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1998 Paper 3 Q8
D: 1700.0 B: 1484.0

  1. [(i)] Consider the sphere of radius \(a\) and centre the origin. %Show that the line through the point with position vector %\({\bf b}\) and parallel to a unit %vector \({\bf m}\) intersects the sphere at two points if %$$ %a^2 > {\bf b}.{\bf b} -({\bf b}.{\bf m})^2 \,. %$$ %What is the corresponding condition for there to be precisely one %point of intersection? %If this point has position vector \({\bf p}\), show that the line %is perpendicular to \({\bf p}\).
  2. Show that the line \({\bf r} ={\bf b} + \lambda {\bf m}\), where \(\bf m\) is a unit vector, intersects the sphere \({\bf r}\cdot {\bf r} = a^2\) at two points if $$ a^2 > {\bf b}\cdot{\bf b} -({\bf b}\cdot{\bf m})^2 \,. $$ Write down the corresponding condition for there to be precisely one point of intersection. If this point has position vector \({\bf p}\), show that \({\bf m}\cdot{\bf p}=0\).
  3. Now consider a second sphere of radius \(a\) and a plane perpendicular to a unit vector~\({\bf n}\). The centre of the sphere has position vector \({\bf d}\) and the minimum distance from the origin to the plane is \(l\). What is the condition for the plane to be tangential to this second sphere?
  4. Show that the first and second spheres intersect at right angles ({\em i.e.\ }the two radii to each point of intersection are perpendicular) if $$ {\bf d}\cdot{\bf d} = 2 a^2 \,. $$

1998 Paper 3 Q9
D: 1700.0 B: 1484.0

A uniform right circular cone of mass \(m\) has base of radius \(a\) and perpendicular height \(h\) from base to apex. Show that its moment of inertia about its axis is \({3\over 10} ma^2\), and calculate its moment of inertia about an axis through its apex parallel to its base. \newline[{\em Any theorems used should be stated clearly.}] The cone is now suspended from its apex and allowed to perform small oscillations. Show that their period is $$ 2\pi\sqrt{ 4h^2 + a^2\over 5gh} \,. $$ \newline[{\em You may assume that the centre of mass of the cone is a distance \({3\over 4}h\) from its apex.}]

1998 Paper 3 Q10
D: 1700.0 B: 1500.0

Two identical spherical balls, moving on a horizontal, smooth table, collide in such a way that both momentum and kinetic energy are conserved. Let \({\bf v}_1\) and \({\bf v}_2\) be the velocities of the balls before the collision and let \({\bf v}'_1\) and \({\bf v}'_2\) be the velocities of the balls after the collision, where \({\bf v}_1\), \({\bf v}_2\), \({\bf v}'_1\) and \({\bf v}'_2\) are two-dimensional vectors. Write down the equations for conservation of momentum and kinetic energy in terms of these vectors. Hence show that their relative speed is also conserved. Show that, if one ball is initially at rest but after the collision both balls are moving, their final velocities are perpendicular. Now suppose that one ball is initially at rest, and the second is moving with speed \(V\). After a collision in which they lose a proportion \(k\) of their original kinetic energy (\(0\le k\le 1\)), the direction of motion of the second ball has changed by an angle \(\theta\). Find a quadratic equation satisfied by the final speed of the second ball, with coefficients depending on \(k\), \(V\) and \(\theta\). Hence show that \(k\le \frac{1}{2}\).


Solution: \begin{align*} \text{COM}: && \mathbf{v}_1+\mathbf{v}_2 &= \mathbf{v}_1'+\mathbf{v}_2' \tag{1}\\ \text{COE}: && \mathbf{v}_1\cdot\mathbf{v}_1+\mathbf{v}_2\cdot\mathbf{v}_2 &= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' \tag{2} \\ \\ (1): && (\mathbf{v}_1+\mathbf{v}_2 )\cdot(\mathbf{v}_1+\mathbf{v}_2 ) &= (\mathbf{v}_1'+\mathbf{v}_2' )\cdot(\mathbf{v}_1'+\mathbf{v}_2' ) \\ \Rightarrow && \mathbf{v}_1 \cdot \mathbf{v}_2 &= \mathbf{v}_1'\cdot \mathbf{v}_2' \\ && \text{Initial relative speed}^2 &= |\mathbf{v}_1 - \mathbf{v}_2|^2 \\ &&&= (\mathbf{v}_1 - \mathbf{v}_2) \cdot (\mathbf{v}_1 - \mathbf{v}_2) \\ &&&= \mathbf{v}_1\cdot \mathbf{v}_1 - 2 \mathbf{v}_1\cdot \mathbf{v}_2 + \mathbf{v}_2\cdot \mathbf{v}_2 \\ &&&= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' -2 \mathbf{v}_1\cdot\mathbf{v}_2\\ &&&= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' -2 \mathbf{v}_1'\cdot\mathbf{v}_2'\\ &&&= | \mathbf{v}_1'-\mathbf{v}_2'|^2 \\ &&&= \text{Final relative speed}^2 \end{align*} Since \(\mathbf{v}_1 \cdot 0 = 0\) we must have \(\mathbf{v}_1'\cdot\mathbf{v}_2' = \mathbf{v}_1\cdot0 = 0\) therefore their final velocities are perpendicular. We now must have \begin{align*} \text{COM}: && \mathbf{v}_1+\mathbf{v}_2 &= \mathbf{v}_1'+\mathbf{v}_2' \tag{3}\\ \Delta\text{E}: && (1-k)(\mathbf{v}_1\cdot\mathbf{v}_1+\mathbf{v}_2\cdot\mathbf{v}_2) &= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' \tag{4} \\ \\ && 0 + \mathbf{v}_2 &= \mathbf{v}_1' + \mathbf{v}_2' \\ \Rightarrow && V^2 &= ( \mathbf{v}_1' + \mathbf{v}_2' ) \cdot ( \mathbf{v}_1' + \mathbf{v}_2' ) \\ &&&= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' +2 \mathbf{v}_1'\cdot \mathbf{v}_2' \\ &&&= (1-k)V^2 + 2 (\mathbf{v}_2-\mathbf{v}_2') \cdot \mathbf{v}_2' \\ &&&= (1-k)V^2 + 2 \mathbf{v}_2 \cdot \mathbf{v}_2'-2\mathbf{v}_2'\cdot \mathbf{v}_2' \\ &&&= (1-k)V^2 + 2Vx \cos \theta - 2x^2 \\ \Rightarrow && 0 &= -kV^2 + 2Vx \cos \theta -2x^2 \\ \Delta \geq 0: && 0 &\leq 4V^2 \cos^2 \theta -8kV^2 \\ \Rightarrow && k &\leq \frac12\cos^2\theta \leq \frac12 \end{align*}

1998 Paper 3 Q11
D: 1700.0 B: 1500.0

Consider a simple pendulum of length \(l\) and angular displacement \(\theta\), which is {\bf not} assumed to be small. Show that $$ {1\over 2}l \left({\d\theta\over \d t}\right)^2 = g(\cos\theta -\cos\gamma)\,, $$ where \(\gamma\) is the maximum value of \(\theta\). Show also that the period \(P\) is given by $$ P= 2 \sqrt{l\over g} \int_0^\gamma \left( \sin^2(\gamma/2)-\sin^2(\theta/2) \right)^{-{1\over 2}} \,\d\theta \,. $$ By using the substitution \(\sin(\theta/2)=\sin(\gamma/2) \sin\phi\), and then finding an approximate expression for the integrand using the binomial expansion, show that for small values of \(\gamma\) the period is approximately $$ 2\pi \sqrt{l\over g} \left(1+{\gamma^2\over 16}\right) \,. $$

1998 Paper 3 Q12
D: 1700.0 B: 1482.8

The mountain villages \(A,B,C\) and \(D\) lie at the vertices of a tetrahedron, and each pair of villages is joined by a road. After a snowfall the probability that any road is blocked is \(p\), and is independent of the conditions of any other road. The probability that, after a snowfall, it is possible to travel from any village to any other village by some route is \(P\). Show that $$ P =1- p^2(6p^3-12p^2+3p+4). $$ %In the case \(p={1\over 3}\) show that this probability is \({208 \over 243}\).

1998 Paper 3 Q13
D: 1700.0 B: 1500.0

Write down the probability of obtaining \(k\) heads in \(n\) tosses of a fair coin. Now suppose that \(k\) is known but \(n\) is unknown. A maximum likelihood estimator (MLE) of \(n\) is defined to be a value (which must be an integer) of \(n\) which maximizes the probability of \(k\) heads. A friend has thrown a fair coin a number of times. She tells you that she has observed one head. Show that in this case there are two MLEs of the number of tosses she has made. She now tells you that in a repeat of the exercise she has observed \(k\) heads. Find the two MLEs of the number of tosses she has made. She next uses a coin biased with probability \(p\) (known) of showing a head, and again tells you that she has observed \(k\) heads. Find the MLEs of the number of tosses made. What is the condition for the MLE to be unique?


Solution: \begin{align*} && \mathbb{P}(k \text{ heads} | n\text{ tosses}) &= \binom{n}k 2^{-n} \\ && \mathbb{P}(1 \text{ head} | n\text{ tosses}) &= n2^{-n} \\ \Rightarrow && \frac{ \mathbb{P}(1 \text{ head} | n+1\text{ tosses}) }{ \mathbb{P}(1 \text{ head} | n\text{ tosses}) } &= \frac{n+1}{2n} \end{align*} Which is less than \(1\) unless \(n \geq 1\). Therefore the MLE is \(n = 1\) or \(n= 2\). \begin{align*} \frac{ \mathbb{P}(k \text{ head} | n+1\text{ tosses}) }{ \mathbb{P}(k \text{ head} | n\text{ tosses}) } &= \frac{\binom{n+1}{k}}{2 \binom{n}{k}} \\ &= \frac{(n+1)!(n-k)!}{2n!(n+1-k)!} \\ &= \frac{n+1}{2(n+1-k)} \end{align*} This is less than or equal to \(1\) if \(n+1 = 2(n+1-k) \Leftrightarrow n= 2k-1\), therefore the MLEs are \(2k-1\) and \(2k\). If the coin is biased, we have \begin{align*} && \frac{ \mathbb{P}(k \text{ head} | n+1\text{ tosses}) }{ \mathbb{P}(k \text{ head} | n\text{ tosses}) } &= \frac{\binom{n+1}{k}p^kq^{n+1-k}}{\binom{n}{k}p^kq^{n-k}} \\ &&&= \frac{n+1}{(n+1-k)}q \\ \\ && 1 & \geq \frac{n+1}{(n+1-k)}q \\ \Leftrightarrow && (n+1)(1-q) &\geq k \\ \Leftrightarrow && n+1 & \geq \frac{k}{p} \end{align*} Therefore the probability is increasing until \(n+1 \geq \frac{k}{p}\). If \(\frac{k}p\) is an integer the MLEs are \(\frac{k}{p}-1\) and \(\frac{k}p\), otherwise it is \(\lfloor \frac{k}{p} \rfloor\) and the MLE is unique.

1998 Paper 3 Q14
D: 1700.0 B: 1500.0

A hostile naval power possesses a large, unknown number \(N\) of submarines. Interception of radio signals yields a small number \(n\) of their identification numbers \(X_i\) (\(i=1,2,...,n\)), which are taken to be independent and uniformly distributed over the continuous range from \(0\) to \(N\). Show that \(Z_1\) and \(Z_2\), defined by $$ Z_1 = {n+1\over n} {\max}\{X_1,X_2,...,X_n\} \hspace{0.3in} {\rm and} \hspace{0.3in} Z_2 = {2\over n} \sum_{i=1}^n X_i \;, $$ both have means equal to \(N\). Calculate the variance of \(Z_1\) and of \(Z_2\). Which estimator do you prefer, and why?

1997 Paper 1 Q1
D: 1484.0 B: 1500.0

Show that you can make up 10 pence in eleven ways using 10p, 5p, 2p and 1p coins. In how many ways can you make up 20 pence using 20p, 10p, 5p, 2p and 1p coins? You are reminded that no credit will be given for unexplained answers.


Solution: Writing out the possibilities in order of the largest coin used (and then second largest and so-on): \begin{align*} && 10 &= 10 \\ &&&= 5 + 5 \\ &&&= 5 + 2 + 2 + 1 \\ &&&= 5 + 2 + 1 + 1 + 1 \\ &&&= 5 + 1 + 1 + 1 + 1 + 1\\ &&&= 2 + 2 + 2 + 2 + 2 = 5 \cdot 2\\ &&&= 4 \cdot 2 + 2 \cdot 1 \\ &&&= 3 \cdot 2 + 4 \cdot 1\\ &&&= 2 \cdot 2 + 6\cdot 1\\ &&&= 1 \cdot 2 + 8\cdot 1 \\ &&&= 10 \cdot 1 \end{align*} For 20p, we have \begin{align*} && 20 &= 20 \\ &&&= 10 + \text{all 11 ways} \\ &&&= 4\cdot 5 \\ &&&= 3\cdot 5 +\text{3 ways} \\ &&&= 2\cdot5 + \text{6 ways} \\ &&&= 1\cdot 5 + \text{8 ways} \\ &&&= k\cdot 2 + (20-2k)\cdot 1 \quad \text{11 ways} \end{align*} ie 41 ways

1997 Paper 1 Q2
D: 1516.0 B: 1484.0

  1. If \[{\mathrm f}(x)=\tan^{-1}x+\tan^{-1}\left(\frac{1-x}{1+x}\right),\] find \({\mathrm f}'(x)\). Hence, or otherwise, find a simple expression for \({\mathrm f}(x)\).
  2. Suppose that \(y\) is a function of \(x\) with \(0 < y < (\pi/2)^{1/2}\) and \[x=y\sin y^{2}\] for \(0 < x < (\pi/2)^{1/2}\). Show that (for this range of \(x\)) \[\frac{{\mathrm d}y}{{\mathrm d}x}= \frac{y}{x+2y^2\sqrt{y^{2}-x^{2}}}.\]


Solution:

  1. \begin{align*} && f(x)&=\tan^{-1}x+\tan^{-1}\left(\frac{1-x}{1+x}\right) \\ \Rightarrow && f'(x) &= \frac{1}{1+x^2} + \frac{1}{1+\l \frac{1-x}{1+x} \r^2} \cdot \l \frac{-2}{(1+x)^2}\r \\ &&&= \frac1{1+x^2}- \frac{2}{(1+x)^2+(1-x)^2} \\ &&&= \frac1{1+x^2} - \frac{2}{2+2x^2} \\ &&&= 0 \end{align*} Therefore $f(x) = \begin{cases} c_1 & \text{if } x < -1 \\ c_2 & \text{if } x > -1 \end{cases}$ \(f(0) = \tan^{-1} 0 + \tan^{-1} 1 = \frac{\pi}{4}\) \(\lim_{x \to \infty} f(x) = -\frac{\pi}{2} + \tan^{-1} -1 = -\frac{3\pi}{4}\) therefore $f(x) = \begin{cases} -\frac{3\pi}{4}& \text{if } x < -1 \\ \frac{\pi}{4} & \text{if } x > -1 \end{cases}$
  2. \begin{align*} && x &= y \sin y^2 \\ \Rightarrow && \frac{\d x}{\d y} &= \sin y^2 + 2y^2 \cos y^2 \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{1}{\sin y^2+2y^2 \cos y^2} \\ &&&=\frac{1}{\frac{x}{y}+2y^2 \sqrt{1-\sin^2y^2}} \\ &&&= \frac{y}{x + 2y^3 \sqrt{1-\frac{x^2}{y^2}}} \\ &&&= \frac{y}{x+2y^2 \sqrt{y^2-x^2}} \end{align*}

1997 Paper 1 Q3
D: 1484.0 B: 1501.4

Let \(a_{1}=3\), \(a_{n+1}=a_{n}^{3}\) for \(n\geqslant 1\). (Thus \(a_{2}=3^{3}\), \(a_{3}=(3^{3})^{3}\) and so on.)

  1. What digit appears in the unit place of \(a_{7}\)?
  2. Show that \(a_{7}\geqslant 10^{100}\).
  3. What is \(\dfrac{a_{7}+1}{2a_{7}}\) correct to two places of decimals? Justify your answer.


Solution:

  1. Notice that \(a_n = 3^{3^{n-1}}\) in particular, \(a_7 = 3^{3^6}\). Using Fermat's little theorem, we can see that \(3^4 \equiv 1 \pmod{5}\) and so we need to figure out \(3^6 \pmod{4}\), which is clearly \(1\). Therefore \(3^{3^6} \equiv 3^{4k+1} \equiv 3 \pmod{5}\). Therefore the units digit is \(3\).
  2. Notice that \(3^5 > 100\) and \(3^3 > 10\). Therefore \begin{align*} a_7 &= 3^{3^6} \\ &= (3^3)^{3^5} \\ &> 10^{3^5} \\ &> 10^{100} \end{align*}
  3. \begin{align*} \frac{a_7+1}{2a_7} &= \frac12 + \frac1{2a_7} \\ &= 0.5 + 0.\underbrace{000\cdots}_{\text{at least }99\text{ zeros}} \\ &= 0.50 \end{align*} Since \(a_7 > 10^{100}, \, \frac{1}{2a_7} < 10^{-100}\)