Irodov 1.169: Elastic Collision Reveals the Mass Ratio

Problem Irodov 1.169

A particle of mass $m_1$ , moving with speed $v_0$ , undergoes a perfectly elastic collision with a stationary particle of mass $m_2$ . Determine the mass ratio $m_1/m_2$ in each of two separate cases: (a) the collision is head-on and, afterward, the two particles move in opposite directions with equal speeds; (b) the two particles fly apart symmetrically about the original direction of particle 1’s motion, the angle between their two paths being $\theta=60^\circ$ .

Case (a): a head-on elastic collision. Before, particle 1 approaches particle 2 at rest; after, they move apart along the same line in opposite directions with equal speeds.

Given

$m_1$ , $m_2$ , elastic collision, particle 2 initially at rest
(a) head-on, equal and opposite outgoing speeds
(b) symmetric divergence, $\theta=60^\circ$

Find

(a)

m_1/m_2

, (b)

m_1/m_2

Solution

Step 1Case (a): head-on elastic collision formulas

With $v_0$ the incoming speed of particle 1 and $v_1'$ , $v_2'$ the signed outgoing velocities along the same line, momentum and kinetic energy are both conserved.

\begin{align*}m_1v_0 &= m_1v_1'+m_2v_2' \\ m_1v_0^2 &= m_1v_1'^2+m_2v_2'^2\end{align*}

Moving the $m_1$ terms across in each line and factoring the difference of squares in the second,

\begin{align*}m_1(v_0-v_1') &= m_2v_2' \\ m_1(v_0-v_1')(v_0+v_1') &= m_2v_2'^2\end{align*}

Dividing the second line by the first (legitimate as long as a collision actually occurs, so that $v_2'\neq0$ ) cancels the shared factor.

v_0+v_1' = v_2'

Substituting that back into the momentum equation and collecting the $v_1'$ terms,

\begin{align*}m_1v_0 &= m_1v_1'+m_2(v_0+v_1') \\ m_1v_0-m_2v_0 &= m_1v_1'+m_2v_1' \\ v_0(m_1-m_2) &= v_1'(m_1+m_2)\end{align*}

v_1' = \dfrac{m_1-m_2}{m_1+m_2}\,v_0

(1)

The relation $v_2'=v_0+v_1'$ then gives the target’s outgoing velocity.

\begin{align*}v_2' &= v_0+\frac{m_1-m_2}{m_1+m_2}v_0 \\ &= \frac{(m_1+m_2)+(m_1-m_2)}{m_1+m_2}v_0\end{align*}

v_2' = \dfrac{2m_1}{m_1+m_2}\,v_0

(2)

Step 2Impose equal and opposite outgoing speeds

The two particles fly apart in opposite directions with equal speeds, so $v_1'=-v_2'$ . Substituting (1) and (2),

\frac{m_1-m_2}{m_1+m_2}v_0 = -\frac{2m_1}{m_1+m_2}v_0

Since $v_0\neq0$ and $m_1+m_2\neq0$ , both may be canceled, and the masses collect on one side.

\begin{align*}m_1-m_2 &= -2m_1 \\ 3m_1 &= m_2\end{align*}

\dfrac{m_1}{m_2} = \dfrac{1}{3}

(3)

Step 3Case (b): symmetric divergence at angle

\theta

Case (b): before, particle 1 approaches particle 2 at rest; after, particle 2 recoils to the side its center lies on and particle 1 continues on the other side, with the shorter arrow. The divergence angle appears in the next figure.

The two outgoing directions on their own, taken from the collision point: particle 2 leaves at θ/2 above particle 1's original line of motion (dashed) and particle 1 at θ/2 below it, so the angle between the two outgoing paths is θ. — The two outgoing directions on their own, taken from the collision point: particle 2 leaves at $\theta/2$ above particle 1’s original line of motion (dashed) and particle 1 at $\theta/2$ below it, so the angle between the two outgoing paths is $\theta$ .

Momentum conservation, resolved along the original direction and transverse to it,

\begin{align*}m_1 v_0 &= m_1 v_1\cos(\theta/2)+m_2 v_2\cos(\theta/2) \\ 0 &= m_1 v_1\sin(\theta/2)-m_2 v_2\sin(\theta/2).\end{align*}

The transverse equation says the two transverse components cancel, so the two outgoing momenta have the same magnitude; name that common magnitude $p$ .

p = m_1v_1 = m_2v_2

(4)

Substituting (4) into the along-axis equation, whose two terms then become equal,

m_1 v_0 = 2p\cos(\theta/2)

and solving for $p$ ,

p = \dfrac{m_1 v_0}{2\cos(\theta/2)}

(5)

Step 4Close with energy conservation

Kinetic energy conservation, written with $v_1=p/m_1$ and $v_2=p/m_2$ from (4),

\begin{align*}\frac{1}{2}m_1v_0^2 &= \frac{1}{2}\frac{p^2}{m_1}+\frac{1}{2}\frac{p^2}{m_2} \\ m_1v_0^2 &= p^2\left(\frac{1}{m_1}+\frac{1}{m_2}\right).\end{align*}

Substituting (5) and simplifying,

\begin{align*}m_1v_0^2 &= \frac{m_1^2v_0^2}{4\cos^2(\theta/2)}\cdot\frac{m_1+m_2}{m_1m_2} \\ 1 &= \frac{m_1+m_2}{4m_2\cos^2(\theta/2)} \\ 4m_2\cos^2(\theta/2) &= m_1+m_2.\end{align*}

Solving for the mass ratio,

\frac{m_1}{m_2} = 4\cos^2(\theta/2)-1.

Using $4\cos^2(\theta/2)-2=2\cos\theta$ and evaluating at $\theta=60^\circ$ ,

\dfrac{m_1}{m_2} = 1+2\cos\theta = 2.0

(6)

Check

In case (a), $m_1/m_2=1/3<1$ correctly makes particle 1 the lighter body, the only way a target can send the incoming particle straight back ((1) is negative exactly when $m_1<m_2$ ). In case (b), the limit $\theta\to0$ gives $m_1/m_2=1+2\cos0^\circ=3$ , the largest ratio the symmetric formula allows, since $\theta\to0$ is as close as the symmetric geometry can come to a head-on hit without leaving it. The other end of the range is not a second head-on limit but a singularity: (5) blows up as $\theta\to180^\circ$ (where $\cos(\theta/2)\to0$ ), so the symmetric formula is valid only on the open range $0<\theta<180^\circ$ and never actually reaches case (a)’s separate, genuinely head-on geometry.

Remark

Case (b)’s figure shows particle 2 recoiling faster than particle 1 continues ( $v_2'>v_1'$ , since $m_1>m_2$ here), and (4) shows why: the transverse-momentum equation alone forces the two outgoing momenta to be equal in magnitude for any symmetric divergence, so whichever particle is lighter must carry the larger speed to keep the two momenta equal in magnitude, a purely geometric consequence of the equal-and-opposite transverse components. This is independent of case (a), where the two outgoing speeds are equal by hypothesis instead, so the momenta there scale with the masses directly rather than inversely.