Classical-Mechanics · Unit 15 · Video 4 · Interactive Practice

Two Bodies, One Equation: Internal Work and the Reduced Mass

IKey Formulas

Formula	Name	What you need
$\dfrac{1}{\mu} = \dfrac{1}{m_1} + \dfrac{1}{m_2} \;\Longrightarrow\; \mu = \dfrac{m_1 m_2}{m_1 + m_2}$	Reduced mass	The two masses
$\vec{F}_{2,1} = \mu\,\dfrac{d^2\vec{r}_{2,1}}{dt^2}$	Effective one-body law	Internal force and $\vec{r}_{2,1} = \vec{r}_1 - \vec{r}_2$
$W = \displaystyle\int_A^B \vec{F}_{2,1}\cdot d\vec{r}_{2,1}$	Work of the internal pair	The relative displacement alone
$W = \Delta K_{\text{rel}} = \tfrac{1}{2}\mu\left(v_B^2 - v_A^2\right)$	Internal work–energy theorem	Relative speeds at $A$ and $B$

Key Insight: Exchanging the labels $1 \leftrightarrow 2$ flips both $\vec{F}$ and $d\vec{r}$ , so $W$ is unchanged; and a rigid translation ( $d\vec{r}_1 = d\vec{r}_2$ ) gives $d\vec{r}_{2,1} = 0$ , so the internal pair does no work at all.

IIVisualization 1 — Two Accelerations, One Reduced Mass

One interaction force, two different accelerations — their relative acceleration obeys a single effective mass $\mu$ .

m₁ (kg) 2 m₂ (kg) 6

IIIVisualization 2 — Two Work Integrals, One Relative Displacement

Each body's own work follows the common drift; the pair's total work follows only the changing separation.

Common drift (m) 0 Final separation (m) 3 Progress A→B 1.00

IVVisualization 3 — Predicting the Final Relative Speed

The same $12\ \text{J}$ of internal work adds less relative speed when the pair is already separating fast.

Initial relative speed (m/s) 0 Your prediction for the final relative speed (m/s) 2

💡 That $12\ \text{J}$ leaves $\tfrac{1}{2}MV_{\text{cm}}^2$ untouched: internal forces cancel in pairs, so in $K = \tfrac{1}{2}MV_{\text{cm}}^2 + \tfrac{1}{2}\mu v_{\text{rel}}^2$ only the second term can change.

VQuiz Questions

Problem 1 · Reduced Mass (Direct)

Given: two bodies with $m_1 = 3\ \text{kg}$ and $m_2 = 6\ \text{kg}$ interacting through a Newton's third law pair — find the reduced mass $\mu$ .

✅ Correct!

\dfrac{1}{\mu} = \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{1}{2}

, so

\mu = 2\ \text{kg}

— smaller than either mass, as it must be.

❌ Close, but…

0.5

is the value of

1/\mu

(in

\text{kg}^{-1}

). The definition gives

1/\mu

first; invert it to get

\mu

❌ Not quite.

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

is always smaller than both masses, so it can be neither their sum nor their average.

Show solution

Add the inverse masses:

\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{1}{2}\ \text{kg}^{-1}

Invert, or use the combined form directly:

\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{3 \cdot 6}{3 + 6} = \frac{18}{9} = 2\ \text{kg}

Sanity check: $2 < 3$ and $2 < 6$ . Equal masses would give $\mu = m/2$ ; a far heavier partner would push $\mu$ up toward $m_1 = 3\ \text{kg}$ but never past it.

Problem 2 · Rigid Translation

Given: a stretched spring joins two bodies, and over some interval both bodies undergo the same displacement $d\vec{r}_1 = d\vec{r}_2 = \vec{d}$ , so the separation never changes — find the total work done by the internal force pair over that interval.

✅ Correct!

W = \displaystyle\int \vec{F}_{2,1}\cdot d\vec{r}_{2,1}

and here

d\vec{r}_{2,1} = d\vec{r}_1 - d\vec{r}_2 = \vec{d} - \vec{d} = 0

❌ Close, but…

d\vec{r}_{2,1} = d\vec{r}_1 - d\vec{r}_2

is the same in every inertial frame, so the pair's total work is frame-independent — it is zero in all of them.

❌ Not quite. Add the two works before evaluating:

\vec{F}_{2,1}\cdot\vec{d} + \vec{F}_{1,2}\cdot\vec{d} = (\vec{F}_{2,1} + \vec{F}_{1,2})\cdot\vec{d}

, and the third law empties that bracket.

Show solution

Route 1 — cancel the forces. The two displacements are equal, so

W = \vec{F}_{2,1}\cdot d\vec{r}_1 + \vec{F}_{1,2}\cdot d\vec{r}_2 = (\vec{F}_{2,1} + \vec{F}_{1,2})\cdot\vec{d} = \vec{0}\cdot\vec{d} = 0

Route 2 — use the collapsed integral. The third law already collapsed the two integrals into one:

W = \int_A^B \vec{F}_{2,1}\cdot d\vec{r}_{2,1}, \qquad d\vec{r}_{2,1} = d\vec{r}_1 - d\vec{r}_2 = 0 \;\Longrightarrow\; W = 0

Each body may individually gain or lose energy while the pair is carried along, but those two amounts are equal and opposite. Internal work exists only when the separation changes.

Problem 3 · From Masses to Final Relative Speed

Given: two carts, $m_1 = 4\ \text{kg}$ and $m_2 = 12\ \text{kg}$ , joined by a stretched spring on a frictionless track with no external forces. Their relative speed at state $A$ is $v_A = 3\ \text{m/s}$ , and between $A$ and $B$ the internal pair does $W = +24\ \text{J}$ of work.

What is the reduced mass?

What is the relative speed at state $B$ ?

✅ Correct!

\mu = 3\ \text{kg}

, so

v_B^2 = v_A^2 + \dfrac{2W}{\mu} = 9 + 16 = 25

❌ Check the reduced mass.

\dfrac{1}{\mu} = \dfrac{1}{4} + \dfrac{1}{12} = \dfrac{1}{3}

— the total mass

M = 16\ \text{kg}

belongs to the centre-of-mass term, not to this one.

❌ Check the relative speed.

v_B^2 = v_A^2 + \dfrac{2W}{\mu}

: the initial relative speed does not drop out, and speeds themselves never add — the squares do.

Show solution

Step 1 — reduced mass:

\frac{1}{\mu} = \frac{1}{4} + \frac{1}{12} = \frac{3+1}{12} = \frac{1}{3} \;\Longrightarrow\; \mu = \frac{4 \cdot 12}{16} = 3\ \text{kg}

Step 2 — internal work–energy theorem:

W = \tfrac{1}{2}\mu\left(v_B^2 - v_A^2\right) \;\Longrightarrow\; 24 = \tfrac{1}{2}(3)\left(v_B^2 - 3^2\right)

v_B^2 - 9 = \frac{2(24)}{3} = 16 \;\Longrightarrow\; v_B^2 = 25 \;\Longrightarrow\; v_B = 5\ \text{m/s}

Common mistakes:

Using $M = 16\ \text{kg}$ instead of $\mu$ : $v_B = \sqrt{9 + 3} \approx 3.5\ \text{m/s}$ .
Dropping $v_A^2$ : $v_B = \sqrt{16} = 4\ \text{m/s}$ .
Adding speeds instead of squares: $3 + 4 = 7\ \text{m/s}$ .

No external work acts, so this $24\ \text{J}$ goes entirely into relative motion; $\tfrac{1}{2}MV_{\text{cm}}^2$ is untouched.

Problem 4 · The Heavy-Partner Limit

Given: a $2.0\ \text{kg}$ ball falls toward the Earth, $m_2 = 6.0\times 10^{24}\ \text{kg}$ . Gravity is the internal force pair of this two-body system, and no external forces act.

What is the reduced mass of the ball–Earth system?

So what does $W = \tfrac{1}{2}\mu\left(v_B^2 - v_A^2\right)$ become here?

✅ Correct!

\mu = \dfrac{m_1}{1 + m_1/m_2} \approx m_1

, and the relative velocity is essentially the ball's velocity — the two-body theorem collapses into the familiar single-particle result.

❌ Check the limit.

\mu = \dfrac{m_1m_2}{m_1+m_2} = \dfrac{m_1}{1 + m_1/m_2}

, and

m_1/m_2 \approx 3\times 10^{-25}

— so

\mu

sits just below the light mass, never near the heavy one.

❌ Close, but… the Earth's own work is negligible, yet the pair's total work is not: it is exactly what speeds the ball up,

\tfrac{1}{2}m_1(v_B^2 - v_A^2)

❌ Not quite. Substitute

\mu \approx m_1 = 2.0\ \text{kg}

, not

m_2

and not

M = m_1 + m_2

Show solution

Step 1 — reduced mass in the heavy-partner limit:

\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m_1}{1 + m_1/m_2}, \qquad \frac{m_1}{m_2} = \frac{2.0}{6.0\times 10^{24}} \approx 3\times 10^{-25}

So $\mu = 2.0\ \text{kg}$ to within about one part in $3\times 10^{24}$ : the light body does all the moving.

Step 2 — substitute:

W = \tfrac{1}{2}\mu\left(v_B^2 - v_A^2\right) \approx \tfrac{1}{2}m_1\left(v_B^2 - v_A^2\right)

and since the Earth barely moves, $\vec{v}_{2,1} = \vec{v}_1 - \vec{v}_2 \approx \vec{v}_1$ . The everyday statement "the work done by gravity equals the change in $\tfrac{1}{2}mv^2$ of the falling object" is this two-body theorem with $\mu \to m_{\text{ball}}$ .

Why not $W \approx 0$ ? The work done on the Earth is negligible because $d\vec{r}_2 \approx 0$ , but the pair's total work depends on $d\vec{r}_{2,1} \approx d\vec{r}_1$ , which is the ball's full fall.

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