Irodov 1.174: Perpendicular Speeds Are a Right Triangle

Problem Irodov 1.174

Two particles of masses $m_1$ and $m_2$ form a closed (isolated) system. At some instant they move with velocities $\vect v_1$ and $\vect v_2$ that are mutually perpendicular. In the reference frame attached to the system’s center of inertia, find (a) the momentum of each particle, and (b) the system’s total kinetic energy.

The two velocities treated as two legs of a right angle at O: since v_1⊥ v_2, the segment closing the triangle has length (v_1^2+v_2^2)^1/2 by the Pythagorean theorem. That segment is exactly the relative velocity v_1- v_2 used below. — The two velocities treated as two legs of a right angle at $O$ : since $\vect v_1\perp\vect v_2$ , the segment closing the triangle has length $(v_1^2+v_2^2)^{1/2}$ by the Pythagorean theorem. That segment is exactly the relative velocity $\vect v_1-\vect v_2$ used below.

Given

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

Find

(a)

\vect p_1'

\vect p_2'

in the C frame, (b)

T'

in the C frame

Solution

Step 1The center of inertia’s velocity

The velocity of the center of inertia (C) is the mass-weighted average of the two particle velocities, by definition of the center of mass.

\vect V = \frac{m_1\vect v_1+m_2\vect v_2}{m_1+m_2} \tag{1}

Step 2Each particle’s momentum in the C frame

Particle 1’s momentum in the C frame is $m_1$ times its C-frame velocity $\vect v_1-\vect V$ .

\begin{align*}\vect p_1' &= m_1(\vect v_1-\vect V) \\ &= m_1\vect v_1-\frac{m_1(m_1\vect v_1+m_2\vect v_2)}{m_1+m_2} \\ &= \frac{m_1(m_1+m_2)\vect v_1-m_1^2\vect v_1-m_1m_2\vect v_2}{m_1+m_2} \\ &= \frac{m_1m_2(\vect v_1-\vect v_2)}{m_1+m_2}\end{align*}

Writing $\mu = m_1m_2/(m_1+m_2)$ for the reduced mass names this compactly, and the total momentum in the C frame is zero by definition of that frame, so particle 2’s momentum is the exact negative.

\vect p_1' = \mu(\vect v_1-\vect v_2), \qquad \vect p_2' = -\vect p_1'

Step 3Magnitude from the perpendicular geometry

Since $\vect v_1\perp\vect v_2$ , the figure’s right triangle gives the length of $\vect v_1-\vect v_2$ directly by the Pythagorean theorem, with no cross term to carry.

|\vect v_1-\vect v_2| = \bigl(v_1^2+v_2^2\bigr)^{1/2}

p_1' = p_2' = \mu\bigl(v_1^2+v_2^2\bigr)^{1/2}, \qquad \mu = \dfrac{m_1m_2}{m_1+m_2}

(2)

Step 4Total kinetic energy in the C frame

Each particle’s C-frame kinetic energy is its momentum squared over twice its own mass; adding the two and using $p_1'=p_2'=p$ from (2) combines the two mass denominators into $1/\mu$ .

\begin{align*}T' &= \frac{p_1'^2}{2m_1}+\frac{p_2'^2}{2m_2} \\ &= \frac{p^2}{2}\left(\frac{1}{m_1}+\frac{1}{m_2}\right) \\ &= \frac{p^2}{2}\cdot\frac{m_1+m_2}{m_1m_2} = \frac{p^2}{2\mu}\end{align*}

Substituting the boxed momentum magnitude (2) gives the final result.

T' = \frac{\mu^2(v_1^2+v_2^2)}{2\mu}

T' = \dfrac{\mu(v_1^2+v_2^2)}{2}

(3)

Check

If $m_2\to0$ with $v_2$ held fixed, $\mu\to m_2\to0$ , so both $p'$ and $T'$ vanish: a massless second particle carries no momentum or energy in any frame, as it must. Also, relabelling the two particles ( $1\leftrightarrow2$ , so $m_1\leftrightarrow m_2$ and $v_1\leftrightarrow v_2$ ) leaves $\mu$ and $v_1^2+v_2^2$ both unchanged, so the boxed $p_1'=p_2'$ and $T'$ correctly do not depend on which particle is called “1,” consistent with the C-frame momenta being equal and opposite by construction.

Remark

$T'$ here is precisely the system’s kinetic energy of relative motion, $\mu v_{\mathrm{rel}}^2/2$ with $v_{\mathrm{rel}}=|\vect v_1-\vect v_2|$ , which is always true in the C frame for any two-particle system, perpendicular or not: the C-frame kinetic energy never depends on $\vect v_1$ and $\vect v_2$ separately, only on their difference. The right angle given here only simplifies the length of that difference to $(v_1^2+v_2^2)^{1/2}$ ; a different angle between $\vect v_1$ and $\vect v_2$ would change that length by a law-of-cosines term without changing the underlying $\mu v_{\mathrm{rel}}^2/2$ structure.