Irodov 1.147: The CM Frame Minimizes Kinetic Energy – Amit & Deepali Sinha Foundation

Problem Irodov 1.147

Two particles move along a common straight line ( $x$ axis) in an inertial frame $K$ : one of mass $m_1$ has velocity $v_1$ , the other of mass $m_2$ has velocity $v_2$ (both signed scalars along $x$ ). A second frame $K'$ moves at some constant velocity $V$ along the same line relative to $K$ .

Find the velocity $V$ of $K'$ for which the two particles’ combined kinetic energy, measured in $K'$ , is as small as possible.
Find that minimum combined kinetic energy.

Top: the two particles in frame K with velocities v_1, v_2 along x. Bottom: the same pair seen from K', which moves at V along x; their velocities there are v_1'= v_1- V and v_2'= v_2- V. — Top: the two particles in frame $K$ with velocities $\vect v_1$ , $\vect v_2$ along $x$ . Bottom: the same pair seen from $K'$ , which moves at $V$ along $x$ ; their velocities there are $\vect v_1'=\vect v_1-\vect V$ and $\vect v_2'=\vect v_2-\vect V$ .

Given

$m_1,\ v_1$
$m_2,\ v_2$ (velocities along $x$ in frame $K$ )

Find

(a)

V

minimizing the combined kinetic energy; (b) that minimum kinetic energy

Solution

Step 1Write the combined kinetic energy in a frame moving at

V

In a frame $K'$ translating at velocity $V$ along $x$ relative to $K$ , each particle’s velocity is $v_i-V$ (Galilean velocity addition), so the combined kinetic energy there is a single function of $V$ .

T(V) = \frac12 m_1(v_1-V)^2+\frac12 m_2(v_2-V)^2 \tag{1}

Step 2Minimize over

V

Differentiating and setting the result to zero,

\begin{align*}\frac{dT}{dV} &= -m_1(v_1-V)-m_2(v_2-V)\\ 0 &= -m_1v_1-m_2v_2+(m_1+m_2)V\end{align*}

V = \dfrac{m_1v_1+m_2v_2}{m_1+m_2}

(2)

Step 3Identify

V

as the center of mass velocity

The right-hand side of (2) is exactly the definition of the velocity of the system’s center of mass, so the frame of minimum kinetic energy is the center-of-mass frame. Expanding the total momentum in that frame and substituting (2) shows it vanishes identically.

m_1(v_1-V)+m_2(v_2-V) = m_1v_1+m_2v_2-(m_1+m_2)V

m_1(v_1-V)+m_2(v_2-V) = 0

(3)

Step 4The velocities in the center-of-mass frame

Substituting (2) into $v_1-V$ and $v_2-V$ ,

\begin{aligned} v_1-V &= v_1-\frac{m_1v_1+m_2v_2}{m_1+m_2} = \frac{m_2(v_1-v_2)}{m_1+m_2}\\ v_2-V &= v_2-\frac{m_1v_1+m_2v_2}{m_1+m_2} = \frac{-m_1(v_1-v_2)}{m_1+m_2} \end{aligned} \tag{4}

Step 5Assemble the minimum kinetic energy

Squaring both velocities (4) and inserting them into (1) collects a common factor.

\begin{aligned} T_{min} &= \frac12 m_1\left(\frac{m_2(v_1-v_2)}{m_1+m_2}\right)^2+\frac12 m_2\left(\frac{m_1(v_1-v_2)}{m_1+m_2}\right)^2\\ &= \frac{(v_1-v_2)^2}{2(m_1+m_2)^2}\Bigl(m_1m_2^2+m_2m_1^2\Bigr)\\ &= \frac{(v_1-v_2)^2}{2(m_1+m_2)^2}\,m_1m_2(m_1+m_2) \end{aligned} \tag{5}

T_{min} = \dfrac{\mu(v_1-v_2)^2}{2}, \qquad \mu = \dfrac{m_1m_2}{m_1+m_2}

(6)

Check

If $m_1=m_2=m$ , $V=(v_1+v_2)/2$ and $\mu=m/2$ , giving $T_{min}=m(v_1-v_2)^2/4$ , the familiar equal-mass center-of-mass kinetic energy. If $v_1=v_2$ the two particles already share one velocity, $T_{min}\to0$ correctly: there is nothing left to average away.

Remark

$T(V)-T_{min}=(m_1+m_2)(V-V_{cm})^2/2$ for any frame, not only $K$ itself (expand (1) around $V_{cm}$ to see the cross term vanish by (3)): the kinetic energy in any frame splits exactly into $T_{min}$ plus the kinetic energy the whole system would have if concentrated at its center of mass and moving at that frame’s relative speed, the two-particle instance of the same theorem used for continuous systems.