Irodov 1.172: The Ordering Rule Picks the Root – Amit & Deepali Sinha Foundation

Problem Irodov 1.172

Particle 1, of mass $m$ and speed $v = 10$ m/s, strikes head on a stationary particle 2 of the same mass $m$ . The collision is not perfectly elastic: the system’s total kinetic energy decreases by a fraction $\eta = 1.0\%$ of its initial value. Find the speed of particle 1 after the collision, and state whether it continues in its original direction or reverses.

The two identical discs immediately before the head on impact, particle 1 approaching at v with particle 2 at rest, and immediately after, both moving off at v_1' and v_2'. The arrow lengths are illustrative. — The two identical discs immediately before the head on impact, particle 1 approaching at $\vect v$ with particle 2 at rest, and immediately after, both moving off at $\vect v_1'$ and $\vect v_2'$ . The arrow lengths are illustrative.

Given

$v = 10\text{ m/s}$
$\eta = 1.0\% = 0.010$
$m_1=m_2=m$

Find

v_1'

, and its direction

Solution

Step 1Momentum ties the two final speeds together

The system is closed along the line of impact, so the total momentum immediately after the collision equals $mv$ , the momentum immediately before it. With both masses equal to $m$ , this fixes one linear relation between the two unknown final speeds $v_1'$ and $v_2'$ .

mv = mv_1'+mv_2' \;\Longrightarrow\; v_1'+v_2' = v \tag{1}

Step 2The stated energy loss gives a second relation

The kinetic energy decreases by the fraction $\eta$ of its initial value $mv^2/2$ , so the final kinetic energy is $(1-\eta)$ times the initial one.

\begin{aligned} \frac12 mv_1'^2+\frac12 mv_2'^2 &= (1-\eta)\frac12 mv^2 \\ v_1'^2+v_2'^2 &= (1-\eta)v^2 \end{aligned} \tag{2}

Step 3Eliminating

v_2'

leaves a quadratic in

v_1'

Substituting $v_2' = v-v_1'$ from (1) into (2) and collecting terms in $v_1'$ produces a quadratic whose two roots are the two candidate values of $v_1'$ .

\begin{align*}v_1'^2+(v-v_1')^2 &= (1-\eta)v^2 \\ 2v_1'^2-2vv_1'+\eta v^2 &= 0 \\ v_1'^2-vv_1'+\frac{\eta}{2}v^2 &= 0\end{align*}

Applying the quadratic formula,

v_1' = \frac{v\pm v\sqrt{1-2\eta}}{2}. \tag{3}

Step 4Discarding the unphysical root

Selecting the minus sign in (3) and substituting the given numbers,

v_1' = \frac{10\text{ m/s}\times\bigl(1-\sqrt{1-0.020}\bigr)}{2}.

v_1' = \dfrac{v\bigl(1-\sqrt{1-2\eta}\bigr)}{2} = 5.0\times10^{-2}\text{ m/s}

(4)

Since $v_1'>0$ in the same sense as $v$ , particle 1 continues moving in its original direction, only much slower.

Check

For $\eta\ll1$ , expanding $\sqrt{1-2\eta}\approx1-\eta$ (binomial, first order) gives $v_1'\approx v\eta/2$ , matching the exact result’s leading behavior; numerically $v\eta/2 = 10\times0.010/2 = 0.050$ m/s, agreeing with the boxed value to two significant figures. The quadratic also requires $\eta\le1/2$ for a real root; the reason is momentum conservation, not merely that energy cannot go negative. With the total momentum $mv$ fixed and the two masses equal, the final kinetic energy can never fall below $(mv)^2/(2\cdot2m)=mv^2/4$ , attained exactly when the two particles move together at $v/2$ (a perfectly inelastic collision), so at most half of the initial kinetic energy $mv^2/2$ can ever be lost, and $\eta=1/2$ is precisely that limiting case.

Remark

The rejected root $v_1'=v(1+\sqrt{1-2\eta})/2$ solves the same momentum and energy equations equally validly; with $\eta$ fixed, those two conservation laws alone leave exactly the two discrete outcomes computed in (3), not a continuum, and it is the physical ordering constraint $v_1'\le v_2'$ , algebraic in form but carrying the physical fact that the trailing particle cannot pass through the one it struck, that picks the realizable root out of the two. Only when $\eta$ itself is left free, parametrized by a coefficient of restitution, does the outcome widen into a genuine one-parameter family.