Problems

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Solution:

1. \(A\) has coordinates \((x-a\cos \theta, a\sin \theta)\). Differentiating with respect to \(t\) the velocity of \(A\) is \((\dot{x}+a\sin \theta \cdot \dot{\theta}, a \cos \theta \cdot \dot{\theta})\)
2. By considervation of momentum \(\rightarrow\) we must have \(mu = m(\dot{x}+a\dot{\theta}\sin \theta) + m\dot{x} + m(\dot{x}+a\dot{\theta}\sin \theta) = m(3\dot{x} + 2a \dot{\theta} \sin \theta)\) and the first equation follows. By conservation of energy, we must have \begin{align*} && \frac12 m u^2 &= \frac12 m \dot{x}^2 + \frac12m((\dot{x}+a\dot{\theta}\sin \theta)^2 + a^2 \dot{\theta}^2 \cos^2\theta ) + \frac12m((\dot{x}+a\dot{\theta}\sin \theta)^2 + a^2 \dot{\theta}^2 \cos^2\theta ) \\ &&&= \frac32m\dot{x}^2 + 2m a\dot{x}\dot{\theta}\sin \theta + ma^2\dot{\theta}^2(\sin^2\theta+\cos^2\theta) \\ \Rightarrow && u^2 &= \dot{x}(3\dot{x} + 4a \dot{\theta} \sin \theta) + 2a^2\dot{\theta}^2 \\ &&&= \left ( \frac{u-2a\dot{\theta}\sin \theta}{3}\right)\left ( 3\left ( \frac{u-2a\dot{\theta}\sin \theta}{3}\right)+ 4a \dot{x}\dot{\theta} \sin \theta \right) + 2a^2\dot{\theta}^2 \\ \Rightarrow && 3u^2 &= (u - 2a\dot{\theta} \sin \theta)^2 + 4a(u - 2 a \dot{\theta} \sin \theta) \dot{\theta}\sin \theta + 6a^2 \dot{\theta}^2 \\ &&&= u^2 + 4a^2\dot{\theta}^2 \sin^2 \theta - 8a^2\dot{\theta}^2\sin^2\theta + 6a^2 \dot{\theta}^2 \\ \Rightarrow && \dot{\theta}^2 &= \frac{u^2}{a^2(3-2\sin^2\theta)} \end{align*}
3. Since \(\dot{\theta}^2 > 0\) \(\theta\) is strictly increasing or decreasing, therefore the first collision will be when \(\theta = 0\), the second when \(\theta = \pi\)
4. If \(v = 0\), from our first equation we have \(2a \dot{\theta} \sin \theta = u \Rightarrow \dot{\theta}^2 = \frac{u^2}{4a^2 \sin^2 \theta} = \frac{u^2}{a^2(3-2\sin^2\theta)}\) so \(4\sin^2 \theta = 3 - 2\sin^2 \theta \Rightarrow \sin^2 \theta = \frac{1}{2}\) therefore the angles are all the multiples of \(\frac{\pi}{4}\).

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Solution:

1. 

   + - ↺
   Since the collision is perfectly elastic, the speed of approach and separation are equal, ie \(v_B = v_A - u\) \begin{align*} \text{COM}: && bmu &= bm(v_A - u) + mv_A \\ \Rightarrow && (b+1)v_A &= 2bu \\ \Rightarrow && v_A &= \frac{2b}{b+1} u \end{align*} Since \(0 < \frac{b}{b+1} < 1\), the largest \(0 < v_A < 2u\)
2. After the first collision with each \(B_i\) we will have \(\displaystyle v_{i+1} = \frac{2\lambda}{\lambda + 1}v_i\), ie \(\displaystyle v_{i+1} = \left (\frac{2\lambda}{\lambda + 1} \right)^i u\) and so \(\displaystyle v_A = \left (\frac{2\lambda}{\lambda + 1} \right)^n u\) which can be arbitrarily large. Suppose \(\lambda = 4\), then \begin{align*} && 20u &< v_A \\ &&&= \left (\frac{8}{5} \right)^n u \\ \Rightarrow && \log_{10} 20 < n \log_{10}(16/10) \\ && \log_{10} 2 + 1 < n 4\log_{10} 2 - n \\ \Rightarrow && n &> \frac{ \log_{10} 2 + 1}{ 4\log_{10} 2 - 1} \\ &&&\approx \frac{0.30103+1}{4 \times 0.30103 -1}\\ &&&= \frac{1.30103}{0.20412} \\ &&&>6 \end{align*} So \(n =7\) is the smallest possible

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Solution:

1. In the direction of \(\mathbf{n}\), Conservation of momentum gives: \(m \mathbf{u} \cdot \mathbf{n} = m v_m + \mu m v_{\mu m}\) Newton's experimental law gives: \(\frac{\mathbf{u} \cdot \mathbf{n}}{v_{\mu m} - v_m} = 1\) Therefore \begin{align*} && \mathbf{u} \cdot \mathbf{n} &= v_m + \mu v_{\mu m} \\ && \mathbf{u} \cdot \mathbf{n} &= v_{\mu m} - v_m \\ \Rightarrow && 2 \mathbf{u} \cdot \mathbf{n} &= (1 + \mu)v_{\mu m} \\ \Rightarrow && v_{\mu m} &= \frac{2 \mathbf{u} \cdot \mathbf{n}}{1 + \mu} \\ \end{align*}
2. Kinetic energy after the collision for the second mass is: \(\frac12 m \mu \frac{4 (\mathbf{u} \cdot \mathbf{n})^2 }{(1 + \mu)^2}\) For the first mass the final speed (in the direction \(\mathbf{n}\) is: \(\mathbf{u} \cdot \mathbf{n}- \frac{2 \mathbf{u} \cdot \mathbf{n}}{1 + \mu} = \frac{(\mu - 1) \mathbf{u} \cdot \mathbf{n}}{1 + \mu}\) It's velocity perpendicular to \(\mathbf{n}\) is unchanged, which is \(\mathbf{u} - (\mathbf{u} \cdot \mathbf{n}) \mathbf{n}\), so it's speed squared is \(\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2\) Therefore the total kinetic energy is: \(\frac12 m \frac{(\mu - 1)^2 (\mathbf{u} \cdot \mathbf{n})^2}{(1 + \mu)} + \frac12 m (\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2)\) Therefore since the kinetic energies are equal we have: \begin{align*} && \frac12 m \frac{(\mu - 1)^2 (\mathbf{u} \cdot \mathbf{n})^2}{(1 + \mu)} + \frac12 m (\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2) &= \frac12 m \mu \frac{4 (\mathbf{u} \cdot \mathbf{n})^2 }{(1 + \mu)^2} \\ \Rightarrow && \frac{(\mu - 1)^2 (\mathbf{u} \cdot \mathbf{n})^2}{(1 + \mu)} + (\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2) &= \mu \frac{4 (\mathbf{u} \cdot \mathbf{n})^2 }{(1 + \mu)^2} \\ \Rightarrow && (\mathbf{u} \cdot \mathbf{n})^2 \l 1 + \frac{4\mu}{(1+ \mu)^2} - \frac{(1-\mu)^2}{(1+ \mu)^2} \r &= (\mathbf{u} \cdot \mathbf{u}) \\ \Rightarrow && (\mathbf{u} \cdot \mathbf{n})^2 \l \frac{(1 + \mu)^2 + 4\mu - (1-\mu)^2}{(1+ \mu)^2} \r &= (\mathbf{u} \cdot \mathbf{u}) \\ \Rightarrow && (\mathbf{u} \cdot \mathbf{n})^2 \l \frac{8\mu}{(1+ \mu)^2} \r &= (\mathbf{u} \cdot \mathbf{u}) \\ \Rightarrow && \cos^2 \theta &=\frac{(1 + \mu)^2}{8\mu} \\ \end{align*} We need \begin{align*} && \frac{(1 + \mu)^2}{8\mu} & \leq 1 \\ \Rightarrow && 1 +2 \mu + \mu^2 \leq 8 \mu \\ \Rightarrow && 1 - 6 \mu + \mu^2 \leq 0 \end{align*} This quadratic has roots at \(3 \pm \sqrt{8}\) and therefore our quadratic inequality is satisfied if: \(\boxed{3 - \sqrt{8} \leq \mu \leq 3 + \sqrt{8}}\)

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Solution: Without loss of generality, let \(m = u = 1\).

1. \begin{align*} \text{COM}: && 1&= v_n + \lambda v_{n+1} \\ && &= v_n + \lambda v_{n+1}\\ \text{COE}: && \frac12 &= \frac12 v_n^2 + \frac12 \lambda v_{n+1}^2 \\ && 1 &= v_n^2 +\lambda v_{n+1}^2 \\ \\ \Rightarrow && v_n^2 + 2\lambda v_n v_{n+1} + \lambda^2 v_{n+1}^2 &= v_n^2 + \lambda v_{n+1}^2 \\ && \lambda v_{n+1}^2 &= v_{n+1}^2 - 2 v_n v_{n+1} \\ && \lambda v_{n+1} &= (v_{n+1} - 2v_n) \\ && (1-\lambda)v_{n+1} &= 2v_n \end{align*} Since \(v_{n+1} > v_n > 0\) this is only possible if \(\lambda < 1\)
2. \begin{align*} \text{COM}: && 1&= v_{n-1}+v_n+\lambda v_{n+1} \\ && 1&= v_{n-1} + v_n + \lambda v_{n+1} \\ \text{COE}: && \frac12 &= \frac12 v_{n-1}^2+\frac12v_n^2+\frac12\lambda v_{n+1}^2 \\ && 1&= v_{n-1}^2 + v_n^2 + \lambda v_{n+1}^2 \\ \\ \Rightarrow && 1 &= v_{n-1}^2 + v_n^2 + \lambda \frac{(1-v_{n-1}-v_n)^2}{\lambda^2} \\ &&&= v_{n-1}^2 + v_n^2 + \frac{(1-v_{n-1}-v_n)^2}{\lambda} \\ \Rightarrow && 1 &< v_{n-1}^2 + v_n^2 + (1-v_{n-1}-v_n)^2 \\ &&&= 2v_{n-1}^2+2v_n^2 + 1-2v_{n-1}-2v_{n-2} +2v_{n-1}v_n\\ \Rightarrow && v_{n-1}+v_n & <(v_{n-1}+v_n)^2 - v_{n-1}v_n \end{align*} but this cannot be true since \(0 < v_{n-1}+v_n < 1\) and \(v_n v_{n-1} > 0\)
3. The only way this is possible is if the first and last marble are moving. \begin{align*} \text{COM}: && 1 &= v_0 +\lambda v_{n+1} \\ \text{COE}: && \frac12 &= \frac12 v_0^2 + \frac12 \lambda^2 v_{n+1} \\ && 1 &= v_0^2 + \lambda v_{n+1}^2 \\ \Rightarrow && 2v_0 + \lambda v_{n+1} &= v_{n+1} \\ \Rightarrow && v_{n+1} &=\frac{2}{1-\lambda} v_0 \\ \Rightarrow && v_0 &= \frac{1-\lambda}{1+\lambda} \\ && v_{n+1} &= \frac{2}{1+\lambda} \end{align*} which will work since \(v_0\) can travel backwards.

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Solution:

As the string wraps around, the total length in contact will be \(a \theta\). The end contact point will be at \((a\cos \theta, a\sin \theta)\) and the string will be tangential to that. The tangent (unit) vector will be \(\binom{-\sin \theta}{\cos \theta}\), and so the particle will be at \(\binom{a\cos \theta - (L-a\theta) \sin \theta}{a \sin \theta + (L-a \theta) \cos \theta}\). The velocity will be: \begin{align*} \frac{\d}{\d t} \binom{a\cos \theta - (L-a\theta) \sin \theta}{a \sin \theta + (L-a \theta) \cos \theta} &= \binom{-a \sin \theta \cdot \dot{\theta} -(L-a \theta) \cos \theta \cdot \dot{\theta} + a \sin \theta \cdot \dot{\theta} }{a \cos \theta \cdot \dot{\theta} + (L-a \theta) \sin \theta \cdot \dot{\theta} - a \cos \theta \cdot \dot{\theta}} \\ &= \binom{-(L-a \theta) \cos \theta \cdot \dot{\theta} }{ (L-a \theta) \sin \theta \cdot \dot{\theta}} \\ \end{align*} Therefore the speed is \((L-a\theta) \dot{\theta}\). By conservation of energy, we must have that speed is constant, ie: \begin{align*} && (L - a \theta)\dot{\theta} &= v \\ \Rightarrow && \int_0^{L/a} (L - a \theta)\d \theta &= \int_0^T v \d t \\ \Rightarrow && vT &= \frac{L^2}{a} - a\frac{L^2}{2a^2} \\ &&&= \frac{L^2}{2a} \\ \Rightarrow && T &= \frac{L^2}{2av} \end{align*} as requried