Irodov 1.173: The Forty Percent That Vanished – Amit & Deepali Sinha Foundation

Problem Irodov 1.173

A particle of mass $m$ , moving with some initial speed, collides with a stationary particle of mass $M$ . After the collision, particle $m$ has turned through $\pi/2$ (a right angle) from its original direction of motion, while particle $M$ moves off at an angle $\theta = 30^\circ$ to that original direction. Given that $M/m = 5.0$ , find by what percentage the total kinetic energy of the two-particle system changed as a result of the collision, and state whether it increased or decreased.

Momentum diagram: the final momenta of m (deflected 90°) and M (recoiling at θ=30° on the other side of the line of motion) compose to the initial momentum m v, drawn to the actual proportions for M/m=5.0. — Momentum diagram: the final momenta of $m$ (deflected $90^\circ$ ) and $M$ (recoiling at $\theta=30^\circ$ on the other side of the line of motion) compose to the initial momentum $m\vect v$ , drawn to the actual proportions for $M/m=5.0$ .

Given

$M/m = 5.0$
$m$ deviates by $\pi/2$
$\theta = 30^\circ$ (recoil of $M$ )

Find

Find

\Delta T/T

, the percentage change in the system’s kinetic energy, and state whether it increases or decreases.

Solution

Step 1Momentum conservation along and across the line of motion

Let $v$ be the initial speed of $m$ , and $v_1$ , $v_2$ the final speeds of $m$ and $M$ , with the initial direction of $m$ as the $x$ axis, $m$ deflected to $+y$ (a $\pi/2$ turn) and $M$ recoiling on the $-y$ side at $\theta$ to $x$ , since the two transverse momenta must cancel. Momentum conservation along each axis reads

Mv_2\cos\theta = mv, \qquad mv_1 = Mv_2\sin\theta.

Step 2Solving for the two final speeds

Dividing the transverse equation by the along-axis one,

\frac{v_1}{v} = \tan\theta \quad\Longrightarrow\quad v_1 = v\tan\theta.

Substituting back into the along-axis equation,

Mv_2\cos\theta = mv \quad\Longrightarrow\quad v_2 = \frac{mv}{M\cos\theta}.

v_1 = v\tan\theta, \qquad v_2 = \dfrac{mv}{M\cos\theta}

(1)

Step 3Kinetic energy before and after

The initial kinetic energy is $T=mv^2/2$ . Using (1), the final kinetic energy is

T' = \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2 = \frac{1}{2}mv^2\tan^2\theta + \frac{1}{2}M\cdot\frac{m^2v^2}{M^2\cos^2\theta} = \frac{1}{2}mv^2\left(\tan^2\theta+\frac{m}{M}\sec^2\theta\right).

Step 4The fractional change

Dividing by $T=mv^2/2$ and using $\sec^2\theta=1+\tan^2\theta$ ,

\frac{T'}{T} = \tan^2\theta+\frac{m}{M}\left(1+\tan^2\theta\right) = \left(1+\frac{m}{M}\right)\tan^2\theta+\frac{m}{M}.

Subtracting 1,

\dfrac{\Delta T}{T} = \left(1+\dfrac{m}{M}\right)\tan^2\theta+\dfrac{m}{M}-1

(2)

Step 5Numerical value

With $M/m=5.0$ , so $m/M=0.20$ , and $\theta=30^\circ$ , so $\tan^2\theta = 1/3$ ,

\frac{\Delta T}{T} = (1+0.20)\cdot\frac{1}{3}+0.20-1 = 0.40+0.20-1 = -0.40.

\Delta T/T = -40\%

(3)

Check

Setting $\Delta T/T=0$ in the boxed formula gives the elastic threshold $\tan^2\theta_{\text{el}} = (M-m)/(M+m)$ ; for $M/m=5.0$ this is $2/3$ , well above the actual $\tan^2 30^\circ=1/3$ . Since $\Delta T/T$ increases with $\tan^2\theta$ , an actual angle smaller than the elastic one must give $\Delta T/T<0$ , confirming the loss found above. (For $m=M$ the threshold collapses to $\theta_{\text{el}}=0$ , the equal-mass elastic case, where the striking particle stops dead, $v_1=v\tan\theta_{\text{el}}=0$ , and its stationary partner carries off all the momentum, recoiling straight along the original line.)

Remark

Given the mass ratio, the two scattering angles alone fix whether the collision lost energy, without any speed or energy measurement: the checkline’s threshold $\tan^2\theta_{\text{el}}=(M-m)/(M+m)$ depends only on the mass ratio, so measuring $90^\circ$ and $\theta$ , once the mass ratio is known, is enough to tell an elastic collision from an inelastic one before any energy is measured directly, which is exactly how real particle detectors infer inelasticity from tracks plus known particle masses.