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1991 Paper 1 Q11
D: 1500.0 B: 1500.1

A piledriver consists of a weight of mass \(M\) connected to a lighter counterweight of mass \(m\) by a light inextensible string passing over a smooth light fixed pulley. By considerations of energy or otherwise, show that if the weights are released from rest, and move vertically, then as long as the string remains taut and no collisions occur, the weights experience a constant acceleration of magnitude \[ g\left(\frac{M-m}{M+m}\right). \] Initially the weight is held vertically above the pile, and is released from rest. During the subsequent motion both weights move vertically and the only collisions are between the weight and the pile. Treating the pile as fixed and the collisions as completely inelastic, show that, if just before a collision the counterweight is moving with speed \(v\), then just before the next collision it will be moving with speed \(mv/\left(M+m\right)\). {[}You may assume that when the string becomes taut, the momentum lost by one weight equals that gained by the other.{]} Further show that the times between successive collisions with the pile form a geometric progression. Show that the total time before the weight finally comes to rest is three times the time from the start to the first impact.

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1991 Paper 1 Q12
D: 1484.0 B: 1500.0

\(\ \)\vspace{-1.5cm} \noindent

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The above diagram illustrates a makeshift stepladder, made from two equal light planks \(AB\) and \(CD\), each of length \(2l\). The plank \(AB\) is smoothly hinged to the ground at \(A\) and makes an angle of \(\alpha\) with the horizontal. The other plank \(CD\) has its bottom end \(C\) resting on the same horizontal ground and makes an angle \(\beta\) with the horizontal. It is pivoted smoothly to \(B\) at a point distance \(2x\) from \(C\). The coefficient of friction between \(CD\) and the ground is \(\mu.\) A painter of mass \(M\) stands on \(CD\), half between \(C\) and \(B\). Show that, for equilibrium to be possible, \[ \mu\geqslant\frac{\cot\alpha\cot\beta}{2\cot\alpha+\cot\beta}. \] Suppose now that \(B\) coincides with \(D\). Show that, as \(\alpha\) varies, the maximum distance from \(A\) at which the painter will be standing is \[ l\sqrt{\frac{1+81\mu^{2}}{1+9\mu^{2}}}. \]

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1991 Paper 1 Q13
D: 1516.0 B: 1484.0

\(\ \)\vspace{-1.5cm} \noindent

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A heavy smooth lamina of mass \(M\) is free to slide without rotation along a straight line on a fixed smooth horizontal table. A smooth groove \(ABC\) is inscribed in the lamina, as indicated in the above diagram. The tangents to the groove at \(A\) and at \(B\) are parallel to the line. When the lamina is stationary, a particle of mass \(m\) (where \(m < M\)) enters the groove at \(A\). The particle is travelling, with speed \(V\), parallel to the line and in the plane of the lamina and table. Calculate the speeds of the particle and of the lamina, when the particle leaves the groove at \(C\). Suppose now that the lamina is held fixed by a peg attached to the line. Supposing that the groove \(ABC\) is a semicircle of radius \(r\), obtain the value of the average force per unit time exerted on the peg by the lamina between the instant that the particle enters the groove and the instant that it leaves it.

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1991 Paper 1 Q14
D: 1516.0 B: 1457.1

A set of \(2N+1\) rods consists of one of each length \(1,2,\ldots,2N,2N+1\), where \(N\) is an integer greater than 1. Three different rods are selected from the set. Suppose their lengths are \(a,b\) and \(c\), where \(a>b>c\). Given that \(a\) is even and fixed, show, by considering the possible values of \(b\), that the number of selections in which a triangle can then be formed from the three rods is \[ 1+3+5+\cdots+(a-3), \] where we allow only non-degenerate triangles (i.e. triangles with non-zero area). Similarly obtain the number of selections in which a triangle may be formed when \(a\) takes some fixed odd value. Write down a formula for the number of ways of forming a non-degenerate triangle and verify it for \(N=3\). Hence show that, if three rods are drawn at random without replacement, then the probability that they can form a non-degenerate triangle is \[ \frac{(N-1)(4N+1)}{2(4N^{2}-1)}. \]

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1991 Paper 1 Q15
D: 1516.0 B: 1484.0

A fair coin is thrown \(n\) times. On each throw, 1 point is scored for a head and 1 point is lost for a tail. Let \(S_{n}\) be the points total for the series of \(n\) throws, i.e. \(S_{n}=X_{1}+X_{2}+\cdots+X_{n},\) where \[ X_{j}=\begin{cases} 1 & \text{ if the }j \text{ th throw is a head}\\ -1 & \text{ if the }j\text{ th throw is a tail.} \end{cases} \]

  1. If \(n=10\,000,\) find an approximate value for the probability that \(S_{n}>100.\)
  2. Find an approximate value for the least \(n\) for which \(\mathrm{P}(S_{n}>0.01n)<0,01.\)
Suppose that instead no points are scored for the first throw, but that on each successive throw, 2 points are scored if both it and the first throw are heads, two points are deducted if both are tails, and no points are scored or lost if the throws differ. Let \(Y_{k}\) be the score on the \(k\)th throw, where \(2\leqslant k\leqslant n.\) Show that \(Y_{k}=X_{1}+X_{k}.\) Calculate the mean and variance of each \(Y_{k}\) and determine whether it is true that \[ \mathrm{P}(Y_{2}+Y_{3}+\cdots+Y_{n}>0.01(n-1))\rightarrow0\quad\mbox{ as }n\rightarrow\infty. \]

Solution: Notice that \(\mathbb{E}(X_i) = 0, \mathbb{E}(X_i^2) = 1\) and so \(\mathbb{E}(S_n) =0, \textrm{Var}(S_n) = n\).

  1. Then by the central limit theorem (or alternatively the normal approximation to the binomial), \begin{align*} && \mathbb{P}(S_n > 100) &\underbrace{\approx}_{\text{CLT}} \mathbb{P} \left (Z > \frac{100}{\sqrt{10\, 000}} \right) \\ &&&= \mathbb{P}(Z > 1) \\ &&&= 1-\Phi(1) \\ &&&\approx 15.9\% \end{align*}
  2. \begin{align*} &&\mathbb{P}(S_n > 0.01n) &\approx \mathbb{P} \left (Z > \frac{0.01n}{\sqrt{n}} \right) \\ &&&= \mathbb{P}(Z > 0.01 \sqrt{n}) \\ &&&= 1-\Phi(0.01\sqrt{n}) \\ &&&< 0.01 \\ && \Phi^{-1}(0.01) &= -2.3263\ldots \\ \Rightarrow && 0.01 \sqrt{n} &= 2.3263\ldots \\ \Rightarrow && n &\approx 233^2 \end{align*}
\begin{array}{cc|cc} 1\text{st throw}& k\text{th throw} & X_1 + X_k & Y_k \\ \hline \text{head} & \text{head} & 1 + 1 & 2 \\ \text{head} & \text{tail} & 1 - 1 & 0 \\ \text{tail} & \text{head} & -1 + 1 & 0 \\ \text{tail} & \text{tail} & -1- 1 & -2 \\ \end{array} Across all possible cases \(Y_k = X_1 + X_k\) so therefore these random variables are equal. \begin{align*} \mathbb{E}(Y_k) &= \mathbb{E}(X_1) + \mathbb{E}(Y_k) \\ &= 0 + 0 = 0 \\ \\ \textrm{Var}(Y_k) &= \textrm{Var}(X_1)+\textrm{Var}(X_k) \\ &= 2 \\ \\ \mathbb{E}\left (\sum_{k=2}^n Y_k \right) &= 0 \\ \textrm{Var}\left (\sum_{k=2}^n Y_k \right) &= 2(n-1) \end{align*} Therefore approximately \(\displaystyle \sum_{k=2}^n Y_k \approx N(0, 2(n-1))\) \begin{align*} \mathbb{P} \left (\sum_{k=2}^n Y_k > 0.01(n-1) \right) &\approx \mathbb{P} \left (Z > \frac{0.01(n-1)}{\sqrt{2(n-1)}} \right) \\ &= \mathbb{P} \left (Z > c \sqrt{n-1} \right) \\ &\to 0 \text{ as } n \to \infty \end{align*}

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1991 Paper 1 Q16
D: 1500.0 B: 1484.0

At any instant the probability that it is safe to cross a busy road is \(0.1\). A toad is waiting to cross this road. Every minute she looks at the road. If it is safe, she will cross; if it is not safe, she will wait for a minute before attempting to cross again. Find the probability that she eventually crosses the road without mishap. Later on, a frog is also trying to cross the same road. He also inspects the traffic at one minute intervals and crosses if it is safe. Being more impatient than the toad, he may also attempt to cross when it is not safe. The probability that he will attempt to cross when it is not safe is \(n/3\) if \(n\leqslant3,\) where \(n\) minutes have elapsed since he firrst inspected the road. If he attempts to cross when it is not safe, he is run over with probability \(0.8,\) but otherwise he reaches the other side safely. Find the probability that he eventually crosses the road without mishap. What is the probability that both reptiles safely cross the road with the frog taking less time than the toad? If the frog has not arrived at the other side 2 minutes after he began his attempt to cross, what is the probability that the frog is run over (at some stage) in his attempt to cross? \textit{[Once moving, the reptiles spend a negligible time on their attempt to cross the road.]}

Solution: Since the toad never crosses when it's not safe, she is certain to cross. (Probability she hasn't crossed after the \(n\)th minute is \(0.9^n \to 0\)). \begin{array}{c|c|c|c|c|c|c|c} \text{will try dangerously} & \text{is safe} & \text{has tried} & \text{tries safely} & \text{tries unsafely} & \text{succeeds} & \text{succeeds unsafely} & \text{fails} \\ \hline 0 & 0.1 & 0 & 0.1 & 0 & 0.1 & 0 & 0\\ \frac13 & 0.1 & 0.1 & 0.09 & 0.27 & 0.144 & 0.054 & 0.216\\ \frac23 & 0.1 & 0.46 & 0.054 & 0.324 & 0.1188 & 0.0648 & 0.2592\\ 1 & 0.1 & 0.838 & 0.0162 & 0.1458 & 0.04536 & 0.02916 & 0.11664\\ \hline & & & & & 0.40816 & 0.14796 & \\ \hline \end{array} So \(\mathbb{P}(\text{frog crosses safely}) = 0.40816\) and \(\mathbb{P}(\text{frog beats toad across}) = 0.14796\). \begin{align*} \mathbb{P}(\text{frog run over} | \text{frog not crossed after 2 minutes}) &= \frac{\mathbb{P}(\text{frog run over and frog not crossed after 2 minutes})}{\mathbb{P}(\text{frog not crossed after 2 minutes})} \\ &= \frac{\mathbb{P}(\text{frog run over within 2 minutes})}{\mathbb{P}(\text{frog not crossed after 2 minutes})} \\ &= \frac{\mathbb{P}(\text{frog run over within 2 minutes})}{1-\mathbb{P}(\text{crossed after 2 minutes})} \\ &= \frac{0.216+0.2592}{1-0.3628} \\ &= 0.7457\ldots \end{align*}

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