 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/14/a/ii/solution

d E = 0

in the linear approximation. Evaluate the coefficients at particle

A

, where

v_{A} = 0.90 c

, and use

Δ v = 0.01 c

. The fractional rest mass change is

\frac{m _{B} - m _{A}}{m _{A}} \approx - \frac{v _{A} Δ v}{c ^{2} - v _{A}^{2}} = - \frac{0.90 \times 0.01}{1 - 0.9 0 ^{2}} = - \frac{9}{190} .

Consequently

m_{A} - m_{B} \approx \frac{9}{190} m_{A} \approx 0.0474 m_{A}, A has the larger rest mass .

As a check beyond the requested first-order approximation, exact equality of relativistic energy gives

m_{B} / m_{A} = (1 - 0.9 1^{2}) / (1 - 0.9 0^{2}) = 1719/1900 \approx 0.95118

, consistent with the sign and approximate size of the difference.

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