= Paper 4 {scope} https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2022/paperia_4_2022.pdf = 1E {parent=Paper 4} {scope} {title2=Numbers and Sets} = Solution {parent=1E} Suppose there were only finitely many congruent to $2$ modulo $3$, and list them as $p_1,\ldots,p_k$. Consider $$ N=3p_1\cdots p_k-1. $$ No $p_i$ divides $N$, and $3$ does not divide $N$. In the of $N$, not every prime factor can be congruent to $1$ modulo $3$, since their product would then also be congruent to $1$, whereas $$ N\equiv2\pmod3. $$ Thus some prime factor is congruent to $2$ modulo $3$, contradicting the completeness of the list. Hence $$ \boxed{\text{there are infinitely many primes of the form }3n+2}. $$ Now let $p$ be prime. If $p\ne3$, then either $p=2$, giving $$ 2p^2+1=9, $$ or $p$ is not divisible by $3$, so $p^2\equiv1\pmod3$. In the latter case $$ 2p^2+1\equiv0\pmod3 $$ and is greater than $3$, hence is composite. For $p=3$ the number is $19$, which is prime. Therefore $$ \boxed{p=3\text{ is the only possibility}}. $$ Solved by gpt-5.6-sol high. = 2D {parent=Paper 4} {scope} {title2=Numbers and Sets} = Solution {parent=2D} Let $$ x=\sqrt[3]2+\sqrt[3]3. $$ If $x$ were [rational], then $$ x^3=5+3x\sqrt[3]6 $$ would make $\sqrt[3]6=(x^3-5)/(3x)$ rational. But if $\sqrt[3]6=a/b$ in lowest terms, then $a^3=6b^3$, which is impossible by : the exponent of $2$ on the two sides is respectively a multiple of three and one more than a multiple of three. Therefore $$ \boxed{\sqrt[3]2+\sqrt[3]3\text{ is irrational}}. $$ Use the given $$ e-e^{-1}=2\sum_{n=0}^{\infty}\frac1{(2n+1)!}. $$ Suppose this number were $a/b\in\mathbb Q$. Choose $N$ large enough that $b\mid(2N+1)!$. Multiplication by $(2N+1)!$ makes both the rational number and the partial sum through $n=N$ integers. Their difference $$ R_N =2(2N+1)!\sum_{n=N+1}^{\infty}\frac1{(2n+1)!} $$ would therefore be an integer. It is positive, while $$ 0 is a complex number that is not a root of any nonzero polynomial with rational, equivalently integer, coefficients. Let $$ y=ae+be^{-1}, \qquad (a,b)\ne(0,0). $$ If $a\ne0$ and $y$ were an , then $e$ would satisfy $$ aX^2-yX+b=0 $$ over the algebraic extension $\mathbb Q(y)$. By , $e$ would be algebraic over $\mathbb Q$, contradicting its transcendence. If $a=0$, then $b\ne0$ and $e=b/y$ would again be algebraic. Thus $$ \boxed{ae+be^{-1}\text{ is transcendental}}. $$ Solved by gpt-5.6-sol high. = 3C {parent=Paper 4} {scope} {title2=Dynamics and Relativity} = a {parent=3c} {scope} = Solution {parent=a} Take the scalar product of the equation with the constant $\mathbf B$. Since $\mathbf E\cdot\mathbf B=0$ and $(\dot{\mathbf x}\times\mathbf B)\cdot\mathbf B=0$, $$ m\frac{d^2}{dt^2}(\mathbf x\cdot\mathbf B)=0. $$ The initial velocity also obeys $\mathbf v\cdot\mathbf B=0$, so $\mathbf x\cdot\mathbf B$ remains equal to $\mathbf x_0\cdot\mathbf B$. The trajectory therefore lies in the plane $$ \boxed{(\mathbf x-\mathbf x_0)\cdot\mathbf B=0}, $$ the plane through $\mathbf x_0$ perpendicular to $\mathbf B$. Solved by gpt-5.6-sol high. = b {parent=3c} {scope} = Solution {parent=b} Let $\mathbf E=E\widehat{\mathbf x}$ and $\mathbf B=B\widehat{\mathbf y}$, and choose $\widehat{\mathbf z}=\widehat{\mathbf x}\times\widehat{\mathbf y}$. Part (a) gives $y=0$. With the origin at $\mathbf x_0$, the remaining Cartesian components are $$ \ddot x=\frac{qE}{m}-\omega\dot z, \qquad \ddot z=\omega\dot x, \qquad \omega=\frac{qB}{m}. $$ The initial conditions are $x=z=\dot x=\dot z=0$. Integrating the second equation gives $\dot z=\omega x$, so $$ \ddot x+\omega^2x=\frac{qE}{m}. $$ Solving this and then integrating $\dot z=\omega x$ yields $$ \boxed{ \begin{aligned} x(t)&=\frac{mE}{qB^2}\bigl(1-\cos\omega t\bigr),\\ y(t)&=0,\\ z(t)&=\frac EB\left(t-\frac{\sin\omega t}{\omega}\right). \end{aligned}} $$ Thus the position vector is $\mathbf x_0+x(t)\widehat{\mathbf x}+z(t)\widehat{\mathbf z}$. The oscillation occurs at the signed , superposed on the usual . Solved by gpt-5.6-sol high. = 4C {parent=Paper 4} {scope} {title2=Dynamics and Relativity} = a {parent=4c} {scope} = i {parent=a} {scope} = Solution {parent=i} The is $$ ct'=ct,\qquad x'=x-ut=x-\beta ct. $$ Therefore $$ \boxed{ A_{\mathrm N} =\begin{pmatrix} 1&0\\ -\beta&1 \end{pmatrix}}. $$ Solved by gpt-5.6-sol high. = ii {parent=a} {scope} = Solution {parent=ii} The one-dimensional gives $$ ct'=\gamma(ct-\beta x), \qquad x'=\gamma(x-\beta ct), \qquad \gamma=\frac1{\sqrt{1-\beta^2}}. $$ Hence $$ \boxed{ A_{\mathrm{SR}} =\gamma \begin{pmatrix} 1&-\beta\\ -\beta&1 \end{pmatrix}}. $$ As $|\beta|\to0$, $\gamma=1+O(\beta^2)$. The spatial transformation is therefore $x'=x-ut+O(\beta^2)$, while $$ t'=t-\frac{ux}{c^2}+O(\beta^2t). $$ Under the stated condition $|x| given by the Galilean transformation. Solved by gpt-5.6-sol high. = b {parent=4c} {scope} = Solution {parent=b} For the special-relativistic matrix, $$ A\binom11=\gamma(1-\beta)\binom11, \qquad A\binom1{-1}=\gamma(1+\beta)\binom1{-1}. $$ Thus the and corresponding are $$ \boxed{ \lambda_+=\sqrt{\frac{1-\beta}{1+\beta}}, \quad v_+=\binom11}, \qquad \boxed{ \lambda_-=\sqrt{\frac{1+\beta}{1-\beta}}, \quad v_-=\binom1{-1}}. $$ The eigenvector equations are $x=ct$ and $x=-ct$. They are the two of the , representing right-moving and left-moving light rays. A Lorentz boost preserves each light-ray direction while rescaling its null coordinate. Solved by gpt-5.6-sol high. = 5E {parent=Paper 4} {scope} {title2=Numbers and Sets} = Solution {parent=5E} The states that for integers $a,b$, not both zero, there are integers $r,s$ such that $$ ra+sb=\gcd(a,b). $$ If the $p$ divides $ab$ but does not divide $a$, then $\gcd(p,a)=1$. Thus $rp+sa=1$ for some integers $r,s$, and multiplication by $b$ gives $$ b=rpb+sab. $$ Both terms on the right are divisible by $p$, so $p\mid b$. Hence $$ \boxed{p\mid ab\Longrightarrow p\mid a\text{ or }p\mid b}. $$ If $\gcd(m,n)=1$, choose $r,s$ with $rm+sn=1$. Then $$ x=a(sn)+b(rm) $$ satisfies $x\equiv a\pmod m$ and $x\equiv b\pmod n$. If $x,x'$ are two simultaneous solutions, both $m$ and $n$ divide $x-x'$. Coprimality implies $mn\mid x-x'$, so the solution is unique modulo $mn$. This proves the two-modulus . Let $p$ be odd and suppose $$ x^2\equiv1\pmod{p^d}. $$ Then $p^d\mid(x-1)(x+1)$. Since $\gcd(x-1,x+1)$ divides $2$, the odd prime $p$ cannot divide both factors. The full power $p^d$ must therefore divide one of them, giving $$ x\equiv1\pmod{p^d} \quad\hbox{or}\quad x\equiv-1\pmod{p^d}. $$ These are distinct, so there are exactly two solutions. For an odd integer $$ n=\prod_{j=1}^kp_j^{d_j}, $$ the Chinese remainder theorem identifies a solution modulo $n$ with independent solutions modulo the $k$ . Each component has two choices, hence $$ \boxed{\#\{x\bmod n:x^2\equiv1\}=2^k}. $$ For $n=2^d$, there is one solution when $d=1$ and two when $d=2$. If $d\geq3$, a solution $x$ is odd, and of the consecutive even integers $x-1,x+1$, exactly one is divisible by $4$. The [2-adic valuations] therefore have minimum one; for their sum to be at least $d$, the other must be at least $d-1$. Thus $x\equiv\pm1\pmod{2^{d-1}}$, giving the four distinct classes $$ 1,\quad -1,\quad1+2^{d-1},\quad-1+2^{d-1}\pmod{2^d}. $$ Consequently the answer is $$ \boxed{ \begin{cases} 1,&d=1,\\ 2,&d=2,\\ 4,&d\geq3. \end{cases}} $$ Solved by gpt-5.6-sol high. = 6D {parent=Paper 4} {scope} {title2=Numbers and Sets} = a {parent=6d} {scope} = Solution {parent=a} For $0\leq k\leq n$, the is $$ \binom nk=\frac{n!}{k!(n-k)!}. $$ When $1\leq k\leq n-1$, $$ \begin{aligned} \binom{n-1}k+\binom{n-1}{k-1} &=\frac{(n-1)!}{k!(n-1-k)!} +\frac{(n-1)!}{(k-1)!(n-k)!}\\ &=\frac{(n-1)![(n-k)+k]}{k!(n-k)!}\\ &=\boxed{\binom nk}. \end{aligned} $$ This is . Solved by gpt-5.6-sol high. = b {parent=6d} {scope} = Solution {parent=b} The special $$ (1+t)^n=\sum_{k=0}^n\binom nk t^k $$ follows by . Multiplication of the formula for $n$ by $1+t$ and collection of the coefficient of $t^k$ uses Pascal's identity from part (a). Termwise integration from $0$ to $1$ gives $$ \boxed{ \sum_{k=0}^n\frac1{k+1}\binom nk =\int_0^1(1+t)^n\,dt =\frac{2^{n+1}-1}{n+1}}. $$ Replacing $t$ by $-t$ gives $$ \boxed{ \sum_{k=0}^n\frac{(-1)^k}{k+1}\binom nk =\int_0^1(1-t)^n\,dt =\frac1{n+1}}. $$ Finally, $$ \sum_{k=1}^n\frac{(-1)^{k+1}}k\binom nk =\int_0^1\frac{1-(1-x)^n}{x}\,dx. $$ Writing $y=1-x$ and using the finite , $$ \frac{1-y^n}{1-y}=1+y+\cdots+y^{n-1}, $$ turns this into $$ \boxed{\sum_{j=0}^{n-1}\int_0^1y^j\,dy =1+\frac12+\cdots+\frac1n}. $$ Solved by gpt-5.6-sol high. = c {parent=6d} {scope} = Solution {parent=c} Put $$ S_n=\sum_{k=0}^{\lfloor(n-1)/2\rfloor} \binom{n-k-1}{k}. $$ The initial values are $S_1=S_2=1$. Using Pascal's identity and the convention that an out-of-range binomial coefficient is zero, $$ \begin{aligned} S_{n+2} &=\sum_{k\geq0}\binom{n+1-k}{k}\\ &=\sum_{k\geq0}\binom{n-k}{k} +\sum_{k\geq1}\binom{n-k}{k-1}\\ &=S_{n+1}+S_n. \end{aligned} $$ Thus $(S_n)$ has the same initial values and as the . gives $$ \boxed{ F_n=\sum_{k=0}^{\lfloor(n-1)/2\rfloor} \binom{n-k-1}{k}}. $$ Solved by gpt-5.6-sol high. = 7F {parent=Paper 4} {scope} {title2=Numbers and Sets} = a {parent=7f} {scope} = Solution {parent=a} Because $P(\mathbf x)\subseteq P(\mathbf x)$, the relation is [reflexive]. If $\mathbf x\preceq\mathbf y$ and $\mathbf y\preceq\mathbf z$, then $$ P(\mathbf x)\subseteq P(\mathbf y)\subseteq P(\mathbf z), $$ so it is [transitive]. It is not [symmetric]: for $$ \mathbf x=(1,0,\ldots,0), \qquad \mathbf y=(1,1,0,\ldots,0), $$ one has $\mathbf x\preceq\mathbf y$ but not $\mathbf y\preceq\mathbf x$. Solved by gpt-5.6-sol high. = b {parent=7f} {scope} = Solution {parent=b} Suppose $\mathbf x\preceq\mathbf y$. Define the nonnegative tuple $\mathbf z$ coordinatewise by $$ z_i= \begin{cases} x_i/y_i,&y_i>0,\\ 0,&y_i=0. \end{cases} $$ When $y_i=0$, support inclusion forces $x_i=0$, so $x_i=y_iz_i$ in every coordinate. Conversely, if $x_i=y_iz_i$ with $y_i,z_i\geq0$, then $x_i>0$ implies $y_i>0$. Hence $P(\mathbf x)\subseteq P(\mathbf y)$ and $\mathbf x\preceq\mathbf y$. This proves the equivalence. Solved by gpt-5.6-sol high. = c {parent=7f} {scope} = Solution {parent=c} The definition says $$ \mathbf x\sim\mathbf y \quad\Longleftrightarrow\quad P(\mathbf x)=P(\mathbf y). $$ Equality of sets is reflexive, symmetric, and transitive, so $\sim$ is an . Its classes are indexed by all subsets of $\{1,\ldots,n\}$, and every subset occurs as the support of its zero-one . The number of classes is therefore $$ \boxed{2^n}. $$ Solved by gpt-5.6-sol high. = d {parent=7f} {scope} = Solution {parent=d} Define $$ y_i= \begin{cases} x_i,&s_i>0,\\ 0,&s_i=0, \end{cases} \qquad z_i=x_i-y_i. $$ Then $\mathbf x=\mathbf y+\mathbf z$, the support of $\mathbf y$ lies in $P(\mathbf s)$, and the support of $\mathbf z$ is disjoint from $P(\mathbf s)$. Thus $\mathbf y\preceq\mathbf s$ and $\mathbf z\perp\mathbf s$. For uniqueness, if $s_i>0$, orthogonality forces $z_i=0$, hence $y_i=x_i$. If $s_i=0$, the condition $\mathbf y\preceq\mathbf s$ forces $y_i=0$, hence $z_i=x_i$. Every coordinate is therefore forced, proving uniqueness. Solved by gpt-5.6-sol high. = 8F {parent=Paper 4} {scope} {title2=Numbers and Sets} = a {parent=8f} {scope} = Solution {parent=a} A set is [countable] when it is finite or its elements can be put in one-to-one correspondence with a subset of $\mathbb N$. Equivalently, there is an injection from the set into $\mathbb N$. Solved by gpt-5.6-sol high. = b {parent=8f} {scope} = i {parent=b} {scope} = Solution {parent=i} List pairs $(m,n)\in\mathbb N^2$ by increasing value of $m+n$, and within each diagonal by increasing $m$: $$ (1,1),\ (1,2),(2,1),\ (1,3),(2,2),(3,1),\ldots. $$ Every pair occurs after finitely many earlier diagonals. More explicitly, the $$ \pi(m,n)=\frac{(m+n-2)(m+n-1)}2+m $$ is an injection $\mathbb N^2\to\mathbb N$. Hence $\mathbb N\times\mathbb N$ is countable. Solved by gpt-5.6-sol high. = ii {parent=b} {scope} = Solution {parent=ii} The integers are countable under the enumeration $$ 0,1,-1,2,-2,\ldots. $$ Part (i) then shows that $\mathbb Z\times\mathbb N$ is countable. The map $$ (a,b)\longmapsto\frac ab $$ is a surjection from $\mathbb Z\times\mathbb N$ onto the . Selecting, for example, the representation in lowest terms with positive denominator gives an injection in the other direction. Thus $\mathbb Q$ is countable. Solved by gpt-5.6-sol high. = iii {parent=b} {scope} = Solution {parent=iii} For a [increasing function] $F$, define its one-sided limits by $$ F(x-)=\sup_{tx}F(t). $$ They exist because of [monotonicity]. The function is discontinuous at $x$ only if the jump interval $$ J_x=(F(x-),F(x+)) $$ is nonempty. By the , choose $q_x\in J_x\cap\mathbb Q$. If $xN$. Then an element of $B$ is determined by its first $N$ coordinates, so $B$ is in bijection with $$ A_1\times\cdots\times A_N. $$ A finite of countable sets is countable by repeated application of the diagonal enumeration in part (b)(i). Hence $B$ is countable. Conversely, suppose infinitely many factors contain at least two elements. Choose increasing indices $n_1 $\varepsilon=(\varepsilon_1,\varepsilon_2,\ldots)$ then defines an element of $B$ by placing $a_j^{\varepsilon_j}$ in coordinate $n_j$ and the fixed element elsewhere. This map is injective. The set $\{0,1\}^{\mathbb N}$ is uncountable by : from any proposed list of binary sequences, form a new sequence whose $j$th digit differs from the $j$th digit of the $j$th listed sequence. It is absent from the list. Thus $B$ contains an uncountable subset and cannot be countable. Therefore $$ \boxed{ B\text{ is countable }\Longleftrightarrow \exists N\ \forall n>N,\ |A_n|=1}. $$ Solved by gpt-5.6-sol high. = 9C {parent=Paper 4} {scope} {title2=Dynamics and Relativity} = a {parent=9c} {scope} = Solution {parent=a} The
is $$ \mathbf R=\frac{m_1\mathbf x_1+m_2\mathbf x_2}{m_1+m_2}. $$ By , $\mathbf F_{21}=-\mathbf F_{12}$, so $$ (m_1+m_2)\ddot{\mathbf R} =m_1\ddot{\mathbf x}_1+m_2\ddot{\mathbf x}_2 =\mathbf F_{12}+\mathbf F_{21}=0. $$ Thus $\dot{\mathbf R}$ is constant. For $\mathbf r=\mathbf x_1-\mathbf x_2$, $$ \ddot{\mathbf r} =\left(\frac1{m_1}+\frac1{m_2}\right)\mathbf F_{12}. $$ Therefore $$ \boxed{\mu\ddot{\mathbf r}=\mathbf F_{12}}, \qquad \boxed{\mu=\frac{m_1m_2}{m_1+m_2}}, $$ where $\mu$ is the . Solved by gpt-5.6-sol high. = b {parent=9c} {scope} = Solution {parent=b} For the proposed , $$ \ddot{\mathbf x}_1=-a\omega^2(\cos\omega t,\sin\omega t,0). $$ The separation is $\mathbf x_1-\mathbf x_2=2a(\cos\omega t,\sin\omega t,0)$, so the on particle 1 is $$ \mathbf F_{12} =-\frac{km^2}{4a^2}(\cos\omega t,\sin\omega t,0). $$ The equation $m\ddot{\mathbf x}_1=\mathbf F_{12}$ holds exactly when $$ \boxed{\omega^2=\frac{km}{4a^3}}. $$ The equation for particle 2 follows by symmetry. Solved by gpt-5.6-sol high. = c {parent=9c} {scope} = i {parent=c} {scope} = Solution {parent=i} At a point $(0,0,z)$, the two particles are at equal distance $\sqrt{a^2+z^2}$. Their force components in the rotating $xy$-plane cancel, while their $z$-components add. Hence a particle initially moving along the $z$-axis remains on it, and $$ m_3\ddot z =-\frac{km_3mz}{(a^2+z^2)^{3/2}} -\frac{km_3mz}{(a^2+z^2)^{3/2}}. $$ Cancelling $m_3$ gives $$ \boxed{\ddot z=-\frac{2mkz}{(z^2+a^2)^{3/2}}}. $$ Solved by gpt-5.6-sol high. = ii {parent=c} {scope} = Solution {parent=ii} The equation has the one-dimensional per unit mass $$ \boxed{\Phi(z)=-\frac{2mk}{\sqrt{z^2+a^2}}}, \qquad \ddot z=-\Phi'(z). $$ Thus the conserved specific energy is $$ \mathcal E=\frac12\dot z^2+\Phi(z). $$ The potential has its minimum $-2mk/a$ at $z=0$ and tends to $0$ from below as $|z|\to\infty$. Therefore $-2mk/a<\mathcal E<0$ gives bounded oscillation between two [turning points]; $\mathcal E=0$ gives marginal escape with asymptotic speed zero; and $\mathcal E>0$ gives escape with nonzero asymptotic speed. The minimum itself is the equilibrium $z=0$. Solved by gpt-5.6-sol high. = iii {parent=c} {scope} = Solution {parent=iii} For $|z|\ll a$, gives $$ \ddot z=-\frac{2mk}{a^3}z+O(z^3). $$ The leading motion is with angular frequency $$ \omega_z=\sqrt{\frac{2mk}{a^3}}. $$ With $z(0)=z_0$ and $\dot z(0)=0$, $$ z(t)\simeq z_0\cos(\omega_zt), $$ and its period is $$ \boxed{T=2\pi\sqrt{\frac{a^3}{2mk}}}. $$ Solved by gpt-5.6-sol high. = iv {parent=c} {scope} = Solution {parent=iv} The initial specific energy is $$ \mathcal E=\frac12u^2-\frac{2mk}{a}. $$ Escape to infinity is possible exactly when $\mathcal E\geq0$. Thus the criterion is $$ \boxed{|u|\geq2\sqrt{\frac{mk}{a}}}. $$ Equality corresponds to marginal escape. Solved by gpt-5.6-sol high. = 10C {parent=Paper 4} {scope} {title2=Dynamics and Relativity} = a {parent=10c} {scope} = Solution {parent=a} The term $-2\boldsymbol\Omega\times\dot{\mathbf x}$ is the , and $-\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf x)$ is the . With $\boldsymbol\Omega=-\Omega\widehat{\mathbf z}$, the Cartesian components are $$ \boxed{ \begin{aligned} \ddot x&=-2\Omega\dot y+\Omega^2x+\frac{N_x}{m},\\ \ddot y&= 2\Omega\dot x+\Omega^2y+\frac{N_y}{m},\\ \ddot z&=g+\frac{N_z}{m}. \end{aligned}} $$ Solved by gpt-5.6-sol high. = b {parent=10c} {scope} = Solution {parent=b} The circular constraint is $$ x=R\sin\theta,\qquad z=R\cos\theta. $$ Its tangent in the $xz$-plane is $(\cos\theta,0,-\sin\theta)$, while the is radial. Projecting the equations from part (a) onto this tangent eliminates the normal force and gives $$ \boxed{ R\ddot\theta =-g\sin\theta+\Omega^2R\cos\theta\sin\theta -2\Omega\dot y\cos\theta}. $$ Because the ramp is translation-invariant along $y$, $N_y=0$; using $\dot x=R\dot\theta\cos\theta$ gives $$ \boxed{\ddot y=\Omega^2y+2\Omega R\dot\theta\cos\theta}. $$ For rest in the rotating frame, $\dot\theta=\dot y=\ddot\theta=\ddot y=0$. Assuming $\Omega\ne0$, the second equation gives $y=0$, and the first gives $$ \sin\theta(\Omega^2R\cos\theta-g)=0. $$ On the semicircular ramp the rest points are therefore $$ \boxed{\theta=0,\ y=0} $$ and, when $\Omega^2R\geq g$, $$ \boxed{\theta=\pm\arccos\frac{g}{\Omega^2R},\ y=0}. $$ Solved by gpt-5.6-sol high. = c {parent=10c} {scope} = Solution {parent=c} Neglecting terms of order $\Omega^2$ and linearizing in $\theta$ gives $$ R\ddot\theta=-g\theta-2\Omega\dot y, \qquad \ddot y=2\Omega R\dot\theta. $$ The second equation and the initial data imply $$ \dot y=2\Omega R(\theta-\theta_0). $$ Its contribution to the first equation is of order $\Omega^2$, so the retained equation is $$ \ddot\theta+\frac gR\theta=0. $$ Thus $$ \boxed{\theta(t)=\theta_0\cos(\omega_0t)}, \qquad \boxed{\omega_0=\sqrt{\frac gR}}. $$ Integrating the associated $y$ velocity and using $y(0)=0$ gives $$ \boxed{ y(t)=2\Omega R\theta_0 \left(\frac{\sin(\omega_0t)}{\omega_0}-t\right)}. $$ Solved by gpt-5.6-sol high. = 11C {parent=Paper 4} {scope} {title2=Dynamics and Relativity} = a {parent=11c} {scope} = i {parent=a} {scope} = Solution {parent=i} The about an axis is $$ I=\int_V\rho\,r_\perp^2\,dV, $$ where $r_\perp$ is the perpendicular distance to the axis. The disc mass is $$ M=\rho\pi r^2\delta. $$ To leading order in $\delta/r$, the distance from the $x$-axis is $|y|$. Using polar coordinates in the disc, $$ \begin{aligned} I_x &=\rho\delta\int_0^r\int_0^{2\pi} (s\sin\phi)^2s\,d\phi\,ds\\ &=\rho\delta\frac{r^4}{4}\pi =\boxed{\frac14Mr^2}. \end{aligned} $$ Solved by gpt-5.6-sol high. = ii {parent=a} {scope} = Solution {parent=ii} The axis $y=0,z=h$ is parallel to the $x$-axis and lies a distance $|h|$ from it. The therefore gives $$ \boxed{I=\frac14Mr^2+Mh^2 =M\left(\frac{r^2}{4}+h^2\right)} $$ to leading order in the thickness. Solved by gpt-5.6-sol high. = b {parent=11c} {scope} = i {parent=b} {scope} = Solution {parent=i} Measure $z$ from the apex along the cone's symmetry axis. A thin disc at height $z$ has radius $$ s(z)=\frac RH z $$ and mass $dM=\rho\pi s^2\,dz$. Its moment about a parallel diameter through its centre is $s^2dM/4$, while its centre is distance $z$ from the required axis. Part (a)(ii) gives $$ dI=\left(z^2+\frac{s^2}{4}\right)dM. $$ Hence $$ \begin{aligned} I &=\rho\pi\int_0^H \frac{R^2z^2}{H^2} \left(z^2+\frac{R^2z^2}{4H^2}\right)dz\\ &=\rho\pi\left(\frac{R^2H^3}{5} +\frac{R^4H}{20}\right). \end{aligned} $$ Since $M=\rho\pi R^2H/3$, $$ \boxed{ I=M\left(\frac3{20}R^2+\frac35H^2\right)}, \qquad \boxed{\alpha=\frac3{20},\quad\beta=\frac35}. $$ Solved by gpt-5.6-sol high. = ii {parent=b} {scope} = Solution {parent=ii} Without friction, is conserved: $$ K_0=\frac12I\omega^2. $$ The constant angular speed is $\omega=\sqrt{2K_0/I}$, so one full rotation takes $$ \boxed{T=\frac{2\pi}{\omega} =2\pi\sqrt{\frac{I}{2K_0}}}. $$ Solved by gpt-5.6-sol high. = iii {parent=b} {scope} = Solution {parent=iii} The work law means that friction exerts a constant opposing of magnitude $W$. Until the cone stops, $$ I\dot\omega=-W. $$ Its initial angular speed is $\omega_0=\sqrt{2K_0/I}$, and uniform angular deceleration gives $$ 0=\omega_0-\frac WI t. $$ Therefore $$ \boxed{t=\frac{I\omega_0}{W} =\sqrt{\frac{2K_0I}{W^2}}}. $$ Solved by gpt-5.6-sol high. = 12C {parent=Paper 4} {scope} {title2=Dynamics and Relativity} = a {parent=12c} {scope} = Solution {parent=a} A is an object whose components transform between inertial frames by a , $$ U'^\mu=\Lambda^\mu{}_\nu U^\nu. $$ Lorentz transformations are defined by $$ \Lambda^T\eta\Lambda=\eta, $$ where $\eta$ is the . Consequently $$ U'\cdot U' =(\Lambda U)^T\eta(\Lambda U) =U^T\Lambda^T\eta\Lambda U =U^T\eta U =U\cdot U. $$ Thus the of a four-vector is the same in every inertial frame. Solved by gpt-5.6-sol high. = b {parent=12c} {scope} = i {parent=b} {scope} = Solution {parent=i} Along the particle's worldline, $$ x=u_xt,\qquad y=u_yt. $$ The boost from $S$ to $S'$ gives $$ x'=\gamma(x-Vt),\qquad y'=y,\qquad t'=\gamma\left(t-\frac{Vx}{c^2}\right). $$ Therefore $$ \frac{dt'}{dt} =\gamma\left(1-\frac{Vu_x}{c^2}\right), $$ and division of the transformed coordinate velocities by this factor gives the $$ \boxed{ u'_x=\frac{u_x-V}{1-Vu_x/c^2}, \qquad u'_y=\frac{u_y}{\gamma(1-Vu_x/c^2)}, \qquad u'_z=0}. $$ Solved by gpt-5.6-sol high. = ii {parent=b} {scope} = Solution {parent=ii} To first order in $V/c$, $\gamma=1+O(V^2/c^2)$ and $$ \frac1{1-Vu_x/c^2} =1+\frac{Vu_x}{c^2}+O(V^2/c^2). $$ Thus $$ u'_x=u_x-V+\frac{Vu_x^2}{c^2}+O(V^2/c^2), \qquad u'_y=u_y+\frac{Vu_xu_y}{c^2}+O(V^2/c^2). $$ Since $\mathbf V=(V,0,0)$ and $\mathbf V\cdot\mathbf u=Vu_x$, these components combine into $$ \boxed{ \mathbf u'=\mathbf u-\mathbf V +\frac{\mathbf V\cdot\mathbf u}{c^2}\mathbf u +O(V^2/c^2)}. $$ Solved by gpt-5.6-sol high. = iii {parent=b} {scope} = Solution {parent=iii} For a photon, $$ \mathbf u=c(\cos\theta,\sin\theta,0). $$ Part (ii) gives the first-order changes $$ \delta u_x=-V\sin^2\theta, \qquad \delta u_y=V\sin\theta\cos\theta. $$ For a vector of fixed leading-order magnitude $c$, its small change of angle is $$ \delta\theta =\frac{u_x\delta u_y-u_y\delta u_x}{c^2}. $$ Substitution yields the formula $$ \boxed{\theta'-\theta=\frac Vc\sin\theta} $$ to leading order. Solved by gpt-5.6-sol high. = c {parent=12c} {scope} = Solution {parent=c} Let the incident and final photon momentum magnitudes be $$ p=\frac h\lambda, \qquad p'=\frac h{\lambda'}. $$ In a nontrivial collinear collision, the photon is backscattered and the electron moves in the original photon direction, so momentum conservation gives electron momentum $P=p+p'$. gives $$ pc+mc^2=p'c+\sqrt{m^2c^4+P^2c^2}. $$ Substitute $P=p+p'$, isolate the square root, and square. Cancellation leaves $$ mc(p-p')=2pp'. $$ Dividing by $pp'$ and using the wavelength definitions gives the shift $$ \boxed{\lambda'=\lambda+\frac{2h}{mc}}. $$ The other collinear possibility is the trivial forward solution $\lambda'=\lambda$, in which the electron remains at rest. Solved by gpt-5.6-sol high.