(\alpha+\Omega t).$$

Linearise

To leading order in $v$ (so to first order in $\Omega t$ and in $u_{F}/l$):

$$\cos(\alpha+\Omega t) \;\approx\; \cos\alpha - \sin\alpha\cdot\Omega t,$$ $$x_{F}(t) \;\approx\; \tfrac{l}{2}\sin\alpha\cdot\Omega t \;+\; u_{F}(t)\cos\alpha.$$

Setting $x_{F}=0$ and substituting $\Omega = 2v\sin\alpha/l$ and $u_{F} = (v\cos\alpha/\omega)\sin(\omega t)$:

$$v\sin^{2}\alpha\,t \;+\; \frac{v\cos^{2}\alpha}{\omega}\sin(\omega t) \;=\; 0.$$

Multiply by $\omega/(v\cos^{2}\alpha)$ and let $s = \omega t$:

$$\boxed{\;\frac{\sin s}{s} \;=\; -\tan^{2}\alpha\;}$$

This is the master equation of part (iv). It asks: at what $s>0$ does the front ball return to the wall?

The threshold condition

Since $\sin s/s$ takes its minimum value $-0.217$ at $s\approx 4.49$ (the first positive root of $s\cos s = \sin s$), the equation has a positive solution iff

$$\tan^{2}\alpha \;\le\; 0.217.$$

For $\tan^{2}\alpha < 0.217$ there are two positive solutions (the curve $\sin s/s$ dips below the horizontal line $-\tan^{2}\alpha$ and rises back through it), and the smaller is the time of the second collision. For $\tan^{2}\alpha > 0.217$ the curve never reaches the line and no second collision of the front ball occurs — the rod has rotated the front ball away from the wall before the breathing could push it back.

The transition therefore takes place at

$$\tan\alpha_{0} \;=\; \sqrt{0.217} \;\approx\; 0.466,$$ $$\boxed{\;\alpha_{0} \;=\; \arctan\sqrt{0.217} \;\approx\; 25^{\circ} \;\approx\; 0.436\,\text{rad}\;}$$

Physical picture

For $\alpha < \alpha_{0}$, the geometry is “head-on enough” that the breathing of the rod can drive the front ball back into the wall before the slow tumble has time to swing it sideways — the front ball undergoes two (or more) collisions, much like part (iii). For $\alpha > \alpha_{0}$, the tumble wins: the front ball heads off into the air and the next wall contact will be by the rear ball, after the rod has rotated through the geometry. That second case is what part (v) addresses.

Sanity checks

• Limit $\alpha\to 0$. $\tan^{2}\alpha\to 0$, and $\sin s/s = 0$ at $s = \pi$; the master equation gives $\omega t = \pi$, i.e. $t = T/2$. This is exactly the head-on result of part (iii), recovered as a check.
• Limit $\alpha\to\pi/2$. $\tan^{2}\alpha\to\infty$, no solution exists — and indeed in the limit the rod is parallel to the wall so the very notion of a “front ball” rebounding is degenerate; the right physical picture is two simultaneous collisions, not a second visit.
• Numerical match. The first positive root of $s\cos s = \sin s$ is $s_{*} = 4.49340945$, with $\sin s_{*}/s_{*} = -0.21723$. Then $\sqrt{0.21723} = 0.46608$ and $\arctan(0.46608) = 24.992^{\circ}$. The approximation $\alpha_{0}\approx 25^{\circ}$ is essentially exact at the precision of the hint.

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Part (v) — CoM departure speed for $\alpha>\alpha_{0}$

State after the rod’s oscillations have decayed

For $\alpha>\alpha_{0}$, the front ball does not revisit the wall. Between the first and second wall contact the rod’s breathing dies away to heat, but rigid-body motion is preserved (energy lost is internal, and the absence of external torque keeps angular momentum about the CoM fixed). So just before the rear ball arrives at the wall:

• CoM velocity: $0$ (lab frame).
• Angular velocity: $\Omega = 2v\sin\alpha/l$ (counterclockwise), unchanged.
• Rod stress: zero (oscillations dissipated).

Energy bookkeeping: the original CoM-frame KE is $mv^{2}$, of which $\tfrac{1}{2}I\Omega^{2} = \tfrac{1}{2}(ml^{2}/2)(2v\sin\alpha/l)^{2} = mv^{2}\sin^{2}\alpha$ is rotational (preserved) and $mv^{2}\cos^{2}\alpha$ is breathing (dissipated). The fraction of dumbbell KE that ends up as heat after the first wall contact is $\cos^{2}\alpha$. (The moment of inertia about the CoM is $I = 2m(l/2)^{2} = ml^{2}/2$ — two point masses on the ends — not a distributed-rod formula, since the balls dominate.)

Geometry of the rear ball’s arrival

Place the lab-frame CoM at $\vec r_{\mathrm{CoM}} = (\tfrac{l}{2}\cos\alpha,\,\tfrac{l}{2}\sin\alpha)$. The rod axis at time $t$ (measured from the first contact, ignoring the breathing decay timescale, which is much shorter than the rotation timescale and irrelevant here) is

$$\hat n(t) \;=\; \bigl(\cos(\alpha+\Omega t),\,\sin(\alpha+\Omega t)\bigr),$$

and the rear ball sits at $\vec r_{R}(t) = \vec r_{\mathrm{CoM}} + \tfrac{l}{2}\hat n(t)$. Its $x$-coordinate is

$$x_{R}(t) \;=\; \tfrac{l}{2}\cos\alpha + \tfrac{l}{2}\cos(\alpha+\Omega t),$$

and the wall is reached when $x_{R} = 0$, i.e.

$$\cos(\alpha+\Omega t) \;=\; -\cos\alpha \;\Longrightarrow\; \alpha + \Omega t \;=\; \pi - \alpha \;\Longrightarrow\; \Omega t_{2} \;=\; \pi - 2\alpha.$$

The time of the second wall contact is

$$t_{2} \;=\; \frac{\pi - 2\alpha}{\Omega} \;=\; \frac{(\pi-2\alpha)\,l}{2v\sin\alpha}.$$

At this moment $\hat n(t_{2}) = (-\cos\alpha,\sin\alpha)$, i.e. the rod has swung from “up-and-right” to “up-and-left”, and the rear ball has just reached the wall at the height $y = l\sin\alpha$.

Velocity of the rear ball just before the second contact

A pure rigid rotation at $\Omega\hat z$ about the CoM gives the rear ball

$$\vec v_{R}(t_{2}^{-}) \;=\; \Omega\hat z\times\tfrac{l}{2}\hat n(t_{2}) \;=\; \tfrac{l\Omega}{2}\bigl(-\sin\alpha,\,-\cos\alpha\bigr) \;=\; v\sin\alpha\bigl(-\sin\alpha,\,-\cos\alpha\bigr).$$

The components are

$$v_{R,x} \;=\; -v\sin^{2}\alpha, \qquad v_{R,y} \;=\; -v\sin\alpha\cos\alpha.$$

Second wall contact: rear ball’s $v_{x}$ reverses

Same impulse approximation as before — only the rear ball is touched by the wall during $\tau\ll l/c$, and the wall’s impulse is purely normal:

$$v_{R,x}\text{:}\;\; -v\sin^{2}\alpha \;\longrightarrow\; +v\sin^{2}\alpha,$$

while $v_{R,y}$ and the front ball’s velocity are unchanged. The front ball’s velocity at that instant (rigid rotation, opposite end) is

$$\vec v_{F}(t_{2}) \;=\; (v\sin^{2}\alpha,\,v\sin\alpha\cos\alpha).$$

CoM velocity after the second contact

Average:

$$\vec v_{\mathrm{CoM}}^{\,\mathrm{after}} \;=\; \tfrac{1}{2}\bigl(\vec v_{F} + \vec v_{R}^{\,\mathrm{after}}\bigr) \;=\; \tfrac{1}{2}\bigl((v\sin^{2}\alpha+v\sin^{2}\alpha,\; v\sin\alpha\cos\alpha-v\sin\alpha\cos\alpha)\bigr) \;=\; (v\sin^{2}\alpha,\,0).$$

The $y$-components cancel — and they had to, since the wall delivered impulse only along $\hat x$ at both contacts, so the total $y$-momentum has been zero all along.

$$\boxed{\;\lvert\vec v_{\mathrm{CoM}}\rvert \;=\; v\sin^{2}\alpha\;}$$

The CoM moves perpendicular to the wall, away from it.

Why the rear ball does not bounce a second time

We should make sure that the rear ball, having now bounced once, doesn’t undergo its own “two-bounce” scenario as in part (iii). After the second contact:

• The rear ball is at the wall, with velocity $(+v\sin^{2}\alpha,\,-v\sin\alpha\cos\alpha)$.
• The CoM moves at $(v\sin^{2}\alpha,0)$.
• The angular momentum has changed by the wall’s torque — a calculation along the same lines as part (iv) gives a new spin rate $\Omega' = 2v\sin\alpha\cos^{2}\alpha/l$.
• New breathing amplitude in velocity along the rod: $v\sin^{2}\alpha\cos\alpha$ on each ball.

Now run the part-(iv) analysis with the rear ball in the role of the front ball and with the rod axis at angle $\alpha$ below $+\hat x$ (the rod makes the same angle $\alpha$ with the wall normal). The relative velocity components decompose exactly as before, with $\Omega'$ and the new breathing amplitude both scaled by the same factor $\sin\alpha\cdot\cos\alpha$. Carrying the algebra through yields the same master equation

$$\frac{\sin s}{s} \;=\; -\tan^{2}\alpha,$$

which has no solution in the regime $\alpha>\alpha_{0}$. So the rear ball departs, the breathing decays again, and the dumbbell flies off at the CoM velocity $v\sin^{2}\alpha$.

This is what makes $\alpha_{0}$ the only relevant transition in the problem: the same threshold gates both “front bounces twice” (small $\alpha$) and “rear bounces twice” (also small $\alpha$). The condition $\alpha>\alpha_{0}$ guarantees a clean two-stage scenario — front, then rear, each just once.

Sanity checks

• Limit $\alpha\to\pi/2$. $v_{\mathrm{CoM}}\to v$. The rod is parallel to the wall, the two balls strike the wall in essentially identical fashion (each reverses its $v_{x}$) and the dumbbell rebounds intact at speed $v$. ✓

• Limit $\alpha\to\alpha_{0}^{+}\approx 25^{\circ}$. $v_{\mathrm{CoM}}\to v\sin^{2}\alpha_{0}\approx 0.179\,v$. So even at the threshold, the dumbbell departs at about $18\%$ of the incoming speed; the rest of the original kinetic energy went into rotation and into heat (the dissipated breathing).

• Below threshold. The formula $v\sin^{2}\alpha$ is not the answer for $\alpha<\alpha_{0}$. There the front ball bounces twice (as in part (iii)), so the wall delivers impulse to the front ball at two separate times, not to the rear ball. Naively extending $v\sin^{2}\alpha$ down to $\alpha=0$ would give $v_{\mathrm{CoM}}=0$, contradicting part (iii)‘s result of $v_{\mathrm{CoM}}=v$.

• Energy budget for $\alpha>\alpha_{0}$. Initial KE: $mv^{2}$. After all dust settles:

  • CoM translational: $\tfrac{1}{2}(2m)v^{2}\sin^{4}\alpha = mv^{2}\sin^{4}\alpha$.
  • Rotational: $\tfrac{1}{2}I\Omega'^{2} = mv^{2}\sin^{2}\alpha\cos^{4}\alpha$.
  • Heat (1st breathing decay): $mv^{2}\cos^{2}\alpha$.
  • Heat (2nd breathing decay): $mv^{2}\sin^{4}\alpha\cos^{2}\alpha$.

  Sum: $mv^{2}\bigl[\sin^{4}\alpha + \sin^{2}\alpha\cos^{4}\alpha + \cos^{2}\alpha + \sin^{4}\alpha\cos^{2}\alpha\bigr] = mv^{2}\bigl[\sin^{4}\alpha(1+\cos^{2}\alpha) + \cos^{2}\alpha(1+\sin^{2}\alpha\cos^{2}\alpha)\bigr]$. A short trig identity (or, easier, just check at $\alpha=\pi/2$ where the sum should be $mv^{2}$) verifies energy conservation at the limits and bookkeeping consistency throughout.

Educational remark — the symmetry-broken bounce

A key conceptual takeaway is that, for the obliquely-hitting elastic dumbbell, the direction of the CoM rebound is purely along $\hat x$ — never deflected sideways — even though the dumbbell has internal angular momentum. The reason is that the wall couples to the dumbbell only through normal impulses, which contribute zero net $y$-momentum. The internal $y$-motion is bookkept entirely as spin and breathing, and it does not leak into the trajectory of the CoM. The dumbbell departs head-on, just slower (by the factor $\sin^{2}\alpha$) than it arrived.

This is the same mechanism that makes a tennis racket bounce: glancing-angle hits transfer relatively little CoM-frame energy to translation, with the rest going into rotation. Our $\sin^{2}\alpha$ is the dumbbell’s analogue of the “slippage” loss in a rolling collision.

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Summary of results

Part	Quantity	Result	Numerics
(i)	Breathing period	$\;T = 2\pi\sqrt{\dfrac{ml}{2YA}} = 2\pi\sqrt{\dfrac{8r^{3}l\rho}{3Yd^{2}}}\;$	$\approx 0.64\,\text{ms}$
(ii)	Wall contact time	$\;\tau \sim \dfrac{2r}{c} = 2r\sqrt{\dfrac{\rho}{Y}}\;$	$\approx 4\,\mu\text{s}$
(iii)	Front-ball trace	$-v$ → (sharp jump) → $v\cos\omega t$ → (sharp jump) → $+v$, with both jumps lasting $\tau$ and the cosine arc lasting $T/2$	—
(iv)	Critical angle	$\;\alpha_{0} = \arctan\sqrt{0.217}\;$ from $\sin s/s = -\tan^{2}\alpha$	$\approx 25^{\circ}$
(v)	CoM departure speed ($\alpha>\alpha_{0}$)	$\;v_{\mathrm{CoM}} = v\sin^{2}\alpha\;$	—

One sentence to take away. Three timescales — wall-contact $\tau$, rod transit $l/c$, and rod breathing $T$ — separate by enough orders of magnitude that wall hits act as instantaneous, single-ball impulses; everything in the problem is then bookkeeping rotation and breathing of a “two heavy beads on a light spring” system, with the critical angle $\alpha_{0}\approx 25^{\circ}$ marking the transition from “front-ball bounces twice” to “front then rear, each once”.