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Use coordinates $s$ down the inclined plane and $n$ along its upward normal, with the launch point as origin. Gravity has components $g\sin \varphi$ and $-g\cos \varphi$. For launch angle $\theta$ above the downhill direction,The nonzero root of $n(t)=0$ is $t_f=2v_0\sin \theta /(g\cos \varphi )$. Substituting into $s(t)$ and combining the two trigonometric terms gives the distance along the slope:This assumes $0\le \varphi <\pi /2$, $\sin \theta >0$, and a downhill landing with $\cos (\theta -\varphi )>0$, as well as the neglected air resistance. It reduces to the horizontal projectile range $v_0^2\sin (2\theta )/g$ at $\varphi =0$. The angle relative to the horizontal is $\theta -\varphi$, which explains the shifted cosine. The original diagram uses stipulated illustrative angles; no examination figure is being reconstructed.

Let the initial velocity be $\mathbf{v}$ and the two final velocities be $\mathbf{u},\mathbf{w}$. In the stated nonrelativistic elastic collision, conservation of momentum and conservation of energy giveSquaring the first equation and comparing it with the second yieldsIf both final velocities are nonzero, their directions are perpendicular. If one is zero, the moving particle retains the initial velocity and a scattering angle for the stationary particle is undefined. This includes the zero-deflection and head-on exchange limits rather than incorrectly assigning a direction to a zero vector. Equal masses and a stationary initial target are essential to the dot-product cancellation.

Measure the vertical displacement $x$ downward from the spring's unstretched position. The kinetic energy is $m{\dot{x}}^2/2$, while the spring and gravitational potential energies are $k{x}^2/2$ and $-mgx$. Thus the Lagrangian isApplying the Euler-Lagrange equation, $d(\partial \mathcal{L}/\partial \dot{x})/dt-\partial \mathcal{L}/\partial x=0$, givesGravity shifts the equilibrium to $x_{\ast }=mg/k$. For displacement $\xi =x-x_{\ast }$ from that equilibrium, the equation of motion becomes $\ddot{\xi }+(k/m)\xi =0$, giving simple harmonic motion with angular frequency $\omega =\sqrt{k/m}$. Choosing an upward coordinate instead reverses the gravitational term, but leaves the equilibrium-relative angular frequency unchanged.

Let the source's proper emission period be $T_0$, and let it approach a stationary observer directly with speed $v=\beta c>0$. In the observer's inertial frame, consecutive emissions are separated by $\gamma T_0$ through time dilation. The second crest starts $v\gamma T_0$ nearer the observer, reducing the arrival interval toSince light travels at $c$ in each inertial frame, the longitudinal relativistic Doppler effect isRecession reverses the sign of $\beta$. More generally, if $\Theta$ is the angle between the source velocity and the photon propagation direction in the observer's frame, a Lorentz transformation of the photon four-momentum gives ${\nu }_0=\gamma {\nu }_{\mathrm{obs}}(1-\beta \cos \Theta )$.

The common Doppler scaling of spectral-line separations follows because every line receives the same wavelength factor $D$. The supplied wavelengths have rest separation $50\mathrm{nm}$; the measured separation gives $D=51/50=1.02$. Defining ${\beta }_{\mathrm{rec}}>0$ for recession, inversion givesSource contradiction: a strictly approaching source has $D<1$ and must reduce the wavelength separation below $50\mathrm{nm}$, so no positive approach speed satisfies the stated $51\mathrm{nm}$. For example, approach at $0.02c$ gives a separation of $49.0098\mathrm{nm}$. The speed inferred from the numerical data is therefore about $0.0198c$ receding; retaining the word “approaching” would require a negative signed approach velocity. The two observed wavelengths consistent with the given separation are $357$ and $408\mathrm{nm}$.

For the attractive central force $\mathbf{F}=-k\mathbf{r}$, with $k>0$, the potential energy is $U(r)=k{r}^2/2$ up to an additive constant. Conservation of angular momentum confines the motion to a plane; writing its magnitude as $J=m{r}^2\dot{\phi }$, the kinetic energy isAt fixed $J$, the radial energy equation becomes $E=m{\dot{r}}^2/2+U_{\mathrm{eff}}(r)$, whereThe effective potential combines the true potential energy with the angular-motion contribution. Its negative derivative gives the radial equation $m\ddot{r}=-U_{\mathrm{eff}}^{\prime }=J^2/(m{r}^3)-kr$; the first term is the centrifugal contribution arising from elimination of the angular coordinate.

A circular orbit is a fixed radial position at a stationary point:For $J\ne 0$, $U_{\mathrm{eff}}^{\prime \prime }(r_0)=4k>0$, so this is a stable minimum. Small radial disturbances have angular frequency $2\sqrt{k/m}$. For $J=0$, there is no nonzero-radius circular orbit.

The parallel axis theorem relates the moment of inertia about an axis through the centre of mass to that about a parallel axis displaced by perpendicular distance $d$:The reference axis must pass through the centre of mass. To see why, write the displacement from it as $\mathbf{r}=\mathbf{R}+\mathbf{d}$ in the plane perpendicular to the axes. Expanding $\int |\mathbf{R}+\mathbf{d}{|}^{2}dm$ gives $I_G+M{d}^2$ because $\int \mathbf{R}dm=0$.

For a thin planar mass distribution, let $x,y$ be perpendicular axes in its plane and $z$ the normal axis through their same intersection. The perpendicular axis theorem statesIndeed, $I_z=\int (x^2+y^2)dm$, while $I_x=\int y^{2}dm$ and $I_y=\int x^{2}dm$. Their common intersection need not be the centre of mass, but an arbitrary three-dimensional body does not satisfy the planar equality.