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Put $M=m_1+m_2$ and let $\mathbf{R}=(m_1{\mathbf{r}}_1+m_2{\mathbf{r}}_2)/M$ be the center of mass. Applying Newton's second law separately to the two point masses givesThus the center of mass has constant velocity, $\mathbf{R}(t)=\mathbf{R}(0)+t\dot{\mathbf{R}}(0)$. For the relative position $\mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2$, subtraction givesHere $\mu$ is the reduced mass; it converts the relative motion into a one-body equation of motion.

For the given linear force, the relative position obeys simple harmonic motion, $\ddot{\mathbf{r}}+{\Omega }^2\mathbf{r}=0$, where ${\Omega }^2=k/\mu$. The initial rest condition givesFor $d>0$, collision first occurs when the relative position vanishes, namely when $\Omega t=\pi /2$. ThereforeThe separation $d$ cancels because the frequency of this simple harmonic motion is independent of its amplitude.

For any vector $\mathbf{a}$, differentiating its components and the rotating basis gives the rotating-frame derivative formulaApply it twice to the position $\mathbf{x}$, with constant angular velocity $\mathbf{\omega }$. The inertial acceleration iswhere dots denote rotating-frame derivatives. Newton's second law therefore becomesThe last terms are the Coriolis force and the force corresponding to centrifugal acceleration.

For the particle on a uniformly rotating inclined plane, choose orthonormal vectors ${\mathbf{e}}_1$ along the horizontal $\xi$ axis and ${\mathbf{e}}_2$ up the slope, and set ${\mathbf{e}}_3={\mathbf{e}}_1\times {\mathbf{e}}_2$, the upward normal vector. Then ${\mathbf{e}}_z=\sin \theta {\mathbf{e}}_2+\cos \theta {\mathbf{e}}_3$ andHere $N$ is the normal reaction. The marble is modeled as the smooth sliding particle specified by the printed equations, without an additional rolling constraint. The rotating-frame Coriolis acceleration and centrifugal acceleration are respectivelyThe normal rotating-frame acceleration is zero, so the normal reaction is not generally $mg\cos \theta$; it satisfiesProjecting the same equation onto the two tangential axes gives exactly

Multiply the equations by $\dot{\xi }$ and $\dot{\eta }$ and add. The Coriolis force terms cancel, as expected because this force does no work relative to the rotating frame. The normal reaction also does no work along the plane. The conserved quantity is rotating-frame kinetic energy plus gravitational and centrifugal potential energy:This need not be the conserved inertial energy, since the externally driven rotating plane can exchange energy with the particle.

For conservative forcing in averaged oscillator equations, $f=f(y)$, the average determining $R_T$ vanishes. If $F^{\prime }=f$, thenwhose period average is zero. Thus $R=R_0$ to this order, while a position-dependent force generally shifts the oscillation phase and frequency. The exact equation also conserves the energy $\frac{1}{2}{\dot{y}}^2+\frac{1}{2}{y}^2-\epsilon F(y)$, consistent with the absence of amplitude drift for bounded orbits.

For the cubic position force, $⟨f\cos \varphi ⟩=R^3⟨{\cos }^4\varphi ⟩=3R^3/8$. ThereforeThe leading oscillation is $R_0\cos [(1-3\epsilon R_0^2/8)t+{\theta }_0]$. For positive $\epsilon$ it is a softening cubic oscillator, obtained by weakly perturbing a harmonic oscillator. The weak-force condition is $\epsilon R_0^2\ll 1$; the constant $R$ is the leading harmonic amplitude, not a claim that the exact waveform remains sinusoidal.

Introduce the slow time $T=\epsilon t$ and treat $t,T$ independently in the method of multiple scales. WriteAt the next order,For fixed $R$ define the period average byHere $f$ inside the averages is evaluated at the leading position and velocity $R\cos \varphi ,-R\sin \varphi$. Orthogonality to both fundamental harmonics is the solvability condition in the method of multiple scales: it removes the resonant forcing that would otherwise generate a secular term. Since the squared sine and cosine averages are $1/2$, the amplitude-phase equations areEquivalently $\dot{R}=-\epsilon ⟨f\sin (t+\theta )⟩$ and $\dot{\theta }=-\epsilon ⟨f\cos (t+\theta )⟩/R$ at first order. These equations describe a bounded, weakly perturbed oscillation over $t=O({\epsilon }^{-1})$ while the amplitude remains in a range where the expansion is ordered. At zero amplitude the oscillation phase coordinate is singular; Cartesian harmonic coefficients or an equilibrium analysis should replace it.

Use an Earth-centred rotating reference frame, with Earth's constant angular velocity vector $\mathbf{\omega }$. The vectors $\mathbf{r}$, $\mathbf{v}$, and $\mathbf{a}$ are the particle's position, velocity, and acceleration measured in that rotating frame. The vector $\mathbf{g}$ is the gravitational acceleration due to the actual gravitational force, expressed in the same axes; it excludes the separate centrifugal acceleration. The remaining term $-2\mathbf{\omega }\times \mathbf{v}$ is the Coriolis acceleration. A constant rotation axis and an unaccelerated Earth-centred origin account for the absence of an Euler term or a translational-origin term.

For the size estimates, Earth's angular speed isWith drop height $H=20\mathrm{m}$, unperturbed free fall gives the fall timeand final speed $gT=20{\mathrm{m}\mathrm{s}}^{-1}$. Thus the gravitational term has size $10{\mathrm{m}\mathrm{s}}^{-2}$, the Coriolis acceleration is at most about $2\omega gT\approx 2.8\times 10^{-3}{\mathrm{m}\mathrm{s}}^{-2}$, and the centrifugal acceleration is at most ${\omega }^2{R}_0\approx 2.9\times 10^{-2}{\mathrm{m}\mathrm{s}}^{-2}$. These are characteristic sizes $10^1$, $10^{-3}$, and $10^{-2}$ in acceleration units; geometric factors can reduce them, and the Coriolis term is zero at release.

For a suitable approximation to the rotational deflection of a falling body, take $\mathbf{g}$ constant over the short fall, start at rest relative to Earth, and use the zeroth-order velocity ${\mathbf{v}}^{(0)}(t)=\mathbf{g}t$ inside the small Coriolis term. In the centrifugal term replace $\mathbf{r}$ by ${\mathbf{R}}_0$, because $H/R_0\ll 1$. Integrating the resulting acceleration twice, with initial position ${\mathbf{r}}_{\mathrm{i}}$, givesSince ${\mathbf{r}}_{\mathrm{i}}+\mathbf{g}{T}^2/2={\mathbf{R}}_0$, evaluating at the unperturbed fall time yields the printed first-iterate estimate

The approximation keeps the leading Coriolis contribution and the leading centrifugal contribution enhanced by Earth's large radius. Feeding the Coriolis velocity correction back into the equation produces terms of size ${\omega }^{2}g{T}^4$, smaller than the retained centrifugal displacement by a factor of order $H/R_0$. Replacing $\mathbf{r}$ by ${\mathbf{R}}_0$ in that term has the same relative error. Here $\omega T\approx 1.4\times 10^{-4}$ and $H/R_0\approx 3.3\times 10^{-6}$, so both approximations are well controlled. Strictly, the displayed vector is the position at time $T$: its small upward component is removed by the slightly longer actual fall time. That arrival-time correction changes the retained horizontal displacements only at higher order. Thus the horizontal components of the printed estimate give the landing point on the locally flat surface; its vertical component should not be interpreted as a landing above the ground.

Let ${\hat{\mathbf{e}}}_E,{\hat{\mathbf{e}}}_N,{\hat{\mathbf{e}}}_U$ point east, north, and vertically upwards. They form a right-handed basis, with ${\hat{\mathbf{e}}}_N\times {\hat{\mathbf{e}}}_U={\hat{\mathbf{e}}}_E$. At latitude $\lambda$, in the spherical-Earth approximation,The required cross products areAt $\lambda =45^{\circ }$, the northerly and easterly displacements are thereforeThe negative northerly component means a southward displacement. With the supplied values, these are approximately $2.9\mathrm{cm}$ south and $1.3\mathrm{mm}$ east. The eastward effect comes from Coriolis acceleration, while the leading southward effect here comes from centrifugal acceleration.

The rotating-frame derivative formula is $(d\mathbf{A}/dt{)}_S=(d\mathbf{A}/dt{)}_{S^{\prime }}+\mathbf{\omega }\times \mathbf{A}$. Applying it twice to the position vector, with constant angular velocity, givesSolving for ${\mathbf{a}}_{S^{\prime }}$ identifies the displayed terms: its left side is the acceleration relative to the rotating reference frame, the first right-hand term is the inertial acceleration, $-2\mathbf{\omega }\times {\mathbf{v}}_{S^{\prime }}$ is the Coriolis acceleration, and $-\mathbf{\omega }\times (\mathbf{\omega }\times \mathbf{r})$ is the outward centrifugal acceleration.

For a particle of mass $m$, replace ${\mathbf{a}}_S$ by the actual force divided by $m$ using Newton's second law. The remaining terms then appear as apparent forces in $S^{\prime }$. There is no Euler acceleration because $\mathbf{\omega }$ is constant and no translational acceleration term because the common origin is fixed in the inertial frame.

Assume the particle masses $m_i$ are fixed and their total mass $M={\sum }_i{m}_i$ is positive. The centre of mass and the relative position vectors areTheir weighted sum is zero, so differentiation gives ${\sum }_i{m}_i{\dot{\mathbf{y}}}_i=0$. Expanding the kinetic energy with ${\dot{\mathbf{x}}}_i=\dot{\mathbf{R}}+{\dot{\mathbf{y}}}_i$ makes the mixed term vanish:This kinetic-energy decomposition about the centre of mass separates translation from internal motion without assuming any particular force law.

For the rigid body, its relative velocity is ${\dot{\mathbf{y}}}_i=\omega \mathbf{n}\times {\mathbf{y}}_i$. The cross product identity for a unit vector gives $|\mathbf{n}\times {\mathbf{y}}_i{|}^2=|{\mathbf{y}}_i{|}^2-(\mathbf{n}\cdot {\mathbf{y}}_i{)}^2$, the squared perpendicular distance to the rotation axis. Hence the moment of inertia of the particle is $I_i=m_i[|{\mathbf{y}}_i{|}^2-(\mathbf{n}\cdot {\mathbf{y}}_i{)}^2]$, and summing the rotational kinetic energy yields

The Lorentz force and Newton's second law giveWriting $\Omega =qB/m$, the component equations areThe initial velocity therefore givesand integration with the initial position givesThis is helical motion in a uniform magnetic field: a helix of radius $v/\Omega$ around the line $x=1+v/\Omega$, $y=0$, with its axis parallel to the magnetic field.

Apply part (a) first to the position $\mathbf{r}$:Applying it again to both terms on the right giveswhere $(\dot{\mathbf{\omega }}{)}_S=(\dot{\mathbf{\omega }}{)}_{S^{\prime }}$ because $\mathbf{\omega }\times \mathbf{\omega }=0$. Substituting the first formula into the last term produces

In the Earth-fixed rotating reference frame, the sideways apparent force is the Coriolis acceleration multiplied by the ski-doo's mass:At the South Pole, $\mathbf{\omega }$ is vertical and antiparallel to the local outward normal, while $\mathbf{v}$ is horizontal, soThe force points to the left of the direction of travel, which is westward when the ski-doo is moving north after crossing the pole.

With $\mathbf{B}=0$ and $\mathbf{E}=(0,0,E)$, the spatial Lorentz force equation givesThe relativistic energy-momentum relation therefore yieldsSince $\mathbf{v}=c^2\mathbf{p}/\mathcal{E}$,Integration from the initial position $z(0)=0$ gives

With $L=1/\sqrt{n^2-1}$, one has $\nu =n$. Since $n\ge 3>\sqrt{3}$, ii proves boundedness. Write the positive rational number $n=p/q$ in lowest terms. The radius and its angular derivative repeat after $\Delta \theta =2\pi /n$. After $p$ such radial oscillations,so both the position and the radial and transverse velocities return to their initial values. The elapsed time iswhich is finite and positive because $r$ is bounded above and below by positive constants. Thus the motion is periodic. The orbit is a closed, generally self-intersecting rosette orbit, with $p$ radial oscillations during $q$ complete revolutions; for integral $n$ it has $n$ evenly spaced lobes around the origin. It never reaches the origin and is not a circular orbit.

After $j$ choices, let $C_j$ be the number of forward moves. The two unwrapped positions are $C_j$ and $F-(j-C_j)$, whose gap is $F-j$. It follows that the first subsequent meeting occurs exactly at step $F$, regardless of the order of the choices. Its position is the number of forward moves in that block.

That count has a binomial distribution with parameters $F,p$. ThereforeThe two endpoint labels $0$ and $F$ represent the same physical seat reached by different complete circuits; the unwrapped convention distinguishes their probabilities.

The meeting position $N$ is the sum of $F$ independent Bernoulli random variables, each of mean $p$ and variance $pq$. Linearity of expectation and variance additivity for independent random variables giveThese are the mean and variance of the unwrapped integer label specified in the question, rather than circular angular statistics.

For the forward-moving object to have been waiting at the meeting position, the last move must be a reverse move. The preceding $F-1$ choices must then contain exactly $n$ forward moves and $F-n-1$ reverse moves. Choosing their order givesAt $n=F$ the probability is zero, since the reverse-moving object was already at that unwrapped position initially. At $n=0$ the displayed formula is $q^F$, consistent with the forward object waiting at its initial position throughout. Equivalently, use the binomial convention $\left(\genfrac{}{}{0ex}{}{F-1}{F}\right)=0$ to include this endpoint in the same formula without a negative power of $q$.

At an intersection, the two position vectors agree, so $a-b=tn-sm$. The scalar triple product with $m,n$ is thereforesince a determinant with repeated columns vanishes. Geometrically the displacement of the two anchor points must lie in the same plane as the two line directions.

This is a necessary condition. If $m,n$ are not parallel, it is also sufficient, because their span is exactly the plane perpendicular to $m\times n$. For parallel directions the triple product vanishes automatically, even for distinct lines, so the condition alone cannot guarantee intersection.

Rewrite the complex position as $z(t)=2+3i/(t-i)$. Differentiating gives the complex velocity and accelerationFor real $t$, $|t-i|=\sqrt{t^2+1}$. Their requested magnitudes are thereforeThere is no pole for real time. The magnitude is the Euclidean length of the two Cartesian components, not the modulus of either component separately.

Uniform sampling without replacement is equivalent to revealing a uniformly random permutation of seven labelled balls. Each draw position is equally likely to contain any label, so its colour has the original proportions. ThusFor example the law of total probability gives $\mathbb{P}(B_2)=(4/7)(3/6)+(3/7)(4/6)=4/7$. Equal marginal probabilities express exchangeability, rather than independence between successive draws.

Distinguish the $N$ balls by labels. For a fixed position $n$, each label appears there in exactly $(N-1)!$ of the $N!$ equally likely permutations. Since $N_B$ labels are black,This is the same marginal law at every position, even the last one. The argument averages over earlier draw outcomes; it does not assert that a conditional probability after a specified history remains $N_B/N$. The original PDF uses $B_n$, correcting the transcription's uppercase subscript $B_N$.

TODO: Biology deferred at the user's request: clinical contrast interpretation, applications, and the head-injury MRI-versus-CT comparison.

The generic physics of magnetic resonance imaging combines a main static magnet, radiofrequency transmit and receive coils, controlled magnetic field gradients, signal electronics and reconstruction. A transverse radiofrequency pulse excites nuclear spins near their Larmor frequency, ${\omega }_0=\gamma B_0$. The evolving transverse magnetization induces a received signal; phase coherence and relaxation determine its time dependence.

Gradients encode position into phase. In a simplified homogeneous receive-sensitivity model,Here $\mathbf{k}$ is in cycles per unit length and $\gamma$ is in radians per second per tesla. Echo sign changes, pulse sequence timing and receive sensitivity belong in a more complete signal model. Sampling this MRI k-space and reconstructing its data produces an image; the measured samples are not already real-space pixel values.

The physical signal depends on spin density, longitudinal recovery and transverse dephasing. Timing changes their weighting. Static field inhomogeneity adds reversible dephasing beyond intrinsic transverse relaxation, motivating echo techniques and distinguishing $T_2^{\ast }$ from $T_2$. Those are physical signal mechanisms; their biological interpretation remains deferred.

TODO: Biology deferred at the user's request: biological tracer behavior and clinical PET applications.

The generic physics of positron emission tomography uses electron-positron annihilation. A slowly moving electron-positron pair predominantly produces two nearly opposite photons with energy close to $m_e{c}^2=511\mathrm{keV}$ each. A ring of detectors records approximately simultaneous interactions; a coincidence identifies a coincidence line of response along which an annihilation could have occurred. Finite Positron travel before annihilation and nonzero pair momentum limit how directly this line identifies the original emission location.

Many such measurements constrain an emission distribution through a tomography model. Attenuation, scattered coincidences, unrelated random coincidences, finite detector resolution and dead time alter the observed counts and require calibration or modeling. With useful time-of-flight information, the arrival-time difference further localizes position along the line: a timing uncertainty $\delta t$ corresponds to a position scale $c\delta t/2$. The coincidence data therefore provide spatial constraints, rather than each event directly supplying a complete three-dimensional image point.