= Paper 4
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2021/paperia_4_2021.pdf

= 1E
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= i
{parent=1e}
{scope}

= Solution
{parent=i}

This can fail. Let $X=\{1\}$ and $Y=\{a,b\}$, put $f(1)=a$, and let $g(a)=g(b)=1$. Then the <function composition> $g\circ f$ is the <identity function> on $X$, hence is a <surjective function>, although $f$ is not surjective.

Solved by gpt-5.6-sol high.

= ii
{parent=1e}
{scope}

= Solution
{parent=ii}

This is always true. If $f(x_1)=f(x_2)$, then
$$
(g\circ f)(x_1)=g(f(x_1))=g(f(x_2))=(g\circ f)(x_2).
$$
The assumed <injective function>[injectivity] of $g\circ f$ gives $x_1=x_2$, so $f$ is injective.

Solved by gpt-5.6-sol high.

= iii
{parent=1e}
{scope}

= Solution
{parent=iii}

This can fail in the same example as in part (i): $g\circ f$ is injective on the singleton $X$, but $g(a)=g(b)$.

For the final count, $g\circ f=\operatorname{id}_X$ forces $f$ to be injective. There are
$$
n(n-1)\cdots(n-m+1)=\frac{n!}{(n-m)!}
$$
such <function>[functions] $f$. Once $f$ is chosen, $g$ is forced on the $m$ elements of $f(X)$ and has $m$ independent choices at each of the other $n-m$ elements of $Y$. Hence the number of pairs is
$$
\frac{n!}{(n-m)!}m^{n-m}.
$$

Solved by gpt-5.6-sol high.

= 2E
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= i
{parent=2e}
{scope}

= Solution
{parent=i}

The <Fibonacci number>[Fibonacci recurrence] gives
$$
a_{n+1}=\frac{F_{n+2}}{F_{n+1}}
=1+\frac{F_n}{F_{n+1}}=1+\frac1{a_n}.
$$
The map $x\mapsto1+1/x$ is strictly decreasing for $x>0$. Since $a_3=2\geq a_1=1$, applying this decreasing map reverses each inequality and proves by <mathematical induction> that
$$
(-1)^na_{n+2}\leq(-1)^na_n.
$$
Taking even $n$ shows $a_{2n+2}\leq a_{2n}$, so $(a_{2n})$ is a <monotone sequence>[decreasing sequence].

Solved by gpt-5.6-sol high.

= ii
{parent=2e}
{scope}

= Solution
{parent=ii}

The stated formula is <Fibonacci determinant identity>[Cassini's identity]. It holds for $n=1$, and the recurrence gives
$$
F_{n+3}F_{n+1}-F_{n+2}^2
=F_{n+1}^2-F_{n+2}F_n,
$$
which is the negative of the expression at index $n$. This proves the identity by induction.

Consequently
$$
a_{n+1}-a_n
=\frac{F_{n+2}F_n-F_{n+1}^2}{F_{n+1}F_n}
=\frac{(-1)^{n+1}}{F_{n+1}F_n}.
$$
The positive <Fibonacci number>[Fibonacci numbers] tend to <infinity>, so this difference tends to zero.

Solved by gpt-5.6-sol high.

= iii
{parent=2e}
{scope}

= Solution
{parent=iii}

Part (i) also shows that the odd subsequence is increasing and the even subsequence is decreasing. Each odd term is below the adjacent even term, so both are <monotone bounded sequence>[bounded monotone sequences] and therefore have limits. Part (ii) says their difference tends to zero, so those limits coincide. Interlacing the two subsequences proves that $(a_n)$ has that common <convergent sequence>[limit].

Solved by gpt-5.6-sol high.

= 3C
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= Solution
{parent=3C}

Let $M(t)=m_0-\alpha t$ be the trolley's remaining <mass>. During $dt$, the expelled mass $-dM$ has ground-frame <velocity> $v+u$. Conservation of <momentum> gives, to first order,
$$
Mv=(M+dM)(v+dv)+(-dM)(v+u),
$$
and hence $M\,dv=u\,dM$. Integration yields the <rocket equation>
$$
v-v_0=u\log\frac{M}{m_0}.
$$
At the stopping time, $M=m_0e^{-v_0/u}$. Therefore the time spent ejecting gas is
$$
\boxed{\frac{m_0}{\alpha}\left(1-e^{-v_0/u}\right)}.
$$

Solved by gpt-5.6-sol high.

= 4C
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= Solution
{parent=4C}

Writing $x_i=(x_i,y_i,z_i)$, rotation about the $z$-axis gives particle $i$ the <speed> $\omega\sqrt{x_i^2+y_i^2}$. Its <kinetic energy> is therefore $m_i\omega^2(x_i^2+y_i^2)/2$, so
$$
T=\frac12I\omega^2,
\qquad
I=\sum_{i=1}^Nm_i(x_i^2+y_i^2).
$$
This is the defining mass sum for the <moment of inertia> about the $z$-axis.

For the uniform cuboid, the <mass density> is $\rho=M/(8abc)$ and
$$
I=\rho\int_{-a}^a\int_{-b}^b\int_{-c}^c(x^2+y^2)\,dz\,dy\,dx
=\boxed{\frac M3(a^2+b^2)}.
$$

Solved by gpt-5.6-sol high.

= 5E
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= a
{parent=5e}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Define the <binomial coefficient> $\binom nr$ as the number of $r$-element <subset>[subsets] of an $n$-element set, equivalently as $n!/[r!(n-r)!]$. Partitioning the $r$-element subsets of $\{1,\ldots,n+1\}$ according to whether they contain $n+1$ proves <Pascal's identity>
$$
\binom{n+1}r=\binom nr+\binom n{r-1}.
$$
Thus the <forward difference operator> satisfies
$$
(\delta f_r)(n)=\binom{n+1}r-\binom nr=\binom n{r-1}=f_{r-1}(n).
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

We use induction on $k$. If $k=0$, then $\delta f=0$, so the integer-valued function $f$ is constant. For $k>0$, the integer-valued function $\delta f$ satisfies $\delta^k(\delta f)=0$. By induction it is an integer linear combination of
$$
\binom n{k-1},\binom n{k-2},\ldots,\binom n1,1.
$$
Part (i) shows that replacing each $\binom nr$ by $\binom n{r+1}$ gives an integer-valued <discrete antiderivative>. Subtracting the resulting integer linear combination from $f$ leaves a function with zero forward difference, hence an integer constant. This gives
$$
f(n)=c_0\binom nk+c_1\binom n{k-1}+\cdots+c_{k-1}\binom n1+c_k
$$
with every $c_j\in\mathbb Z$.

Solved by gpt-5.6-sol high.

= b
{parent=5e}
{scope}

= Solution
{parent=b}

The <binomial theorem> states
$$
(x+y)^n=\sum_{r=0}^n\binom nrx^ry^{n-r}.
$$
The required sum is the coefficient of $x^n$ in
$$
(1-x)^n(1+x)^n=(1-x^2)^n.
$$
There is no $x^n$ term when $n$ is <odd number>[odd]. When $n$ is <even number>[even], the relevant term has exponent $2(n/2)=n$ and coefficient $(-1)^{n/2}\binom n{n/2}$. Hence
$$
\sum_{r=0}^n(-1)^r\binom nr^2
=\begin{cases}
0,&n\text{ odd},\\
(-1)^{n/2}\binom n{n/2},&n\text{ even}.
\end{cases}
$$

Solved by gpt-5.6-sol high.

= 6E
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= a
{parent=6e}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The <Euclidean algorithm> repeatedly replaces a pair $(a,b)$ by $(b,a-qb)$ without changing its <greatest common divisor>. Reversing the divisions expresses $\gcd(a,b)$ as an integer linear combination $\alpha a+\beta b$, which is <Bezout identity>[Bézout's identity].

Here $3=3(15)-2(21)$, and therefore
$$
1=12\cdot3-35=36(15)-24(21)-35.
$$
One solution is $(x,y,z)=(36,-24,-1)$. It is not unique: for every $t\in\mathbb Z$, $(x,y,z)=(36+7t,-24-5t,-1)$ is another solution.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Because $\gcd(m,n)=1$, choose $u,v\in\mathbb Z$ with $um+vn=1$ by <Bezout identity>[Bézout's identity]. Then
$$
c=avn+bum
$$
satisfies $c\equiv a\pmod m$ and $c\equiv b\pmod n$. Any two simultaneous solutions differ by a multiple of both coprime integers and hence by a multiple of $mn$. This proves the two-modulus <Chinese remainder theorem>.

The three congruences reduce to
$$
x\equiv2\pmod5,
\qquad x\equiv3\pmod7,
\qquad x\equiv1\pmod3.
$$
The first two give $x\equiv17\pmod{35}$; imposing the last gives
$$
\boxed{x\equiv52\pmod{105}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=6e}
{scope}

= Solution
{parent=b}

The <inclusion-exclusion principle> says that the size of a finite union is the alternating sum of the sizes of all nonempty intersections of its constituent sets.

For each <prime number> $p\mid n$, let $A_p$ consist of the tuples for which $p$ divides every $x_i$. Exactly $(n/p)^r$ tuples lie in $A_p$, and for distinct primes $p_1,\ldots,p_j\mid n$, exactly $(n/(p_1\cdots p_j))^r$ lie in their intersection. A tuple has greatest common divisor greater than one with $n$ exactly when it belongs to some $A_p$. Inclusion-exclusion therefore gives the <Jordan totient function>
$$
\phi_r(n)
=n^r\sum_{S\subseteq\{p:p\mid n\}}(-1)^{|S|}\prod_{p\in S}p^{-r}
=\boxed{n^r\prod_{p\mid n}(1-p^{-r})}.
$$

Solved by gpt-5.6-sol high.

= 7E
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= a
{parent=7e}
{scope}

= Solution
{parent=a}

This is an infinite <binary expansion> with digits in $\{0,1\}$. If the usual expansion of $\alpha$ has infinitely many $1$ digits, list their positions as $b_1<b_2<\cdots$. If it terminates with a final term $2^{-k}$, replace that term by
$$
2^{-k}=\sum_{j=k+1}^{\infty}2^{-j};
$$
this also covers every <dyadic rational>. For $\alpha=1$, use $b_n=n$. Thus every $\alpha\in(0,1]$ has the required form.

The representation by an infinite strictly increasing sequence is unique. If two sequences first differ at exponent $k$, one sum contains $2^{-k}$ and the other does not. The latter's entire possible tail is at most $\sum_{j>k}2^{-j}=2^{-k}$, while the former has $2^{-k}$ plus its own nonempty tail, so the sums cannot agree.

Solved by gpt-5.6-sol high.

= b
{parent=7e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

For $M=\max\{|s|,|t|,1\}$, factorization of a difference of powers gives
$$
|s^j-t^j|
=|s-t|\left|\sum_{l=0}^{j-1}s^{j-1-l}t^l\right|
\leq jM^{j-1}|s-t|.
$$
Consequently the <polynomial> $f$ obeys the <Lipschitz bound>
$$
|f(s)-f(t)|\leq A M^{d-1}|s-t|,
\qquad
A=\sum_{j=1}^dj|\alpha_j|.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Since $p/q$ is not a root, $q^df(p/q)$ is a nonzero <integer>, and hence
$$
|f(p/q)|\geq q^{-d}.
$$
Moreover $0<\theta-p/q<1$ implies $|p/q|<|\theta|+1$. Applying part (i) to $s=\theta$, $t=p/q$, and using $f(\theta)=0$ gives
$$
q^{-d}\leq A(|\theta|+1)^{d-1}\left|\theta-\frac pq\right|.
$$
Rearrangement proves
$$
\boxed{\theta-\frac pq\geq
\frac1A\left(\frac1{|\theta|+1}\right)^{d-1}\frac1{q^d}}.
$$
This is the core estimate behind the <Liouville approximation theorem>.

Solved by gpt-5.6-sol high.

= c
{parent=7e}
{scope}

= Solution
{parent=c}

Let
$$
\alpha_N=\sum_{n=1}^N2^{-n!}=\frac{p_N}{q_N},
\qquad q_N=2^{N!}.
$$
The tail satisfies
$$
0<\alpha-\alpha_N<2\,2^{-(N+1)!}.
$$
For every fixed positive <integer> $d$, this is smaller than $q_N^{-d}$ for all sufficiently large $N$.

The binary expansion has $1$ digits at factorial positions and arbitrarily long blocks of zeros, but it is neither eventually zero nor eventually periodic. Since a <rational number> has an <eventually periodic positional expansion of a rational number>[eventually periodic binary expansion], $\alpha$ is <irrational number>[irrational]. If it were an <algebraic number> of degree $d$, part (b), or equivalently the <Liouville approximation theorem>, would give a fixed positive multiple of $q^{-d}$ as a lower bound for all rational approximations. The fractions $p_N/q_N$ violate that bound. Thus $\alpha$ is a <transcendental number>.

Solved by gpt-5.6-sol high.

= d
{parent=7e}
{scope}

= i
{parent=d}
{scope}

= Solution
{parent=i}

This can be false. For any nonzero <transcendental number> $\beta$, the number $\gamma=1/\beta$ is also transcendental, since otherwise its reciprocal would be <algebraic number>[algebraic]. Yet $\beta\gamma=1$ is algebraic.

Solved by gpt-5.6-sol high.

= ii
{parent=d}
{scope}

= Solution
{parent=ii}

This is always true for every positive integer $n$. If $\beta^n$ were algebraic, then $\beta$ would be a root of the polynomial $X^n-\beta^n$ over the field of <algebraic number>[algebraic numbers]. By <transitivity of algebraic extensions>, $\beta$ would then be algebraic over $\mathbb Q$, contradicting its transcendence.

Solved by gpt-5.6-sol high.

= 8E
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= a
{parent=8e}
{scope}

= Solution
{parent=a}

Write the sets as $A_1,A_2,\ldots$, and list each nonempty countable set as $A_i=\{a_{i1},a_{i2},\ldots\}$. Enumerate the pairs $(i,j)\in\mathbb N^2$ along successive diagonals, as in the <Cantor pairing function>, and output $a_{ij}$ when it has not appeared before. Every member of $\bigcup_iA_i$ eventually appears, proving that a <countable union of countable sets> is countable.

Solved by gpt-5.6-sol high.

= b
{parent=8e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Suppose all functions $\mathbb N\to\mathbb N$ could be listed as $f_1,f_2,\ldots$. Define
$$
g(n)=f_n(n)+1.
$$
Then $g$ differs from $f_n$ at $n$ for every $n$, so it is absent from the list. This <Cantor's diagonal argument> proves that $\mathbb N^{\mathbb N}$ is an <uncountable set>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

$A$ is uncountable. For each <binary sequence> $(\varepsilon_j)$, define
$$
f(n)=1+\sum_{j=1}^n\varepsilon_j.
$$
This is a nondecreasing function, and different binary sequences give different functions, so the uncountable set of binary sequences injects into $A$.

$B$ is countable. Every nonincreasing sequence of positive integers can decrease only finitely many times and is therefore eventually constant. Such a sequence is specified by a finite sequence of positive integers, and the set of all finite integer sequences is a <countable union of countable sets>. Hence $B$ is countable.

Solved by gpt-5.6-sol high.

= c
{parent=8e}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

For each $N$, only finitely many <permutation>[permutations] fix every $j>N$: they are precisely the permutations of $\{1,\ldots,N\}$. Thus
$$
C=\bigcup_{N\geq1}S_N
$$
is a countable union of finite sets and is countable.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

For every binary sequence $(\varepsilon_n)$, independently fix the pair $(2n-1,2n)$ when $\varepsilon_n=0$ and swap its two entries when $\varepsilon_n=1$. The resulting map is a <permutation> with $|\sigma(j)-j|\leq1$, and different binary sequences produce different permutations. Therefore $D$ is uncountable.

In fact these are all the possibilities: if $\sigma(j)=j+1$, bijectivity and the displacement bound force $\sigma(j+1)=j$, while a point not in such an adjacent transposition is fixed.

Solved by gpt-5.6-sol high.

= 9C
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= Solution
{parent=9C}

The <work> done along the trajectory is
$$
W=\int_{t_a}^{t_b}F(x(t),t)\dot x(t)\,dt
=\int_a^bF(x,t(x))\,dx.
$$
By <Newton's second law>, $F=m\ddot x$, so
$$
W=m\int_{t_a}^{t_b}\ddot x\dot x\,dt
=\left[\frac12m\dot x^2\right]_{t_a}^{t_b}.
$$
This is the <work-energy theorem>: the work equals the change in <kinetic energy>.

A force is <conservative force>[conservative] when its work between two points is independent of the path, equivalently when it is the negative gradient of a <potential energy>. Here a potential with $V(0)=0$ is
$$
V(x)=F_0\lambda(1-e^{-|x|/\lambda}).
$$
Conservation of <energy> for $x>0$ gives
$$
\frac12mv^2+F_0\lambda(1-e^{-x/\lambda})=\frac12mv_0^2.
$$
Thus
$$
v(x)=\sqrt{v_0^2+v_e^2(e^{-|x|/\lambda}-1)},
\qquad
\boxed{v_e=\sqrt{\frac{2F_0\lambda}{m}}}.
$$

If $v_0>v_e$, the particle escapes with asymptotic speed $\sqrt{v_0^2-v_e^2}$; the graph decreases from $v_0$ to this horizontal asymptote. At the <escape velocity> $v_0=v_e$, it decreases as $v_e e^{-x/(2\lambda)}$ and approaches zero only at infinity. If $v_0<v_e$, it reaches the <turning point>
$$
x_{\max}=-\lambda\log\left(1-\frac{v_0^2}{v_e^2}\right),
$$
where $v=0$, then returns on the negative branch of the same curve and oscillates symmetrically about the origin.

For oscillation put $w=1-v_0^2/v_e^2$. One quarter of the <period of an oscillation>[period] is
$$
\int_0^{x_{\max}}\frac{dx}{v(x)}
=\frac{\lambda}{v_e}\int_w^1\frac{du}{u\sqrt{u-w}}
=\frac{2\lambda}{v_e\sqrt w}\cos^{-1}\sqrt w,
$$
where $u=e^{-x/\lambda}$. Therefore
$$
\boxed{T=\frac{8\lambda}{v_e\sqrt{1-v_0^2/v_e^2}}
\cos^{-1}\sqrt{1-v_0^2/v_e^2}}.
$$

Solved by gpt-5.6-sol high.

= 10C
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= a
{parent=10c}
{scope}

= Solution
{parent=a}

The <torque> about the origin is
$$
\tau=r\times F=-\frac{km}{r^3}r\times r=0.
$$
Since $\dot L=\tau$, the <angular momentum> $L=mr\times\dot r$ is conserved. The force is the negative gradient of the <inverse-square potential>
$$
V(r)=-\frac{km}{r},
$$
so the conserved <energy> is
$$
E=\frac12m|\dot r|^2-\frac{km}{r}.
$$
Directly, $\dot E=\dot r\mathbin{\cdot}(m\ddot r+km r/r^3)=0$.

Solved by gpt-5.6-sol high.

= b
{parent=10c}
{scope}

= Solution
{parent=b}

Motion under a <central force> is planar. In <plane polar coordinates>, the radial component of the <acceleration> is $\ddot r-r\dot\theta^2$, so
$$
\ddot r-r\dot\theta^2=-\frac{k}{r^2}.
$$
The conserved specific <angular momentum> is $h=r^2\dot\theta$. Substitution of $\dot\theta=h/r^2$ yields
$$
\boxed{\ddot r=-\frac{k}{r^2}+\frac{h^2}{r^3}}.
$$
The last term is the centrifugal part of the radial <effective potential>.

Solved by gpt-5.6-sol high.

= c
{parent=10c}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

At the <periapsis> $r_1$ and <apoapsis> $r_2$, the given <Kepler orbit> satisfies
$$
r_1=\frac{h^2}{k(1+e)},
\qquad
r_2=\frac{h^2}{k(1-e)}.
$$
Hence
$$
e=\frac{r_2-r_1}{r_1+r_2},
\qquad
h^2=\frac{2kr_1r_2}{r_1+r_2}.
$$
Therefore
$$
\boxed{h=\sqrt{\frac{2kr_1r_2}{r_1+r_2}}},
\qquad \boxed{C=2}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

A circular orbit of radius $r$ has speed $v_c(r)=\sqrt{k/r}$. On the transfer <ellipse>, the velocity is tangential at both apsides and has magnitude $h/r$. The first prograde burst therefore changes the speed by
$$
\boxed{\Delta v_1
=\sqrt{\frac{k}{r_1}}
\left(\sqrt{\frac{2r_2}{r_1+r_2}}-1\right)}.
$$
At $r_2$ the satellite must again accelerate in the direction of motion to circularize, by
$$
\boxed{\Delta v_2
=\sqrt{\frac{k}{r_2}}
\left(1-\sqrt{\frac{2r_1}{r_1+r_2}}\right)}.
$$
These are the two impulses of a <Hohmann transfer>.

Solved by gpt-5.6-sol high.

= iii
{parent=c}
{scope}

= Solution
{parent=iii}

The transfer ellipse has <semi-major axis> and <semi-minor axis>
$$
a=\frac{r_1+r_2}{2},
\qquad
b=a\sqrt{1-e^2}=\sqrt{r_1r_2}.
$$
Conservation of angular momentum gives constant <areal velocity> $h/2$. The satellite sweeps half the ellipse, whose area is $\pi ab/2$, so
$$
t_2-t_1=\frac{\pi ab/2}{h/2}=\frac{\pi ab}{h}.
$$
Using part (i),
$$
\boxed{t_2-t_1
=\frac{\pi(r_1+r_2)^{3/2}}{2\sqrt{2k}}}
=\pi\sqrt{\frac{a^3}{k}}.
$$
This is half the orbital period from <Kepler third law>[Kepler's third law].

Solved by gpt-5.6-sol high.

= 11C
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= i
{parent=11c}
{scope}

= Solution
{parent=i}

Let $e_i'(t)$ be a <Cartesian basis> fixed in the rotating frame. A vector $A=A_i'e_i'$ then satisfies
$$
\left(\frac{dA}{dt}\right)_S
=\dot A_i'e_i'+A_i'\dot e_i'
=\left(\frac{dA}{dt}\right)_{S'}+\omega\times A,
$$
because a <derivative of a body-fixed basis vector> is $\dot e_i'=\omega\times e_i'$. Applying this transport formula twice to $r$ gives, for constant $\omega$,
$$
a_S=a_{S'}+2\omega\times v_{S'}+\omega\times(\omega\times r).
$$
The second and third terms correspond to <Coriolis acceleration> and <centrifugal acceleration> in the rotating description.

For the bead, use cylindrical unit vectors $(e_r,e_\phi,e_z)$ with $e_z$ upward and
$$
r=R(\sin\theta\,e_r-\cos\theta\,e_z),
\qquad \dot e_r=\omega e_\phi.
$$
Projection of $m\ddot r=-mg e_z+N$ along the wire tangent $\cos\theta\,e_r+\sin\theta\,e_z$ eliminates the smooth-wire <normal force> $N$ and gives
$$
\boxed{\ddot\theta=(\omega^2\cos\theta-g/R)\sin\theta}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=11c}
{scope}

= Solution
{parent=ii}

The downward and upward positions $\theta=0,\pi$ are equilibria. Linearization about $\theta=0$ gives
$$
\ddot\eta=(\omega^2-g/R)\eta,
$$
so the downward equilibrium is stable for $\omega^2<g/R$ and unstable for $\omega^2>g/R$; at equality it is marginal at linear order. At $\theta=\pi$, the linear coefficient is $\omega^2+g/R>0$, so the upward equilibrium is unstable.

When $\omega^2\geq g/R$, there are also the two equilibria
$$
\theta=\pm\cos^{-1}\left(\frac{g}{R\omega^2}\right)
\pmod{2\pi}.
$$
At either nonvertical equilibrium, differentiation of the right-hand side gives $-\omega^2\sin^2\theta<0$, so both are stable when distinct. This is a <pitchfork bifurcation of a rotating hoop bead> at $\omega^2=g/R$.

Solved by gpt-5.6-sol high.

= iii
{parent=11c}
{scope}

= Solution
{parent=iii}

The inertial <acceleration> obtained by differentiating the position used in part (i) is
$$
\ddot r
=R(\ddot\theta\cos\theta-\dot\theta^2\sin\theta-\omega^2\sin\theta)e_r
+2R\omega\dot\theta\cos\theta e_\phi
+R(\ddot\theta\sin\theta+\dot\theta^2\cos\theta)e_z.
$$
Since $N=m\ddot r+mg e_z$, substitution of the equation of motion gives the wire's force on the bead as
$$
\boxed{
N=m\bigl(R\dot\theta^2+R\omega^2\sin^2\theta+g\cos\theta\bigr)
(-\sin\theta\,e_r+\cos\theta\,e_z)
+2mR\omega\dot\theta\cos\theta\,e_\phi }.
$$
The first component is normal to the circle within its plane; the second is perpendicular to the rotating plane.

Solved by gpt-5.6-sol high.

= 12C
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= Solution
{parent=12C}

With <Lorentz factor> $\gamma=(1-v^2/c^2)^{-1/2}$, the relativistic <momentum> and <four-momentum> are
$$
p=\gamma mv,
\qquad
P^\mu=(\gamma mc,\gamma mv).
$$
The Newtonian formula $a=F/m$ fails because $p$ is not $mv$: both the magnitude and direction of $v$ affect $dp/dt$. Since
$$
\dot\gamma=\frac{\gamma^3}{c^2}v\mathbin{\cdot}a,
$$
the <relativistic force> is
$$
F=\frac{dp}{dt}
=m\gamma\left(a+\frac{\gamma^2}{c^2}(v\mathbin{\cdot}a)v\right).
$$
Taking the dot product with $v$ gives $F\mathbin{\cdot}v=m\gamma^3v\mathbin{\cdot}a$, and substitution yields the inverse relation
$$
\boxed{a=\frac{F}{m\gamma}-\frac{(F\mathbin{\cdot}v)v}{m\gamma c^2}}.
$$
This is the requested sum of a vector parallel to $F$ and one parallel to $v$.

In the constant <electric field>, $dp/dt=qE$ and the particle starts with $p(0)=0$, so $p=qEt$. The <relativistic energy-momentum relation> implies
$$
v=\frac{c^2p}{\sqrt{m^2c^4+c^2|p|^2}}
=\boxed{\frac{qEt/m}{\sqrt{1+q^2|E|^2t^2/(m^2c^2)}}}.
$$
Thus the speed increases monotonically but remains <subluminal speed>[subluminal], and $v$ tends to $c$ in the direction of $qE$ as $t\to\infty$.

Solved by gpt-5.6-sol high.