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1989 Paper 1 Q11
A shot-putter projects a shot at an angle \(\theta\) above the horizontal, releasing it at height \(h\) above the level ground, with speed \(v\). Show that the distance \(R\) travelled horizontally by the shot from its point of release until it strikes the ground is given by \[ R=\frac{v^{2}}{2g}\sin2\theta\left(1+\sqrt{1+\frac{2hg}{v^{2}\sin^{2}\theta}}\right). \] The shot-putter's style is such that currently \(\theta=45^{\circ}\). Determine (with justification) whether a small decrease in \(\theta\) will increase \(R\). [Air resistance may be neglected.]
Solution: Notice that \(u_x = v \cos \theta, u_y = v \sin \theta\). We are interested in the time taken for the shot to hit the ground. \(-h = u_y t -\frac12 g t^2\) since our distance will be \(v \cos \theta \cdot t\). Solving this quadratic for \(t\) we obtain: \begin{align*} && 0 &= h + v \sin \theta \cdot t - \frac12 g \cdot t^2 \\ \Rightarrow && t_\pm &= \frac{-v \sin \theta \pm \sqrt{v^2 \sin^2 \theta+2hg}}{-g} \\ \Rightarrow && t_- &= \frac{v \sin \theta+v\sin \theta \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}}}{g} \\ \Rightarrow && v \cos \theta t_{-} &= \frac{v^2}{g} \cos \theta \sin \theta \left( 1 + \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}} \right) \\ &&&= \frac{v^2}{2g} \sin 2 \theta \left( 1 + \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}} \right) \end{align*} Differentiating \(R\) wrt to \(\theta\) at \(\frac{\pi}{4}\) we obtain: \begin{align*} \frac{\d R}{\d \theta} &= \frac{v^2}{2g} \left (2 \cos 2 \theta + 2 \cos 2 \theta \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}} + \sin 2\theta \left ( 1 + \frac{2hg}{v^2 \sin^2 \theta}\right)^{-\frac12} \frac12 \frac{2hg} {v^2}(-2) \frac{\cos \theta}{\sin^3 \theta}\right) \\ \frac{\d R}{\d \theta} \biggr \rvert_{\theta = \frac{\pi}{4}} &=\frac{v^2}{2g}\left(0+0- 4\left ( 1 + \frac{4hg}{v^2 }\right)^{-\frac12} \frac{hg} {v^2} \right) \\ &< 0 \end{align*} Therefore, since \(R\) is locally decreasing in \(\theta\) he should reduce his angle of projection slightly.
1989 Paper 1 Q12
A regular tetrahedron \(ABCD\) of mass \(M\) is made of 6 identical uniform rigid rods, each of length \(2a.\) Four light elastic strings \(XA,XB,XC\) and \(XD\), each of natural length \(a\) and modulus of elasticity \(\lambda,\) are fastened together at \(X\), the other end of each string being attached to the corresponding vertex. Given that \(X\) lies at the centre of mass of the tetrahedron, find the tension in each string. The tetrahedron is at rest on a smooth horizontal table, with \(B,C\) and \(D\) touching the table, and the ends of the strings at \(X\) attached to a point \(O\) fixed in space. Initially the centre of mass of the tetrahedron coincides with \(O.\) Suddenly the string \(XA\) breaks, and the tetrahedron as a result rises vertically off the table. If the maximum height subsequently attained is such that \(BCD\) is level with the fixed point \(O,\) show that (to 2 significant figures) \[ \frac{Mg}{\lambda}=0.098. \]
Solution:
1989 Paper 1 Q13
A uniform ladder of mass \(M\) rests with its upper end against a smooth vertical wall, and with its lower end on a rough slope which rises upwards towards the wall and makes an angle of \(\phi\) with the horizontal. The acute angle between the ladder and the wall is \(\theta\). If the ladder is in equilibrium, show that \(N\) and \(F\), the normal reaction and frictional force at the foot of the ladder are given by \[ N=Mg\left(\cos\phi-\frac{\tan\theta\sin\phi}{2}\right), \] \[ F=Mg\left(\sin\phi+\frac{\tan\theta\cos\phi}{2}\right). \] If the coefficient of friction between the ladder and the slope is \(2\), and \(\phi=45^{\circ}\), what is the largest value of \(\theta\) for which the ladder can rest in equilibrium?
Solution: \begin{align*} \overset{\curvearrowleft}{X}: && 0&= \frac{l}{2} Mg\sin \theta - l R_1 \cos \theta \\ \Rightarrow && R_1 &= \frac12 \tan \theta Mg \\ \text{N2}(\uparrow): && 0 &= R\cos \phi +F \sin \phi - Mg \\ \text{N2}(\rightarrow):&& 0&=R_1-F \cos \phi + R \sin \phi \\ \Rightarrow && \frac12 \tan \theta Mg &= F \cos \phi- R \sin \phi \\ && Mg &= F \sin \phi +R \cos \phi \\ \Rightarrow && F &= Mg \left ( \sin \phi + \frac12 \tan \theta \cos \phi \right) \\ && N &= Mg \left (\cos \phi - \frac12 \tan \theta \sin \phi \right ) \end{align*} If \(\mu = 2\) and \(\phi = 45^{\circ}\), we must have \(F \leq 2 N\), so: \begin{align*} && Mg \left ( \sin \phi + \frac12 \tan \theta \cos \phi \right) &\leq 2 Mg \left (\cos \phi - \frac12 \tan \theta \sin \phi \right ) \\ \Rightarrow && 1 + \frac12 \tan \theta \leq 2-\tan \theta \\ \Rightarrow && \frac 32 \tan \theta \leq 1 \\ \Rightarrow && \tan \theta \leq \frac23 \\ \Rightarrow && \theta \leq \tan^{-1} \frac23 \end{align*}
1989 Paper 1 Q14
The prevailing winds blow in a constant southerly direction from an enchanted castle. Each year, according to an ancient tradition, a princess releases 96 magic seeds from the castle, which are carried south by the wind before falling to rest. South of the castle lies one league of grassy parkland, then one league of lake, then one league of farmland, and finally the sea. If a seed falls on land it will immediately grow into a fever tree. (Fever trees do not grow in water). Seeds are blown independently of each other. The random variable \(L\) is the distance in leagues south of the castle at which a seed falls to rest (either on land or water). It is known that the probability density function \(\mathrm{f}\) of \(L\) is given by \[ \mathrm{f}(x)=\begin{cases} \frac{1}{2}-\frac{1}{8}x & \mbox{ for }0\leqslant x\leqslant4,\\ 0 & \mbox{ otherwise.} \end{cases} \] What is the mean number of fever trees which begin to grow each year?
- The random variable \(Y\) is defined as the distance in leagues south of the castle at which a new fever tree grows from a seed carried by the wind. Sketch the probability density function of \(Y\), and find the mean of \(Y\).
- One year messengers bring the king the news that 23 new fever trees have grown in the farmland. The wind never varies, and so the king suspects that the ancient tradition have not been followed properly. Is he justified in his suspicions?
Solution: \begin{align*} \mathbb{P}(\text{fever tree grows}) &= \mathbb{P}(0 \leq L \leq 1) + \mathbb{P}(2 \leq L \leq 3) \\ &= \int_0^1 \frac12 -\frac18 x \d x + \int_2^3 \frac12 - \frac18 x \d x \\ &= \left [\frac12 x - \frac1{16}x^2 \right]_0^1+ \left [\frac12 x - \frac1{16}x^2 \right]_2^3 \\ &= \frac12 - \frac1{16}+\frac32-\frac9{16} - 1 + \frac{4}{16} \\ &= \frac58 \end{align*} The expected number of fever trees is just \(96 \cdot \frac58 = 60\).
- \(f_Y(t)\) must match the distribution for \(L\), but limited to the points we care about, therefore it should be:
$f_Y(t) = \begin{cases} ( \frac45 - \frac15t ) & \text{if } t \in [0,1]\cup[2,3] \\
0 & \text{otherwise} \end{cases}$
\begin{align*} \mathbb{E}(Y) &= \frac12 \cdot \frac15 (4 - \frac12)+\frac52 \cdot (1 - \frac15 (4 - \frac12)) \\ &= \frac12 \cdot \frac7{10} + \frac52 \cdot \frac3{10} \\ &= \frac{22}{20} \\ &= \frac{11}{10} \end{align*}
- Given the seeds are blown independently and the wind hasn't changed, it is reasonable to model the number of fever trees as \(B(96, \frac{5}{8})\), it is also acceptable to approximate this using a Normal distribution, ie \(N(60, 22.5)\), \(23\) is \(\frac{23-60}{\sqrt{22.5}}\) is a very negative number, so he should be extremely suspicious.
1989 Paper 1 Q15
I can choose one of three routes to cycle to school. Via Angle Avenue the distance is 5\(\,\)km, and I am held up at a level crossing for \(A\) minutes, where \(A\) is a continuous random variable uniformly distributed between \(0\) and 10. Via Bend Boulevard the distance is 4\(\,\)km, and I am delayed, by talking to each of \(B\) friends for 3\(\,\)minutes, for a total of \(3B\) minutes, where \(B\) is a random variable whose distribution is Poisson with mean 4. Via Detour Drive the distance should be only 2\(\,\)km, but in addition, due to never-ending road works, there are five places at each of which, with probability \(\frac{4}{5},\) I have to make a detour that increases the distance by 1\(\,\)km. Except when delayed by talking to friends or at the level crossing, I cycle at a steady 12\(\,\)km\(\,\)h\(^{-1}\). For each of the three routs, calculate the probability that a journey lasts at least 27 minutes. Each day I choose one of the three routes at random, and I am equally likely to choose any of the three alternatives. One day I arrive at school after a journey of at least 27 minutes. What is the probability that I came via Bend Boulevard? Which route should I use all the time: \begin{questionparts} \item if I wish my average journey time to be as small as possible; \item if I wish my journey time to be less than 32 minutes as often as possible? \end{questionpart} Justify your answers.
Solution: \(A \sim 5\cdot 5 + U[0,10]\) \(B \sim 4 \cdot 5 + 3 \textrm{Po}(4)\) \(C \sim 2 \cdot 5 + B(5, \frac{4}{5}) \cdot 5\) \begin{align*} && \mathbb{P}(A \leq 27) &= \mathbb{P}(U \leq 2) = 0.2 \\ && \mathbb{P}(B \leq 27) &= \mathbb{P}(3 \textrm{Po}(4) \leq 7) \\\ &&&= \mathbb{P}(Po(4) \leq 2) \\ &&&= e^{-4}(1 + 4 + \frac{4^2}{2}) \\ &&&= 0.23810\ldots \\ && \mathbb{P}(C \leq 27) &= \mathbb{P}(5 \cdot B(5,\tfrac45) \leq 17) \\ &&&= \mathbb{P}(B(5,\tfrac45) \leq 3) \\ &&&= \binom{5}{0} (\tfrac15)^5 + \binom{5}{1} (\tfrac45)(\tfrac 15)^4+ \binom{5}{2} (\tfrac45)^2(\tfrac 15)^3 + \binom{5}3 (\tfrac45)^3(\tfrac 15)^2+\\ &&&= 0.26272 \end{align*} \begin{align*} \mathbb{P}(\text{came via B} | \text{at least 27 minutes}) &= \frac{\mathbb{P}(\text{came via B and at least 27 minutes})}{\mathbb{P}(\text{at least 27 minutes})} \\ &= \frac{\frac13 \cdot 0.23810\ldots }{\frac13 \cdot 0.2 + \frac13 \cdot 0.23810\ldots + \frac13 \cdot 0.26272} \\ &= 0.3397\ldots \\ &= 0.340 \, \, (3\text{ s.f.}) \end{align*}
- \begin{align*} \mathbb{E}(A) &= 25 + 5 &= 30 \\ \mathbb{E}(B) &= 20 + 3\cdot 4 &= 32 \\ \mathbb{E}(C) &= 10 + 5 \cdot 4 &= 30 \end{align*} \(A\) and \(C\) are equally good.
- \begin{align*} \mathbb{P}(A \leq 32) &= \mathbb{P}(U \leq 7) &= 0.7 \\ \mathbb{P}(B \leq 32) &= \mathbb{P}(Po(4) \leq 4) \\ &= e^{-4}(1 + 4 + 8 + \frac{4^3}{6}) &= 0.4334\ldots \\ \mathbb{P}(C \leq 32) &= \mathbb{P}(B(5,\tfrac45) \leq 4) \\ &= 1 - \mathbb{P}(B(5,\tfrac45) = 5) \\ &= 1 - \frac{4^5}{5^5} &=0.67232 \end{align*} So you should choose route \(A\).
1989 Paper 1 Q16
A and B play a guessing game. Each simultaneously names one of the numbers \(1,2,3.\) If the numbers differ by 2, whoever guessed the smaller pays the opponent £\(2\). If the numbers differ by 1, whoever guessed the larger pays the opponent £\(1.\) Otherwise no money changes hands. Many rounds of the game are played.
- If A says he will always guess the same number \(N\), explain (for each value of \(N\)) how B can maximise his winnings.
- In an attempt to improve his play, A announces that he will guess each number at random with probability \(\frac{1}{3},\) guesses on different rounds being independent. To counter this, B secretly decides to guess \(j\) with probability \(b_{j}\) (\(j=1,2,3,\, b_{1}+b_{2}+b_{3}=1\)), guesses on different rounds being independent. Derive an expression for B's expected winnings on any round. How should the probabilities \(b_{j}\) be chosen so as to maximize this expression?
- A now announces that he will guess \(j\) with probability \(a_{j}\) (\(j=1,2,3,\, a_{1}+a_{2}+a_{3}=1\)). If B guesses \(j\) with probability \(b_{j}\) (\(j=1,2,3,\, b_{1}+b_{2}+b_{3}=1\)), obtain an expression for his expected winnings in the form \[ Xa_{1}+Ya_{2}+Za_{3}. \] Show that he can choose \(b_{1},b_{2}\) and \(b_{3}\) such that \(X,Y\) and \(Z\) are all non-negative. Deduce that, whatever values for \(a_{j}\) are chosen by A, B can ensure that in the long run he loses no money.
Solution:
- Suppose A always plays \(1\), then B should always play \(2\) and every time they will win 1. Suppose A always plays \(2\) then B should always play \(3\) and every time they will win 1. If A always plays \(3\) then B should always play \(1\) and every time they will win 2.
- \begin{array}{cccc} & b_1 & b_2 & b_3 \\ \frac13 & (0, \frac{b_1}{3}) & (1, \frac{b_2}{3}) & (-2, \frac{b_3}{3}) \\ \frac13 & (-1, \frac{b_1}{3}) & (0, \frac{b_2}{3}) & (1, \frac{b_3}{3}) \\ \frac13 & (2, \frac{b_1}{3}) & (-1, \frac{b_2}{3}) & (0, \frac{b_3}{3}) \\ \end{array} Therefore the expected value is: \(\frac{b_1}{3} - \frac{b_3}{3}\) and to maximise this he should always guess \(1\) (ie \(b_1 = 1, b_2 = 0, b_3 = 0\).)
- \begin{array}{cccc} & b_1 & b_2 & b_3 \\ a_1 & (0, a_1b_1) & (1, a_1b_2) & (-2, a_1b_3) \\ a_2 & (-1, a_2b_1) & (0, a_2b_2) & (1, a_2b_3) \\ a_3 & (2, a_3b_1) & (-1, a_3b_2) & (0, a_3b_3) \\ \end{array} Therefore the expected value is: \((b_2-2b_3)a_1 + (b_3-b_1)a_2 + (2b_1-b_2)a_3\) We need \(b_2 \geq 2b_3, b_3 \geq b_1, 2b_1 \geq b_2\) so \(b_1 \leq b_3 \leq \frac12 b_2 \leq b_1\) so we could take \(b_1 = b_3 = \frac12 b_2\) or \(b_1 = b_3 = \frac14, b_2 = \frac12\) and all values would be \(0\). Therefore by choosing these values \(B\) can guarantee his expected value is \(0\) and therefore shouldn't expect to lose money in the long run.