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2015 Paper 2 Q11
D: 1600.0 B: 1484.0

Three particles, \(A\), \(B\) and \(C\), each of mass \(m\), lie on a smooth horizontal table. Particles \(A\) and \(C\) are attached to the two ends of a light inextensible string of length \(2a\) and particle~\(B\) is attached to the midpoint of the string. Initially, \(A\), \(B\) and \(C\) are at rest at points \((0,a)\), \((0,0)\) and \((0,-a)\), respectively. An impulse is delivered to \(B\), imparting to it a speed \(u\) in the positive \(x\) direction. The string remains taut throughout the subsequent motion.

\psset{xunit=1.0cm,yunit=1.0cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-2.18,-3.26)(6.26,3.18) \psline[linewidth=2pt](2.,1.5)(3.,0.) \psline[linewidth=2pt](3.,0.)(2.,-1.5) \psline(-2.,0.)(6.,0.) \psline(0.,3.)(0.,-3.) \parametricplot{2.1587989303424644}{3.1415926535897936}{0.7*cos(t)+3.|0.7*sin(t)+0.} \rput[tl](-0.25,2.96){\(y\)} \rput[tl](5.76,-0.1){\(x\)} \rput[tl](2.5,0.38){\(\theta\)} \rput[tl](2.1,1.94){\(A\)} \rput[tl](3.2,0.4){\(B\)} \rput[tl](2.16,-1.46){\(C\)} \begin{scriptsize} \psdots[dotsize=8pt 0,dotstyle=*](2.,1.5) \psdots[dotsize=8pt 0,dotstyle=*](3.,0.) \psdots[dotsize=8pt 0,dotstyle=*](2.,-1.5) \end{scriptsize} \end{pspicture*}
  1. At time \(t\), the angle between the \(x\)-axis and the string joining \(A\) and \(B\) is \(\theta\), as shown in the diagram, and \(B\) is at \((x,0)\). Write down the coordinates of \(A\) in terms of \(x,a\) and \(\theta\). Given that the velocity of \(B\) is \((v,0)\), show that the velocity of \(A\) is \((\dot x + a\sin\theta \,\dot \theta\,,\, a\cos\theta\, \dot\theta)\), where the dot denotes differentiation with respect to time.
  2. Show that, before particles \(A\) and \(C\) first collide, \[ 3\dot x + 2a \dot\theta \sin\theta =v \text{ \ \ \ \ \ \ and \ \ \ \ \ \ } \dot \theta^2 = \frac{v^2}{a^2(3-2\sin^2\theta)} \,. \]
  3. When \(A\) and \(C\) collide, the collision is elastic (no energy is lost). At what value of \(\theta\) does the second collision between particles \(A\) and \(C\) occur? (You should justify your answer.)
  4. When \(v=0\), what are the possible values of \(\theta\)? Is \(v =0\) whenever \(\theta\) takes these values?

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2015 Paper 2 Q12
D: 1600.0 B: 1500.0

Four players \(A\), \(B\), \(C\) and \(D\) play a coin-tossing game with a fair coin. Each player chooses a sequence of heads and tails, as follows: \noindent Player A: HHT; \ \ Player B: THH; \ \ Player C: TTH; \ \ Player D: HTT. \noindent The coin is then tossed until one of these sequences occurs, in which case the corresponding player is the winner.

  1. Show that, if only \(A\) and \(B\) play, then \(A\) has a probability of \(\frac14\) of winning.
  2. If all four players play together, find the probabilities of each one winning.
  3. Only \(B\) and \(C\) play. What is the probability of \(C\) winning if the first two tosses are~TT? Let the probabilities of \(C\) winning if the first two tosses are HT, TH and HH be \(p\), \(q\) and \(r\), respectively. Show that \(p=\frac12 +\frac12q\). Find the probability that \(C\) wins.

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2015 Paper 2 Q13
D: 1600.0 B: 1516.0

The maximum height \(X\) of flood water each year on a certain river is a random variable with probability density function \(\f\) given by \[ \f(x) = \begin{cases} \lambda \e^{-\lambda x} & \text{for \(x\ge0\)}\,, \\ 0 & \text{otherwise,} \end{cases} \] where \(\lambda\) is a positive constant. It costs \(ky\) pounds each year to prepare for flood water of height \(y\) or less, where \(k\) is a positive constant and \(y\ge0\). If \(X \le y\) no further costs are incurred but if \(X> y\) the additional cost of flood damage is \(a(X - y )\) pounds where \(a\) is a positive constant.

  1. Let \(C\) be the total cost of dealing with the floods in the year. Show that the expectation of \(C\) is given by \[\mathrm{E}(C)=ky+\frac{a}{\lambda}\mathrm{e}^{-\lambda y} \, . \] How should \(y\) be chosen in order to minimise \(\mathrm{E}(C)\), in the different cases that arise according to the value of \(a/k\)?
  2. Find the variance of \(C\), and show that the more that is spent on preparing for flood water in advance the smaller this variance.

Solution:

  1. \(\,\) \begin{align*} && \mathbb{E}(C) &= \int_0^\infty \text{cost}(x) f(x) \d x \\ &&&= ky + \int_y^{\infty} a(x-y) \lambda e^{-\lambda x} \d x\\ &&&= ky + \int_0^{\infty} a u \lambda e^{-\lambda u -\lambda y} \d x \\ &&&= ky + ae^{-\lambda y} \left( \left [ -ue^{-\lambda u} \right]_0^\infty -\int_0^\infty e^{-\lambda u} \d u\right) \\ &&&= ky + \frac{a}{\lambda}e^{-\lambda y} \\ \\ && \frac{\d \mathbb{E}(C)}{\d y} &= k - ae^{-\lambda y} \\ \Rightarrow && y &= \frac{1}{\lambda}\ln \left ( \frac{a}{k} \right) \end{align*} Since \(\mathbb{E}(C)\) is clearly increasing when \(y\) is very large, the optimal value will be \(\frac{1}{\lambda}\ln \left ( \frac{a}{k} \right)\), if \(\frac{a}{k} > 1\), otherwise you should spend nothing on flood defenses.
  2. \begin{align*} && \mathbb{E}(C^2) &= \int_0^{\infty} \text{cost}(x)^2 f(x) \d x \\ &&&= \int_0^{\infty}(ky + a(x-y)\mathbb{1}_{x > y})^2 f(x) \d x \\ &&&= k^2y^2 + \int_y^{\infty}2kya(x-y)f(x)\d x + \int_y^{\infty}a^2 (x-y)^2 f(x) \d x \\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{- \lambda y}+a^2e^{-\lambda y}\int_{u=0}^\infty u^2 \lambda e^{-\lambda u} \d u \\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y}+a^2e^{-\lambda y}(\textrm{Var}(Exp(\lambda)) + \mathbb{E}(Exp(\lambda))^2\\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y} + a^2e^{-\lambda y} \frac{2}{\lambda^2} \\ && \textrm{Var}(C) &= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y} + a^2e^{-\lambda y} \frac{2}{\lambda^2} - \left ( ky + \frac{a}{\lambda} e^{-\lambda y}\right)^2 \\ &&&= a^2e^{-\lambda y} \frac{2}{\lambda^2} - a^2 e^{-2\lambda y}\frac{1}{\lambda^2} \\ &&&= \frac{a^2}{\lambda^2} e^{-\lambda y}\left (2 - e^{-\lambda y} \right) \\ \\ && \frac{\d \textrm{Var}(C)}{\d y} &= \frac{a^2}{\lambda^2} \left (-2\lambda e^{-\lambda y} +2\lambda e^{-2\lambda y} \right) \\ &&&= \frac{2a^2}{\lambda} e^{-\lambda y}\left (e^{-\lambda y}-1 \right) \leq 0 \end{align*} so \(\textrm{Var}(C)\) is decreasing in \(y\).

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