= Solution
The incident particle obeys the <energy–momentum relation> $E^2=p^2c^2+m_1^2c^4$. The stationary target contributes <energy> $m_2c^2$ and zero <momentum>. A single product inherits the total <four-momentum>, so its <invariant mass> satisfies
$$
M^2c^4=(E+m_2c^2)^2-p^2c^2=m_1^2c^4+m_2^2c^4+2Em_2c^2.
$$
Hence
$$
\boxed{M=\sqrt{m_1^2+m_2^2+\frac{2m_2E}{c^2}}.}
$$
Here $E$ is the incident total <energy>, including rest <energy>. In a one-particle final state, exact production of a specified <mass> fixes the incident <energy>; there is no adjustable final relative motion to absorb excess <energy>. A resonance with a finite <decay width> relaxes that ideal exact equality over its line shape.
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