Irodov 1.175: Tangent to the Momentum Circle – Amit & Deepali Sinha Foundation

Problem Irodov 1.175

A particle of mass $m_1$ , moving with some speed $v$ , collides elastically with a stationary particle of mass $m_2$ , with $m_1>m_2$ . Over all possible elastic outcomes of this collision (every scattering angle the internal collision force can produce), find the largest angle by which the incident particle’s velocity direction can be deflected from its original direction.

The setup: particle 1, of mass m_1, approaches the stationary particle 2, of mass m_2<m_1, with velocity v. The collision is elastic, and how far it can turn particle 1 is what the solution finds. — The setup: particle 1, of mass $m_1$ , approaches the stationary particle 2, of mass $m_2<m_1$ , with velocity $\vect v$ . The collision is elastic, and how far it can turn particle 1 is what the solution finds.

Given

$m_1$
$m_2$
$m_1>m_2$

Find

\theta_{\max}

Solution

Step 1The center-of-mass velocity and particle 1’s speed relative to it

The center of mass moves at $\vect V=m_1\vect v/(m_1+m_2)$ (particle 2 starts at rest), and particle 1’s velocity in the C frame is its lab velocity minus $\vect V$ .

\begin{align*}\vect V &= \frac{m_1\vect v}{m_1+m_2} \\ \vect v_{1,C} &= \vect v-\vect V\end{align*}

\vect v_{1,C} = \dfrac{m_2\vect v}{m_1+m_2}

(1)

Step 2This C-frame speed is unchanged by an elastic collision

Write $\vect u_1$ and $\vect u_2$ for the two C-frame velocities before the collision, and $\vect u_1'$ and $\vect u_2'$ for the same two after it. The C frame is by definition the one in which the total momentum vanishes, and it is inertial throughout, so that vanishing holds immediately before the collision,

m_1\vect u_1+m_2\vect u_2 = \vect 0 \;\Longrightarrow\; \vect u_2 = -\frac{m_1}{m_2}\vect u_1, \tag{2}

and again immediately after it.

m_1\vect u_1'+m_2\vect u_2' = \vect 0 \;\Longrightarrow\; \vect u_2' = -\frac{m_1}{m_2}\vect u_1' \tag{3}

The collision is elastic, so the kinetic energy measured in this same frame is the same at the two instants. Before it that energy is

K_C = \frac12 m_1u_1^2+\frac12 m_2u_2^2, \tag{4}

and after it, in the primed speeds,

K_C' = \frac12 m_1u_1'^2+\frac12 m_2u_2'^2. \tag{5}

Setting $K_C'=K_C$ and eliminating $u_2$ with (2) and $u_2'$ with (3), the mass ratio collects into one bracket that is common to the two sides.

\begin{align*}\frac12 m_1u_1^2\Bigl(1+\frac{m_1}{m_2}\Bigr) &= \frac12 m_1u_1'^2\Bigl(1+\frac{m_1}{m_2}\Bigr) \\ u_1'^2 &= u_1^2\end{align*}

A speed cannot be negative, so $u_1'=u_1$ , and (3) carries the same conclusion to particle 2, $u_2'=u_2$ . An elastic collision therefore leaves both C-frame speeds exactly as they were and can only turn the pair. Particle 1’s C-frame speed after the collision is again the $v_{1,C}$ of (1), in some direction that the internal collision law leaves free.

Step 3The lab-frame velocity’s tip sweeps a circle

The lab-frame velocity after the collision is $\vect v_1' = \vect V+\vect v_{1,C}'$ , a fixed vector $\vect V$ plus a vector of the fixed length $v_{1,C}$ set by (1) and free direction. As the C-frame scattering angle ranges over every value, the tip of $\vect v_1'$ therefore sweeps the full circle of radius $v_{1,C}$ centered at the tip of $\vect V$ .

Step 4Maximizing the deflection: the tangent line

Drawn for an illustrative mass ratio. The dashed circle of radius v_1,C, centered at the tip of the center-of-mass velocity V, is the locus of every possible v_1'; the tangent from the origin marks θ_. — Drawn for an illustrative mass ratio. The dashed circle of radius $v_{1,C}$ , centered at the tip of the center-of-mass velocity $\vect V$ , is the locus of every possible $\vect v_1'$ ; the tangent from the origin marks $\theta_{\max}$ .

In the right triangle formed by the origin, the circle’s center (at distance $V$ ), and the tangent point (at the circle’s radius $v_{1,C}$ , meeting the radius at a right angle), the sine of the angle at the origin is the opposite side over the hypotenuse.

\sin\theta_{\max} = \frac{v_{1,C}}{V} = \frac{m_2v/(m_1+m_2)}{m_1v/(m_1+m_2)}

\sin\theta_{\max} = \dfrac{m_2}{m_1}

(6)

Check

The construction requires $v_{1,C}<V$ , i.e. $m_2<m_1$ , for the origin to lie outside the circle and a genuine tangent (a genuine maximum angle) to exist, matching exactly the condition $m_1>m_2$ given in the problem. In the limit $m_2\to m_1$ (from below), $\sin\theta_{\max}\to1$ , so $\theta_{\max}\to90^\circ$ , the supremum over every glancing elastic collision between equal masses (never more than a right-angle deflection).

Remark

When $m_1<m_2$ instead, the origin sits inside the circle, no ray from it is excluded, and every angle from $0$ to $180^\circ$ , including a full backward bounce, becomes reachable: this is why a light particle striking a heavy stationary one has no upper bound on its scattering angle, while a heavy particle striking a light one always does. The two regimes are the same construction; only whether the origin falls outside or inside the momentum circle changes between them.