Irodov 1.149: Relative Velocity Alone Carries Internal Energy

Problem Irodov 1.149

Two small discs of masses $m_1$ and $m_2$ , connected by a weightless spring, lie at rest on a smooth horizontal plane (so the spring is momentarily unstretched). The discs are then set moving with initial velocities $\vect v_1$ and $\vect v_2$ , directions mutually perpendicular, both in the horizontal plane. Find the total energy $\tilde E$ of the two-disc system as measured in the reference frame of its center of inertia.

The two discs joined by the spring, launched with mutually perpendicular velocities v_1 and v_2; the spring is at its natural length at this instant. — The two discs joined by the spring, launched with mutually perpendicular velocities $\vect v_1$ and $\vect v_2$ ; the spring is at its natural length at this instant.

Given

$m_1$
$m_2$
$\vect v_1\perp\vect v_2$

Find

\tilde E

Solution

Step 1Energy in the center-of-mass frame is purely kinetic at release

The plane is smooth and the spring is the only internal force, so the total mechanical energy is conserved and equal in every inertial frame’s bookkeeping once the frame’s own kinetic term is subtracted correctly; evaluate it at the launch instant, where the spring is still at its natural length and carries no potential energy. There $\tilde E$ is just the kinetic energy of the two discs relative to the center of mass, whose velocity is the momentum-weighted average.

\begin{align*}\vect V_{cm} &= \frac{m_1\vect v_1+m_2\vect v_2}{m_1+m_2} \\ \tilde E &= \frac12 m_1|\vect v_1-\vect V_{cm}|^2+\frac12 m_2|\vect v_2-\vect V_{cm}|^2\end{align*}

Step 2Reducing the two-body kinetic energy to the reduced mass form

Expanding both squares and collecting terms turns the two-body expression into one depending only on the relative velocity $\vect v_1-\vect v_2$ .

\begin{align*}\tilde E &= \frac12 m_1v_1^2-m_1\vect v_1\cdot\vect V_{cm}+\frac12 m_1V_{cm}^2+\frac12 m_2v_2^2-m_2\vect v_2\cdot\vect V_{cm}+\frac12 m_2V_{cm}^2 \\ &= \frac12\bigl(m_1v_1^2+m_2v_2^2\bigr)-(m_1\vect v_1+m_2\vect v_2)\cdot\vect V_{cm}+\frac12(m_1+m_2)V_{cm}^2 \\ &= \frac12\bigl(m_1v_1^2+m_2v_2^2\bigr)-\frac12(m_1+m_2)V_{cm}^2 \\ &= \frac{(m_1+m_2)(m_1v_1^2+m_2v_2^2)-(m_1\vect v_1+m_2\vect v_2)^2}{2(m_1+m_2)} \\ &= \frac{m_1m_2\bigl(v_1^2+v_2^2-2\vect v_1\cdot\vect v_2\bigr)}{2(m_1+m_2)}\end{align*}

\tilde E = \frac{\mu}{2}\left|\vect v_1-\vect v_2\right|^2,\quad \mu=\frac{m_1m_2}{m_1+m_2}

(1)

Step 3Perpendicular velocities add in quadrature

Since $\vect v_1\perp\vect v_2$ , their dot product vanishes, so (1) loses its cross term entirely.

|\vect v_1-\vect v_2|^2 = v_1^2+v_2^2-2\vect v_1\cdot\vect v_2 = v_1^2+v_2^2

\tilde E = \dfrac{\mu(v_1^2+v_2^2)}{2}

(2)

Check

Setting $v_2=0$ collapses the problem to the textbook case of one disc launched at a stationary second one, where the kinetic energy relative to the center of mass is the standard $\frac12\mu v_1^2$ ; the boxed result reduces to exactly that. Dimensionally $[\mu][v^2]=\mathrm{kg}\cdot\mathrm{m^2/s^2}=\mathrm J$ , an energy.

Remark

The general two-body result is $\tilde E=\frac{\mu}{2}(v_1^2+v_2^2-2v_1v_2\cos\theta)$ for an angle $\theta$ between the launch velocities; perpendicular launches are the special case where the interference term $2v_1v_2\cos\theta$ vanishes on its own, not a case where it was ever assumed away. Parallel launches ( $\theta=0$ ) would instead subtract, giving $\frac{\mu}{2}(v_1-v_2)^2$ , smaller than the perpendicular value whenever both discs move at all.