= Solution
\b[The <magnetic rigidity> gives the <momentum>, the bending direction gives the sign of <electric charge>, and the calorimetric <energy> then gives the <mass>.] For planar tracks perpendicular to a uniform <magnetic field>, the <Lorentz force> and circular motion give
$$
p=|q|Br,\qquad
p\,[\mathrm{GeV}/c]=0.2998\,|q/e|\,B\,[\mathrm T]r\,[\mathrm m].
$$
Trace each track away from its production or <decay vertex>. With the <magnetic field> into the page, a positive track bends counterclockwise and a negative track bends clockwise. For example, a positive particle initially travelling left accelerates downwards. The known incoming negative <pion> fixes the sign convention experimentally. A radius alone does not distinguish <electric charge> signs.
For a calibrated total <energy> $E$, the <relativistic energy-momentum relation> gives
$$
\boxed{m^2c^4=E^2-p^2c^2.}
$$
If the calorimeter instead supplies only <kinetic energy> $T$, solve $(T+mc^2)^2=p^2c^2+m^2c^4$ to obtain
$$
\boxed{mc^2=\frac{p^2c^2-T^2}{2T}.}
$$
Compare with candidate <pion>, <kaon>, and <proton> <masses>, together with the <electric charge> sign and shower behavior. An <electron> makes an <electromagnetic shower>; a <muon> is penetrating; a <hadron> generally requires a hadronic calorimeter and its own calibration. If either <energy> escapes or the response is uncalibrated, simply substituting a deposited signal for $E$ need not identify the particle correctly. The absent figure prevents attaching identities to individual drawn tracks, but the requested identification procedure does not require its geometry.
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