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The parallel axis theorem states $I=I_G+M{d}^2$ for parallel axes separated by $d$, with the reference axis through the centre of mass. The perpendicular axis theorem states $I_z=I_x+I_y$ for a planar lamina in the $xy$ plane, about mutually perpendicular axes through one point. It does not apply directly to a three-dimensional solid.

For the transverse moment of inertia of a uniform solid cylinder, divide the cylinder into thin disks at axial positions $z\in [-h/2,h/2]$. Symmetry and the perpendicular axis theorem give the diametral moment of inertia $r^{2}dM/4$ of each disk; the parallel axis theorem adds $z^{2}dM$. HenceThe thin-disk limit is $M{r}^2/4$, while the slender-rod limit is $M{h}^2/12$.

Treat the carriage as translating along the circular track, neglecting wheel rotational kinetic energy and its finite size. At the top, gravity and the inward normal force supply the centripetal acceleration:For unilateral contact, $N\ge 0$, so the limiting speed is $v_{\mathrm{top}}^2=gr$. Conservation of energy from rest at height $h$ then givesThis top contact condition is the most restrictive: at angle $\varphi$ measured from the bottom, $N/(mg)=2h/r-2+3\cos \varphi$, whose minimum occurs at the top. A mechanically captive carriage able to sustain an outward rail reaction has a different constraint; the result here uses the usual requirement of positive normal contact.

Let $\mathbf{a},\mathbf{b}$ be the outgoing velocities. In the nonrelativistic elastic collision, equal masses giveby conservation of momentum and conservation of energy. Squaring the first relation and comparing with the second yieldsIf both particles move, their paths are perpendicular. If either velocity vanishes, its direction is undefined and the other particle carries all the incident momentum and kinetic energy. This conclusion depends on the stated $v\ll c$ approximation.

The Lorentz factor is $\gamma =E/(m_{\mu }{c}^2)=20000/106=188.6792$ and $\beta =\sqrt{1-{\gamma }^{-2}}$. Traversal of the height $H$ takes proper time $H/(\beta \gamma c)$. Using the exponential decay law with the given muon lifetime,The laboratory flight lasts about $83.39\mu \mathrm{s}$, while the elapsed proper time is about $0.4420\mu \mathrm{s}$. This is the probability of surviving decay to the surface, with atmospheric energy loss and detector efficiency neglected.