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

Equal energies mean $dE=0$ in the linear approximation. Evaluate the coefficients at particle $A$, where $v_A=0.90c$, and use $\Delta v=0.01c$. The fractional rest mass change isConsequentlyAs a check beyond the requested first-order approximation, exact equality of relativistic energy gives $m_B/m_A=\sqrt{(1-0.91^2)/(1-0.90^2)}=\sqrt{1719/1900}\approx 0.95118$, consistent with the sign and approximate size of the difference.