Irodov 1.176: Two Simultaneous Contacts, One Symmetry

Problem Irodov 1.176

Three identical discs of equal mass, $A$ , $B$ and $C$ , lie on a smooth (frictionless) horizontal plane. Discs $B$ and $C$ are at rest, their centers a distance $\eta$ times each disc’s diameter $d$ apart. Disc $A$ , set moving with velocity $\vect v$ along the perpendicular bisector of the line joining $B$ ’s and $C$ ’s centers, strikes both simultaneously in one elastic collision. Find disc $A$ ’s velocity $v'$ after the collision, as a function of $v$ and $\eta$ , and determine the ranges of $\eta$ for which $A$ recoils, stops, or continues moving forward.

Redrawn from the book's figure: the three identical discs at the instant of simultaneous contact, disc A arriving along v on the perpendicular bisector of BC and touching B and C together. — Redrawn from the book’s figure: the three identical discs at the instant of simultaneous contact, disc $A$ arriving along $\vect v$ on the perpendicular bisector of $BC$ and touching $B$ and $C$ together.

Given

$v$
$\eta$ (with $1\le\eta\le2$ : $\eta\ge1$ since $B$ and $C$ must not overlap, $\eta\le2$ since $\sin\theta\le1$ )

Find

v'

(magnitude and sign along

\vect v

); the ranges of

\eta

for recoil, stop, continue

Solution

Step 1The contact angle from the geometry

The same instant as a triangle of centers: isosceles, with AB=AC=d and base BC=η d. The dashed axis of symmetry, along which v points, bisects the apex angle at A and meets BC square, fixing the half-angle θ. — The same instant as a triangle of centers: isosceles, with $AB=AC=d$ and base $BC=\eta d$ . The dashed axis of symmetry, along which $\vect v$ points, bisects the apex angle at $A$ and meets $BC$ square, fixing the half-angle $\theta$ .

Touching discs have centers one diameter apart, which is what fixes $AB=AC=d$ . Reading the right triangle above (the altitude from $A$ bisects the isosceles apex and lands at $BC$ ’s midpoint) gives $\sin\theta$ directly in terms of $\eta$ , and the identity $\cos^2\theta=1-\sin^2\theta$ converts it to the form the momentum and energy equations below will need.

\begin{aligned} \sin\theta &= \frac{\eta}{2} \\ \cos^2\theta &= 1-\frac{\eta^2}{4} \end{aligned} \tag{1}

Step 2Momentum and energy along the line of motion

The outgoing velocities this step establishes: B leaves at the angle θ above the axis and C at θ below it, at the same speed by symmetry, while A stays on the axis. — The outgoing velocities this step establishes: $B$ leaves at the angle $\theta$ above the axis and $C$ at $\theta$ below it, at the same speed by symmetry, while $A$ stays on the axis.

That same symmetry makes the two struck discs interchangeable, so their outgoing speeds are one unknown and not two.

|\vect v_B'| = |\vect v_C'| = u \tag{2}

Each of $B$ and $C$ leaves along its own line of centres, at the angle $\theta$ to the axis, so each carries $u\cos\theta$ of momentum along the axis and its full speed $u$ into the energy, while the two transverse components cancel against each other. With every mass equal to $m$ , momentum along the axis reads

mv = mv'+2mu\cos\theta \;\Longrightarrow\; v = v'+2u\cos\theta, \tag{3}

and the collision being elastic, kinetic energy reads

\frac12 mv^2 = \frac12 mv'^2+2\cdot\frac12 mu^2 \;\Longrightarrow\; v^2 = v'^2+2u^2. \tag{4}

Step 3Solving the pair for

u

Substituting $v'=v-2u\cos\theta$ from (3) into (4) and collecting terms in $u$ leaves a quadratic whose nonzero root is the common recoil speed $u$ of (2).

\begin{align*}v^2 &= (v-2u\cos\theta)^2+2u^2 \\ 0 &= -4uv\cos\theta+u^2\bigl(4\cos^2\theta+2\bigr) \\ 0 &= u\Bigl[-4v\cos\theta+u\bigl(4\cos^2\theta+2\bigr)\Bigr]\end{align*}

Discarding the trivial root $u=0$ (no collision at all) leaves

u = \frac{4v\cos\theta}{4\cos^2\theta+2} = \frac{2v\cos\theta}{2\cos^2\theta+1}. \tag{5}

Step 4Disc

A

’s velocity in terms of

\theta

, then

\eta

Substituting (5) back into (3) and simplifying over the common denominator gives $v'$ in terms of $\cos^2\theta$ alone; (1) then converts it to $\eta$ .

\begin{align*}v' &= v-2u\cos\theta = v-\frac{4v\cos^2\theta}{2\cos^2\theta+1} \\ &= \frac{v(1-2\cos^2\theta)}{2\cos^2\theta+1}\end{align*}

\begin{align*}1-2\cos^2\theta &= 1-2\Bigl(1-\frac{\eta^2}{4}\Bigr) = \frac{\eta^2-2}{2} \\ 2\cos^2\theta+1 &= 2\Bigl(1-\frac{\eta^2}{4}\Bigr)+1 = \frac{6-\eta^2}{2}\end{align*}

v' = -\dfrac{v(2-\eta^2)}{6-\eta^2}

(6)

Step 5The sign of

v'

across the allowed range of

\eta

On $1\le\eta\le2$ the denominator $6-\eta^2$ stays positive throughout, so the sign of $v'$ follows the sign of $-(2-\eta^2)=\eta^2-2$ alone. Disc $A$ recoils (moves backward, $v'<0$ ) for $\eta<\sqrt2$ , stops exactly ( $v'=0$ ) at $\eta=\sqrt2$ , and continues forward ( $v'>0$ ) for $\eta>\sqrt2$ .

Check

At the grazing limit $\eta\to2$ , $\theta\to90^\circ$ and $\cos\theta\to0$ : the lines of centers become perpendicular to $\vect v$ , so a contact impulse along either one can carry no component along $\vect v$ at all, and the boxed formula gives $v'\to-v(2-4)/(6-4)=v$ , disc $A$ passing through untouched, consistent with a contact that (in this limit) exerts no force along the direction of approach.

Remark

The whole problem collapses to two scalar unknowns, $v'$ and the common recoil speed $u$ , only because the mirror symmetry between $B$ and $C$ removes the sideways component of $A$ ’s motion before any equation is written down; a $B$ and $C$ placed asymmetrically about the line of $\vect v$ would need a transverse momentum equation as well, and disc $A$ would leave along a genuinely different line than the one it arrived on.