Problems

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Solution:

Whilst the particle is on the surface of the sphere consider the energy. Letting the height of centre of the sphere by our \(0\) GPE level, the initial energy is \(mgr\) (assuming that the initial speed is so close to \(0\) as to make no difference). When it makes an angle \(\theta\) with the horizontal it's energy will be \(mgr \sin \theta + \frac12 m v^2\). By conservation of energy: \(mgr \sin \theta + \frac12 m v^2 = mgr \Rightarrow v^2 = 2gr(1-\sin \theta)\) \begin{align*} \text{N2}(\text{radially}): && mg \sin \theta - R &= m \frac{v^2}{r} \\ \Rightarrow && R &= mg\sin \theta - \frac{m}{r} 2gr(1-\sin \theta) \\ &&&=mg \l 3\sin \theta - 2 \r \end{align*} Since \(R\) must be positive whilst the particle is in contact with the sphere, the angle \(\theta\) makes with the horizontal when it leaves the sphere is \(\sin^{-1} \frac{2}{3}\). At this point \(v^2 = 2gr(1-\sin \theta) = \frac{2}{3}gr\) Again, considering energy, the initial energy is \(mgr\). The final energy is \(-mgr + \frac12mu_x^2 + \frac12mu_y^2\) When the particle leaves the surface it has speed \(v= \frac23 gr\), so the component \(u_x = \sqrt{v}\sin \theta\). By conservation of energy therefore: \begin{align*} && mgr &= -mgr + \frac12mu_x^2 + \frac12mu_y^2 \\ \Rightarrow && \frac12 u_y^2 &= 2gr - \frac12 u_x^2 \\ &&&= 2gr - \frac12 (\sqrt{v} \sin \theta)^2 \\ &&&= 2gr - \frac12 \frac23gr \sin^2 \theta \\ &&&= 2gr - \frac13gr \frac{4}{9} \\ &&&= \frac{50}{27}gr \\ \Rightarrow && u_y &= \frac{10}{3\sqrt{3}}\sqrt{gr} \end{align*} Therefore vertical component of momentum is \(\displaystyle \frac{10}{3\sqrt{3}}\sqrt{gr}m\)

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Solution: Considering the horizontal component, this will be constant as there are no forces acting in that direction. The first step will take the particle \(t = \sqrt{\frac{2h}g}\) to reach. At which point it will be travelling with speed \(v = \sqrt{2gh} \) (by energy considerations, \(mgh = \frac12 mv^2\)). To reach the second step must take twice as long (since the ball has to travel \(2d\) horizontally, rather than \(d\)). Since \(t = 2\sqrt{\frac{2h}g}\) we must have that: \begin{align*} && s &= ut + \frac12 gt^2 \\ \Rightarrow && h &= u 2\sqrt{\frac{2h}g} + \frac12 g \frac{8h}g \\ \Rightarrow && u &= -\frac{3}{2} h\sqrt{\frac{g}{2h}} \\ &&&= -\frac{3}{2\sqrt{2}} \sqrt{gh} \end{align*} Therefore, using Newton's experimental law, we must have that \(e = \frac{\frac{3}{2 \sqrt{2}} \sqrt{gh}}{\sqrt{2} \sqrt{gh}} = \frac{3}{4}\). Again by conservation of energy \(mgh + \frac12 \frac{9}{8} mgh = \frac12 mv^2 \Rightarrow v = \frac{5}{2\sqrt{2}} \sqrt{gh}\) when it lands on the next step. Therefore we would need the coefficient of restitution for the second (and subsequent steps) to be: \(\displaystyle \frac{\frac{3}{2\sqrt{2}} \sqrt{gh}}{\frac{5}{2\sqrt{2}} \sqrt{gh}} = \frac35\)

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Solution: Using Newton's second law in the form, \(\F(x) = m \ddot{x}\). Our two different differential equations can be solved as follows: When \(x \geq 0\) \(-\mu^2x = \ddot{x} \Rightarrow x = A\sin \mu t + B \cos \mu t\) when \(x \geq 0\). And when \(x < 0\) \(-\kappa \dot{x} = \ddot{x} \Rightarrow \dot{x} = Ce^{-\kappa t} \Rightarrow x = De^{-\kappa t} + E\) when \(x < 0\) Following the trajectory of the particle: At \(t = 0, x = 0, \dot{x} = v_0 > 0\), so \(x = \frac{v_0}{\mu} \sin \mu t\) until \(t = \frac{\pi}{\mu}\). When \(t = \frac{\pi}{\mu}\) the particle will head into the negative \(x\)-axis with velocity \(-v_0\). At which point our initial conditions for our differential equations give us that \(De^{-\frac{\pi\kappa}{\mu}} + E = 0, -\kappa De^{-\frac{\pi\kappa}{\mu}} = -v_0 \Rightarrow De^{-\frac{\pi\kappa}{\mu}} = \frac{v_0}{\kappa}, E = -\frac{v_0}{\kappa}\). To summarise: \[ x(t) = \begin{cases} \frac{v_0}{\mu} \sin \mu t & 0 \leq t \leq \frac{\pi}{\mu} \\ -\frac{v_0}{\kappa} \l 1-e^{-\kappa(t-\frac{\pi}{\mu})}\r & t > \frac{\pi}{\mu}\end{cases}\]

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Solution:

1. \(ACB\) is not a complete tournament since no-one has won two matches.
2. \(ABB\) is not a possible complete tournament since it implies \(B\) won game 2, which is between \(A\) (winner of game 1) and \(C\) (referee of game 1).
3. \(ACBB\) is a valid tournament, \(A\) beat \(B\), then \(C\) beat \(A\), then \(B\) beat \(C\) and finally \(B\) beat \(A\) to win.

After the first game there is always someone playing for the tournament, so for there to be no result after 5 games, 4 games must have gone against the leader, so the probability is \(\frac{1}{2^4} = \frac{1}{16}\). If \(A\) wins their first game, they can either win in two games (WW) or in five games (WLRWW). The probability of this is \(\frac14 + \frac1{16} = \frac{5}{16}\). Similarly \(B\) has exactly the same chance as \(A\) since everything about them is symmetric, ie a probability of \(\frac5{16}\) of winning. Since there is a \(\frac{15}{16}\) chance the tournament is decided after 5 games, the remaining \(\frac{5}{16}\) must be \(C\)'s chance of winning. After the first game is played, there's \(3\) states for each player. King (about to win if they win, becomes Ref if they lose), Challenger (needs to win to become king) and Ref (who becomes Challenger if Challenger wins). \begin{align*} \P(\text{King wins}) &= \frac{1}{2} + \frac{1}{2}\P(\text{Ref wins})\\ \P(\text{Challenger wins}) &= \frac{1}{2} \P(\text{King wins}) \\ \P(\text{Ref wins}) &= \frac{1}{2} \P(\text{Challenger wins}) \\ \end{align*} \(p_K = \frac12 + \frac12 (\frac12 \frac12 p_K) \Rightarrow \frac78 p_K = \frac12 \Rightarrow p_K = \frac47, p_C = \frac27, p_R = \frac17\). \(A\) has \(\frac12\) of being king, \(\frac12\) of being ref after the first match, so \(\frac12 \frac47 + \frac12 \frac17 = \frac{5}{14}\). Similarly \(B\) has \(\frac5{14}\) chance of winning, but unfortunately \(C\) must be the challenger after the first match and only has \(\frac27 = \frac4{14}\) chances of winning.

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Solution: Consider the angle from the origin, then \(P = (\cos \theta, \sin \theta)\) where \(\theta \sim U(0, 2\pi)\), and \(X = \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta}\) \begin{align*} \mathbb{E}[X] &= \int_0^{2\pi} \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta} \frac1{2\pi} \d \theta \\ &= \frac1{2\pi}\int_0^{2\pi} \sqrt{2 - 2\cos \theta} \d \theta \\ &= \frac{1}{2\pi}\int_0^{2\pi} \sqrt{4\sin^2 \frac{\theta}{2}} \d \theta \\ &= \frac{1}{\pi}\int_0^{2\pi} \left |\sin \frac{\theta}{2} \right| \d \theta \\ &= \frac{1}{\pi} \left [ -2\cos \frac{\theta}{2} \right]_0^{2\pi} \\ &= \frac1{\pi} \l 2 + 2\r \\ &= \frac{4}{\pi} \end{align*} \begin{align*} \mathbb{E}(X^2) &= \frac1{2\pi}\int_0^{2\pi} (\cos \theta - 1)^2 + \sin^2 \theta \d \theta \\ &= \frac1{2\pi}\int_0^{2\pi} 2 - 2 \cos \theta \d \theta \\ &= \frac{4\pi}{2\pi} \\ &= 2 \\ \end{align*} \(\Rightarrow\) \(\mathrm{Var}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2 = 2 - \frac{16}{\pi^2} = \frac{2\pi^2 - 16}{\pi^2}\).

Where the line makes a length longer than \(\frac{4}{\pi}\) it will make an angle at the origin of \(2\sin^{-1} \frac{2}{\pi}\). Therefore the probability of being larger than this is \(\frac{2\pi - 2 \times 2\sin^{-1} \frac{2}{\pi}}{2 \pi} = 1 - \frac{2}{\pi} \sin^{-1} \frac{2}{\pi} \approx 0.560\)

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Solution:

1. The vote is will now be \(60(1-\alpha)\) for, \(60\alpha\) against and \(X \sim B(40, \frac12)\) at random between those. For a majority, they need \(60(1-\alpha) + X > 50\), ie \(\P(X > 60\alpha - 10) \geq 0.8\). Using a normal approximation to the binomial, we need \(X \approx N(20, 10)\), so \begin{align*} \P(X > 60 \alpha - 10) &= 1- \P(X \leq 60 \alpha - 10) \\ &\approx 1 - \P(\sqrt{10}Z+20 \leq 60\alpha - 10.5) \\ &\approx 1 - \P(Z \leq \frac{60\alpha - 30.5}{\sqrt{10}}) \end{align*} If we want this to be less than \(0.2\) we need \( \frac{60\alpha - 30.5}{\sqrt{10}} < -0.8416 \Rightarrow \alpha < 0.4639\). This would correspond to 27 or fewer exiles or 33 or more remaining preservatives. [Actual computations using Binomial distribution shows we should expect at least 17 to randomly join 20% of the time, so 34 preservatives are required]

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Solution:

1. \begin{align*} \left(x-a\right)^{2}+\left(2x-b\right)^{2}+\left(3x-c\right)^{2} &= x^2-2ax+a^2 + 4x^2 -4bx+b^2 + 9x^2-6cx + c^2 \\ &= (1^2 + 2^2 + 3^2)x^2 - 2x(a+2b+3c) +a^2+b^2 + c^2 \\ &= 14x^2 - 2x(7x) + a^2 + b^2 + c^2 \\ &= a^2 + b^2 + c^2 \end{align*}
2. \begin{align*} \left(2x-a\right)^{2}+\left(3x-b\right)^{2}+\left(3x-c\right)^{2} &= (2^2+3^2+3^2)x^2 - 2x(2a+3b+3c) + (a^2 + b^2+c^2) \\ &= 22x^2 - 2x(11x) + a^2+b^2+c^2 \\ &= a^2+b^2+c^2 \end{align*}

The general result is: If \(\frac{A^2+B^2+C^2}{2}x =Aa+Bb+Cc\) then \((Ax-a)^2 + (Bx-b)^2 + (Cx-c)^2 = a^2+b^2+c^2\) Alternatively, if \(\lambda = \frac{2\mathbf{x} \cdot \mathbf{y}}{\Vert x \Vert^2}\) then \(\Vert \lambda \mathbf{x} - \mathbf{y}\Vert^2 = \Vert \mathbf{y} \Vert^2\) which is easy to see is true.

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Solution:

1. If we can show that \(0 < a - \sqrt{a^2-b} < 1\) then we will be done, since raising a number in \([0,1)\) to a positive integer power will always remain in the same interval. Clearly \(\sqrt{a^2-b} < \sqrt{a^2} = a\) so we have \(a-\sqrt{a^2-b} > 0\) We also have that \(\sqrt{a^2-b} > \sqrt{a^2-(2a-1)} = (a-1)\). Therefore \(a - \sqrt{a^2-b} < a - (a-1) = 1\) as required.
2. If we can show that \(\l a + \sqrt{a^2-b} \r^n + \l a - \sqrt{a^2-b} \r^n = N -r + s\) is an integer we will be done, since the only integer value \(-r+s\) can be is \(0\). This is easy to see, since \begin{align*} \l a + \sqrt{a^2-b} \r^n + \l a - \sqrt{a^2-b} \r^n &= \sum_{k=0}^n \binom{n}{k}a^{n-k}(\sqrt{a^2-b})^k +\sum_{k=0}^n \binom{n}{k}a^{n-k}(-\sqrt{a^2-b})^k \\ &= \sum_{k=0}^n \binom{n}{k}a^{n-k}\l (\sqrt{a^2-b})^k +(-\sqrt{a^2-b})^k \r \\ \end{align*} But every term where \(k\) is odd in this sum is \(0\) (since they cancel) and ever term where \(k\) is even in this sum is an integer. Therefore the sum is an integer and we're done.
3. \begin{align*} -r^2+rN &= -r(r-N) \\ &= s(r-N) \\ &=- \l a - \sqrt{a^2-b} \r^n \l a + \sqrt{a^2-b} \r^n \\ &= -\l \l a - \sqrt{a^2-b} \r \l a + \sqrt{a^2-b} \r\r^n \\ &= - \l a^2 - a^2+b\r^n \\ &= b^n \end{align*} Therefore \(r^2-rN + b^n = 0\)

Looking at \(\left(8+3\sqrt{7}\right)^{n}\) we have \(a = 8, b = 1\) (since \(8^2 - 1 = 9 \cdot 7\). So we can apply the result of the previous question to see that: \(\left(8+3\sqrt{7}\right)^{n}\) differs from an integer by \(\left(8-3\sqrt{7}\right)^{n}\). \begin{align*} 8-3\sqrt{7} &= \frac{1}{8+3\sqrt{7}} \\ &\approx \frac{1}{8 + 8} \\ &\approx 2^{-4} \end{align*} Therefore it differs by approximation \((2^{-4})^n = 2^{-4n}\)

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Solution: The point \(ze^{i\theta}\) is obtained by rotating the point \(z\) about \(0\) by an angle \(\theta\) anticlockwise. The complex numbers \(\alpha, \beta\) and \(\gamma\) will form an equilateral triangle iff the angles between each side are \(\frac{\pi}{3}\), ie \begin{align*} \begin{cases}{\gamma - \beta} &= e^{i \frac{\pi}{3}}({\beta - \alpha}) \\ {\alpha- \gamma} &= e^{i \frac{\pi}{3}}({\gamma- \beta}) \\ {\beta- \alpha} &= e^{i \frac{\pi}{3}}({\alpha- \gamma})\end{cases} \end{align*} We don't need all these equations, since the first two are equivalent to the third. Combining the first two equations, we have \begin{align*} && \frac{\gamma - \beta}{\beta-\alpha} &= \frac{\alpha-\gamma}{\gamma - \beta} \\ \Leftrightarrow && (\gamma - \beta)^2 &= (\alpha-\gamma)(\beta-\alpha) \\ \Leftrightarrow && \gamma^2 +\beta^2 - 2\gamma \beta &= \alpha\beta-\alpha^2-\gamma\beta+\gamma\alpha \\ \Leftrightarrow && \alpha^{2}+\beta^{2}+\gamma^{2}-\beta\gamma-\gamma\alpha-\alpha\beta&=0 \end{align*} as required. If the roots of \(z^{3}+az^{2}+bz+c=0\) are \(\alpha, \beta, \gamma\) then \(\alpha+\beta+\gamma = -a\) and \(\beta\gamma+\gamma\alpha+\alpha\beta = b\). We also have that \(a^2 - 2b = \alpha^2+\beta^2+\gamma^2\). Therefore there roots will lie at the vertices of an equilateral triangle iff \(a^2-3b = 0\)