Problems

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Solution:

1. \begin{align*} \text{N2}(\uparrow): && R_1 + R_2\sin(\frac{\pi}{6})-2mg &= 0 \\ \text{N2}(\rightarrow): && R_2 \cos (\frac{\pi}{6})-F_r &= 0 \\ \overset{\curvearrowleft}{X}: && lmg \cos \tfrac{\pi}{6} - l R_2 \cos \tfrac{\pi}{6} &= 0 \\ \\ \Rightarrow && R_2 &= mg \\ \Rightarrow && R_1 &= 2mg - \frac12mg \\ &&&=\frac32mg \\ \Rightarrow && \frac{\sqrt{3}}2mg - \mu\frac32mg &= 0 \\ \Rightarrow && \mu &= \frac{1}{\sqrt{3}} \end{align*}
2. \begin{align*} \text{N2}(\uparrow): && R_1 + R_2\sin(\frac{\pi}{6})-2mg &= 0 \\ \overset{\curvearrowleft}{X}: && \frac12 lmg \cos \tfrac{\pi}{6}+xmg \cos \tfrac{\pi}{6} - l R_2 \cos \tfrac{\pi}{6} &= 0 \\ \\ \Rightarrow && R_2 &= mg(\frac{1}2+\frac{x}{l}) \\ \Rightarrow && R_1 &= 2mg - \frac12mg(\frac{1}2+\frac{x}{l}) \\ &&&=(\frac74 - \frac{x}{2l})mg \\ &&&\geq \frac{5}{4}mg\\ \text{N2}(\rightarrow): && R_2 \cos (\frac{\pi}{6})-\mu R_1& \leq 0 \\ \Rightarrow && \frac{\sqrt{3}}2mg - \mu\frac54mg &\leq 0 \\ \Rightarrow && \mu &\geq \frac{2\sqrt{3}}{5} \end{align*}

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Solution:

Set up the coordinate frame so that the \(x\)-direction is the line of centres of the spheres. Then if the initial velocities are \(\displaystyle \binom{u_x}{u_y}\) and \(\displaystyle \binom{0}{0}\). Then the final velocities must be: \(\displaystyle \binom{v_{x1}}{u_y}\) and \(\displaystyle \binom{v_{x2}}{0}\) where \(mu_x = mv_{x1}+mv_{x2}\) by conservation of energy and \(\frac{v_{x1}-v_{x2}}{u_x} = -e\). \begin{align*} && \begin{cases} v_{x1}+v_{x2} &= u_x \\ v_{x1}-v_{x2} &= -eu_x \\ \end{cases} \\ \Rightarrow && 2v_{x1} &= (1-e)u_x \\ \Rightarrow && v_{x1} &= \frac{(1-e)}{2} u_x \\ && v_{x2} &= \frac{1+e}{2} u_x \end{align*} Notice that since \(0 < e < 1\) we must have \(v_{x1} > 0\) and so the ball on the left is still continuing in the positive direction, therefore the angle will be less than \(\frac12 \pi\). The angle the first ball is deflected through is the angle between: \(\displaystyle \binom{u_x}{u_y}\) and \(\displaystyle \binom{\frac{1-e}{2}u_x}{u_y}\). We can scale the velocities so \(u_y = 1\). So we are interested in the angle between \(\displaystyle \binom{x}{1}\) and \(\displaystyle \binom{\frac{1-e}{2}x}{1}\). To maximise \(\theta\) we can maximise \(\tan \theta\), so: \begin{align*} && \tan \theta &= \frac{\frac{2}{(1-e)x-\frac{1}{x}}}{1+\frac{2}{(1-e)x^2}} \\ &&&= \frac{2x-(1-e)x}{(1-e)x^2+2} \\ &&&= \frac{(1+e)x}{(1-e)x^2+2} \\ \\ \frac{\d}{\d t}: &&&= \frac{(1+e)((1-e)x^2+2)-2(1+e)(1-e)x^2}{\sim} \\ &&&= \frac{2(1+e)-(1+e)(1-e)x^2}{\sim}\\ \frac{\d}{\d t} = 0: &&0 &= 2(1+e)-(1+e)(1-e)x^2 \\ \Rightarrow && x &= \pm \sqrt{\frac{2}{1-e}} \\ \\ \Rightarrow && \tan \theta &= \frac{\pm(1+e)\sqrt{\frac{2}{1-e}}}{2+2} \\ &&&= \pm \frac{\sqrt{2}(1+e)}{4\sqrt{1-e}} \\ \Rightarrow && \cot^2 \theta &= \frac{8(1-e)}{(1+e)^2} \\ \Rightarrow && \cosec^2 \theta &= \frac{8(1-e)}{(1+e)^2} + 1 \\ &&&= \frac{8-8e+1+2e+e^2}{(1+e)^2} \\ &&&= \frac{9-6e+e^2}{(1+e)^2} \\ &&&= \frac{(3-e)^2}{(1+e)^2} \\ \Rightarrow && \theta &= \sin^{-1} \left ( \frac{1+e}{3-e}\right) \end{align*}

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Solution: Applying \(\mathbf{F} = m\mathbf{a} = m \frac{\d \mathbf{v}}{dt}\) we have: \begin{align*} && m \frac{\d \mathbf{v}}{d t} &= m\mathbf{g} - mk(\mathbf{v} - \mathbf{w}) \\ \Rightarrow && \frac{\d \mathbf{v}}{d t} +k \mathbf{v} &= \mathbf{g} + k \mathbf{w} \\ \\ \Rightarrow && e^{k t} \l \frac{\d \mathbf{v}}{d t} +k \mathbf{v} \r &= e^{kt} ( \mathbf{g} + k \mathbf{w}) \\ \Rightarrow && \frac{\d}{\d t} \l e^{kt} \mathbf{v} \r &= e^{kt}( \mathbf{g} + k \mathbf{w}) \\ \Rightarrow && e^{kt} \mathbf{v} &= \frac{1}ke^{kt}( \mathbf{g} + k \mathbf{w}) + c \\ \Rightarrow && \mathbf{v}_0 &= \frac{1}{k} ( \mathbf{g} + k \mathbf{w})+c \\ \Rightarrow && \mathbf{v} &= e^{-kt} \l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + \frac{1}{k} \mathbf{g} + \mathbf{w} \\ \Rightarrow && \mathbf{x} &= -\frac{1}{k}e^{-kt} \l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + \frac{1}{k} \mathbf{g}t + \mathbf{w}t+C \\ \Rightarrow && \mathbf{0} &= -\frac{1}{k} \l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + C \\ \Rightarrow && \mathbf{x} &= \frac1{k}\l 1- e^{-kt} \r\l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + \frac{1}{k} \mathbf{g}t + \mathbf{w}t \end{align*} Position at time \(t\) is: \begin{align*} && x_x &= \frac1{k} ( 1-e^{-kt})(u_x - w)+wt \\ && x_y &= \frac1{k} ( 1-e^{-kt})(u_x \frac{g}{kw} - \frac{g}{k})+\frac{1}{k}gt \\ &&&= \frac{g}{kw} \left ( ( 1-e^{-kt})(u_x - w)+wt \right) \\ &&&= \frac{g}{kw} x_x \end{align*} Therefore if \(x_x\) is ever \(0\) then \(x_y\) will also be zero. But the ball must eventually hit the ground, and when it does, it will be in the process of scoring an own goal.

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Solution:

1. The condition that the bead exerts no pressure on the wire is equivalent to the condition that the wire exerts no force on the bead. (Newton's Third Law). This is equivalent to the bead being projected under gravity. Notice that the initial projection is at \(45^{\circ}\) since \(\frac{dy}{dx}|_{x=0} = 1\). The position of the particle (under gravity) at time \(t\) is \(x = \frac{1}{\sqrt{2}}Vt\) and \(y = \frac{1}{\sqrt{2}}Vt - \frac12 gt^2 = x - \frac{1}{2}g \frac{2x^2}{V^2} = x - \frac{g}{V^2}x^2\). Therefore they follow the same trajectory if \(\frac{g}{V^2} = \frac{1}{4a} \Leftrightarrow V = 2\sqrt{ag}\)
2. First note that the wire does no work as it is perpendicular to the velocity, so it is fine to use conservation of momentum. If we take our \(0\) GPE level to be be \(x = 0\), then we notice the initial energy is \(\frac12mV^2\). Secondly, notice that \(\tan \theta = \frac{\d y}{\d x} = 1- \frac{x}{2a} \Rightarrow x = 2a - 2a \tan \theta\) \begin{align*} y &= 2a(1-\tan \theta) - \frac{4a^2(1-\tan \theta)^2}{4a}\\ &= (1-\tan \theta)(2a-a(1-\tan \theta)) \\ &= a(1-\tan \theta)(1+\tan \theta) \\ &= a(1-\tan^2 \theta) \end{align*} GPE \(mga(1-\tan^2 \theta)\). To calculate the kinetic energy, notice that \(\dot{x} = v \cos \theta \dot{\theta}\) and \(\dot{x} = -2a\sec^2 \theta\dot{\theta} \Rightarrow v = -\frac{2a\dot{\theta} }{\cos^{3} \theta}\). Therefore, energy at time \(t\) is: \begin{align*} && \frac12 m V^2 &= \frac12 m \l - \frac{2a\dot{\theta}}{\cos^3 \theta} \r^2 + mga(1-\tan^2 \theta) \\ \Rightarrow && V^2 &= \frac{4a^2\dot{\theta}^2}{\cos^6 \theta} + 2ag(1-\tan^2 \theta) \end{align*}

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Solution:

1. \(M(0) = 0\) since if there's no space on the shelf, we wont be able to put any books on the shelf.
2. If the shelf has length \(1\) it can only fit a thin book. For a thin book to be placed on the shelf, the very first book which comes to be placed must be thin. But this happens with probability \(q\). Therefore \(M(1) = q\).
3. Suppose no books have been placed on the shelf, then with probability \(p\) a large book gets placed on the shelf, and the expected number of books to be placed on the shelf is equivalent to how many books will be placed on the shelf if the shelf only had \(n-2\) spaces. This is \(M(n-2)\). Similar if the book which arrives first is thin (with probability \(q\)) then there will be \(M(n-1)\) more books placed on the shelf in expectation. We've just added \(1\) more book, therefore \(M(n) = 1+pM(n-2) + qM(n-1)\) or rearranging \(M(n) - qM(n-1) - pM(n-2) = 1\).

Suppose \(M(n) = (-p)^n\), notice that: \begin{align*} M(n) - qM(n-1) - pM(n-2) &= (-p)^n - (1-p)(-p)^n - p(-p)^{n-2} \\ &= (-p)^{n-2}(p^2+(1-p)p-p) \\ &= 0 \end{align*} Suppose \(M(n) = B\), notice that: \begin{align*} M(n) - qM(n-1) - pM(n-2) &= B - (1-p)B - pB \\ &= 0 \end{align*} Finally, if \(M(n) = Cn\) notice that: \begin{align*} M(n) - qM(n-1) - pM(n-2) &= Cn - (1-p)C(n-1) - pC(n-2) \\ &= C(n(1-(1-p)+p)+(1-p)+2p) \\ &= C(1+p) \end{align*} Therefore if \(C = \frac{1}{1+p}\) we have that: \(M(n) = A(-p)^n + B + Cn\) satisfies our recurrence. We also need \(M(0) = 0\) and \(M(1) = q\) \begin{align*} 0 &= M(0) \\ &= A + B \\ 1-p &= M(1) \\ &= -pA+B \end{align*} \((1+p)A = p-1 \Rightarrow A = \frac{p-1}{1+p}, B = \frac{1-p}{1+p}\). Therefore: \[ M(n) = -\frac{1-p}{1+p}(-p)^n + \frac{1-p}{1+p} + \frac{n}{1+p} \] is a possible solution to this equation

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Solution: There are \(\displaystyle \binom{m}{k} \binom{M-m}{n-k}\) ways to choose \(k\) red and and \(n-k\) green balls out of a total \(\displaystyle \binom{M}{n}\) ways to choose balls. Therefore the probability is: \[ \mathbb{P}(\text{exactly }k\text{ red balls in }n\text{ choices}) = \frac{\binom{m}{k} \binom{M-m}{n-k}}{ \binom{M}{n}}\]

1. Note that there is nothing special about the \(i\)th ball chosen. (We could consider all draws look at the \(i\)th ball, or consider all draws apply a permutation to make the \(i\)th ball the first ball, and both would look like identical sequences). Therefore \(\mathbb{P}(X_i = 1) = \mathbb{P}(X_1 = 1) = \frac{m}{M}\).
2. Similarly we could apply a permutation to all sequences which takes the \(i\)th ball to the first ball and the \(j\)th ball to the second ball, therefore: \begin{align*} \mathbb{P}(X_i = 1, X_j = 1) &= \mathbb{P}(X_1 = 1, X_2 = 1) \\ &= \mathbb{P}(X_1 = 1) \cdot \mathbb{P}(X_2 = 1 | X_1 = 1) \\ &= \frac{m}{M} \cdot \frac{m-1}{M-1} \\ &= \frac{m(m-1)}{M(M-1)} \end{align*}

So: \begin{align*} \mathbb{E}(X) &= \mathbb{E}(\sum_{i=1}^{n}X_{i}) \\ &= \sum_{i=1}^{n}\mathbb{E}(X_{i}) \\ &= \sum_{i=1}^{n} 1\cdot\mathbb{P}(X_i = 1) \\ &= \sum_{i=1}^{n} \frac{m}{M} \\ &= \frac{mn}{M} \end{align*} and \begin{align*} \mathbb{E}(X^2) &= \mathbb{E}\left[\left(\sum_{i=1}^{n}X_{i} \right)^2 \right] \\ &= \mathbb{E}\left[\sum_{i=1}^n X_i^2 + 2 \sum_{i < j} X_i X_j \right] \\ &= \sum_{i=1}^n \mathbb{E}(X_i^2) + 2 \sum_{i < j} \mathbb{E}(X_i X_j) \\ &= \frac{nm}{M} + n(n-1) \frac{m(m-1)}{M(M-1)} \\ \textrm{Var}(X) &= \mathbb{E}(X^2) - (\mathbb{E}(X))^2 \\ &= \frac{nm}{M} + n(n-1) \frac{m(m-1)}{M(M-1)} - \frac{n^2m^2}{M^2} \\ &= \frac{nm}{M} \left (1-\frac{nm}{M}+(n-1)\frac{m-1}{M-1} \right) \\ &= \frac{nm}{M} \left ( \frac{M(M-1)-(M-1)nm+(n-1)(m-1)M}{M(M-1)} \right) \\ &= \frac{nm}{M} \frac{(M-m)(M-n)}{M(M-1)} \\ &= n \frac{m}{M} \frac{M-m}{M} \frac{M-n}{M-1} \end{align*} Note: This is a very nice way of deriving the mean and variance of the hypergeometric distribution