Problems

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Solution: The particle has initial velocity \(\displaystyle \binom{v \cos \frac{\pi}{3}}{v \sin \frac{\pi}{3}}\) and acceleration \(\displaystyle \binom{0}{-g}\). It will have zero vertical speed (ie be at the top of its trajectory) when \(t = \frac{\sqrt{3}v}{2g}\). Since \(0 = v^2-u^2 + 2as\) the height achieved will be \(\frac{3v^2}{8g}\) At this point it will need to travel the same distance again, but this time the initial speed is \(\frac{3v}{2}\) so: \begin{align*} && \frac{3v^2}{8g} &= \frac{3v}{2} t + \frac12 g t^2 \\ \Rightarrow && 0 &= 4g^2t^2+12vgt - 3v^2 \\ \Rightarrow && t &= \l \frac{-3+2\sqrt{3}}{2} \r \frac{v}{g} \end{align*} Therefore the total time is: \begin{align*} \l \frac{\sqrt{3}}{2} - \frac32 + \sqrt{3} \r \frac{v}{g} &= \frac{3\sqrt{3}-3}{2}\frac{v}{g} \end{align*} \begin{align*} COM(\uparrow): && 0 &= 2m v_y - m \frac{3}{2}v \\ \Rightarrow &&v_y &= \frac34 v \\ COM(\rightarrow): && 3mV &= 2m (3v) +m \frac{v}{2} \\ \Rightarrow && V &= \frac{13}6\\ \end{align*} Therefore superman is now travelling at a vector of \(\displaystyle \binom{\frac{13}6}{\frac34}v\) ie an angle of \(\tan^{-1} \frac 9{26}\) to the horizontal, approximately \(19^\circ\)

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Solution:

\(|AB| = l \cos \tfrac{\pi}{6} = \frac{\sqrt{3}}{2}l\) therefore \(|AC| = \sqrt{3}l\) and the compression is \((2l - \sqrt{3}l)\) and so \(T_2 = \frac{\lambda}{2l} (2l - \sqrt{3}l) = \frac12\lambda(2- \sqrt{3})\) \begin{align*} \text{N2}(\rightarrow, A): && T_1 \cos \tfrac{\pi}{6} - T_2 &= 0 \\ \Rightarrow && T_1 &= \frac12 \frac{2\lambda(2-\sqrt{3})}{\sqrt{3}} \\ &&&= \lambda \left ( \frac{2}{\sqrt{3}} - 1 \right) \\ \\ \text{N2}(\uparrow, D): && 2T_1 \cos \frac{\pi}{3} - mg &= 0 \\ \Rightarrow && mg &= \lambda \left ( \frac{2}{\sqrt{3}} - 1 \right) \end{align*} Suppose \(|AC| = 2xl\), then: \begin{array}{c|c} \text{energy} & \\ \hline \text{GPE} & -mg \sqrt{l^2 - x^2l^2} \\ \text{EPE} & \frac12 \frac{\lambda (2l - 2lx)^2}{2l} \\ \text{KE} & \frac12 m v^2 \end{array} Therefore \[ E = \frac12 mv^2 + \lambda l (1-x)^2-mgl \sqrt{1-x^2}\] Initially, \(E = 0 + 0 + 0 = 0\). When the particle first comes to rest: \begin{align*} \text{COE}: && 0 &= E \\ &&&= \lambda l^2 (1-x)^2 - mgl \sqrt{1-x^2} \\ &&&= \lambda l (1-x)^2 - l \lambda \left ( \frac{2}{\sqrt{3}} - 1 \right) \sqrt{1-x^2} \\ \Rightarrow && (1-x)^2 &= \sqrt{1-x^2} \left ( \frac{2}{\sqrt{3}} - 1 \right) \\ \Rightarrow && (1-x)^2(1-x^2)^{-1/2} &= \left ( \frac{2}{\sqrt{3}} - 1 \right) \\ \Rightarrow && (1-2x+x^2)(1+\frac12 x^2+\cdots) &= \left ( \frac{2}{\sqrt{3}} - 1 \right) \\ \end{align*} If \(x = \frac23\) then \((1-x)^2(1-x^2)^{-1/2} = \frac19 \cdot \left ( \frac{5}{9} \right)^{-1/2} = \frac{\sqrt{5}}{15}\) If \(2\sqrt{3}-3 \approx \frac{\sqrt{5}}5\) we're done.

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Solution:

\begin{align*} \text{N2}(\nwarrow): && R -mg \cos \theta &= 0 \\ \text{N2}(\rightarrow): && -R \sin \theta + F \cos \theta &= 0 \\ \\ \Rightarrow && F &= \tan \theta R \\ \\ && F & \leq \mu R \\ \Rightarrow && \tan \theta R &\leq \mu R \\ \Rightarrow && \tan \theta &\leq \mu \end{align*} (Notice also \(F = mg \sin \theta\)) Once everything is released, we have the following situation. (Red forces act on the cylinder, blue forces on the particle). \begin{align*} \text{N2}(\uparrow): && 0 &= R_g - Mg - \underbrace{mg}_{R_p \text{ and } F_p} + \frac{1}{\sqrt{2}}R_r \\ \text{N2}(\rightarrow): && 0 &= \frac{1}{\sqrt{2}}R_r - F_g \\ \overset{\curvearrowleft}{O}: && 0 &= aF_p - aF_g \\ \Rightarrow && F_g &= mg \sin \theta \\ \Rightarrow && R_r &= \sqrt{2} mg \sin \theta \\ \Rightarrow && R_g &=(M+m)g + mg \sin \theta \\ \\ && F_g &\leq \mu R_g \\ \Rightarrow && mg \sin \theta &\leq \mu (M+m(1+\sin \theta))g \\ \Rightarrow && \mu &\geq \frac{m \sin \theta}{M+m(1+\sin \theta)} \end{align*} If \(m = 7M\) and \(\mu = \frac34\) we have: \begin{align*} && \tan \theta &\leq \frac34 \\ && 3(M+7M(1 + \sin \theta)) &\geq 4 \cdot 7 M \sin \theta \\ \Rightarrow && 10 + 7 \sin \theta & \geq 28 \sin \theta \\ \Rightarrow && 10 &\geq 21 \sin \theta \\ \Rightarrow && \sin \theta &\leq \frac{10}{21} \end{align*} If \(\tan \theta = \frac{3}{4}, \sin \theta = \frac35 > \frac{10}{21}\), so the critical bound is \(\sin \theta \leq \frac{10}{21}\), ie \( \theta \leq \sin^{-1} \frac{10}{21} \approx 30^{\circ}\)

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Solution: The probability Juggin's has a non-zero score is the probability he never draws a black ball in his three goes. This is \((1-\frac15)^3 = \frac{64}{125}\). Let's consider the \(\frac{61}{125}\) probability world where he never draws a black ball. In this conditional probability space, he has \(\frac{5}{8}\) chances of pulling out white balls and \(\frac38\) or pulling out red. His expected score per pull is \(\frac58 \cdot 1 + \frac38 \cdot 2 = \frac{11}{8}\). Therefore his expected score in this universe is \(\frac{33}8\) and his expected score is \(\frac{33}{8} \cdot \frac{61}{125} = \frac{2013}{1000} = 2.013\) . The expected score after drawing another ball is \(( N + 1)\frac{5}{10} + (N+2) \frac{3}{10} + 0 \cdot \frac{2}{10} = \frac{8}{10}N + \frac{11}{10}\). A sensible strategy would be to only draw if \(\frac{8}{10}N + \frac{11}{10} > N \Rightarrow N < \frac{11}{2}\), ie keep drawing until \(N \geq 6\) or we bust out. [The expected score for this strategy is: \begin{array}{ccc} \text{score} & \text{route} & \text{count} & \text{prob} \\ \hline 6 & \text{6 1s} & 1 & \left ( \frac12 \right)^6 \\ 6 & \text{4 1s, 1 2} & 5 & 5 \cdot \left ( \frac12 \right)^4 \cdot \frac{3}{10} \\ 6 & \text{2 1s, 2 2s} & 6 & 6 \cdot \left ( \frac12 \right)^2 \cdot \left ( \frac{3}{10} \right)^2 \\ 6 & \text{3 2s} & 1 & 1 \cdot \left ( \frac{3}{10} \right)^3 \\ 7 & \text{5 1s, 1 2} & 1 &\left ( \frac12 \right)^5 \cdot \frac{3}{10} \\ 7 & \text{3 1s, 2 2s} & 4 & 4\cdot \left ( \frac12 \right)^3 \cdot \left ( \frac{3}{10} \right)^2 \\ 7 & \text{1 1, 3 2s} & 3 & 3\cdot \left ( \frac12 \right) \cdot \left ( \frac{3}{10} \right)^3 \\ \end{array} For an expected value of \(\frac{2171}{8000} \cdot 6 + \frac{759}{8000} \cdot 7 = \frac{18\,339}{8000} = 2.29 \quad (3\text{ s.f.})\)]

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Solution: We must have a sequence consisting of \(\underbrace{HTT\cdots TH}_{k-1\text{ heads and }n-k\text{ tails}}\underbrace{H}_{k\text{th head}}\). There are \(\binom{n-1}{k-1}\) ways to chose how to place the \(k-1\) heads in the first \(n-1\) flips, and each sequence has probability \(p^{k-1}(1-p)^{n-k}p\) which gives a probability of \(\displaystyle \binom{n-1}{k-1} p^k (1-p)^{n-k}\). Given that it took an even number of tosses to achieve \(k-1\) heads, this is equivalent to the problem of what is the probability that the first head occurs on an even flip, ie \begin{align*} \mathbb{P}(\text{even flip}) &= \mathbb{P}(2\text{nd flip}) +\mathbb{P}(4\text{th flip}) +\mathbb{P}(6\text{th flip}) + \cdots \\ &= (1-p)p + (1-p)^3p + (1-p)^5p + \cdots \\ &= (1-p)p \left ( \sum_{r=0}^\infty (1-p)^{2r}\right) \\ &= \frac{p(1-p)}{1-(1-p)^2} \\ &= \frac{p(1-p)}{2p-p^2} \\ &= \frac{1-p}{2-p} \end{align*} The ways to achieve \(2\) heads on even tosses are \(EEO\), \(EOE\), \(OEE\). The probability of going from \(O\) to \(E\) is the same as the initial probability of an \(O\) flip, etc. Therefore \begin{align*} \mathbb{P}(EEO) &=\left( \frac{1-p}{2-p} \right)^2 \left ( 1- \frac{1-p}{2-p} \right) \\ &= \left( \frac{1-p}{2-p} \right)^2 \left ( \frac{1}{2-p} \right) \\ \mathbb{P}(EOE) &= \left( \frac{1-p}{2-p} \right) \left ( \frac{1}{2-p} \right)^2 \\ \mathbb{P}(OEE) &= \left ( \frac{1}{2-p} \right)^2 \left( \frac{1-p}{2-p} \right)\\ \mathbb{P}(2 \text{ heads on even tosses}) &= \frac{(1-p)^2 + 2(1-p)}{(2-p)^3} \\ &= \frac{(1-p)(2-p)}{(2-p)^3} \\ &= \frac{1-p}{(2-p)^2} \end{align*}

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Solution: The probability density function will look like a triangle with base \(20\) minutes and therefore height \(\frac{1}{10}\) per minute, ie: \begin{align*} f_T(t) &= \begin{cases} \frac{1}{100}(t+10) & \text{if } -10 \leq t \leq 0 \\ \frac{1}{100}(10-t) & \text{if } 0 \leq t \leq 10 \\ 0 & \text{otherwise} \end{cases} \end{align*} \begin{align*} \mathbb{P}(\text{catch bus}) &=0.5 \mathbb{P}(\text{bus arrives after 7:55})+0.4 \mathbb{P}(\text{bus arrives after 7:58}) + 0.1 \mathbb{P}(\text{bus arrives after 8:02}) \\ &= \frac12 \cdot \left (1 - \frac18 \right) + \frac{2}{5} \cdot \left ( 1 - \frac{4^2}{5^2} \cdot \frac{1}{2} \right) + \frac{1}{10} \cdot \frac{4^2}{5^2} \cdot \frac12 \\ &= \frac{1\,483}{2\,000} \\ &\approx 74\% \end{align*} \begin{align*} \mathbb{P}(\text{catch bus}) &= \mathbb{P}(\text{bus arrives after 7:55}) \mathbb{P}(\text{catch next bus by 9:00}) \\ &= \frac78 + \frac18 \cdot \frac12 \\ &= \frac{15}{16} \end{align*} He should expect to miss \(6.25\) buses, so missing \(5\) seems about right. (Using a binomial calculation, seeing 5 or fewer buses is ~\(40\%\) which isn't suspicious).