A small block of mass \(km\) is initially at rest on a smooth horizontal surface. Particles \(P_1\), \(P_2\), \(P_3\), \(\ldots\) are fired, in order, along the surface from a fixed point towards the block. The mass of the \(i\)th particle is \(im\) (\(i = 1, 2, \ldots\))and the speed at which it is fired is \(u/i\,\). Each particle that collides with the block is embedded in it. Show that, if the \(n\)th particle collides with the block, the speed of the block after the collision is \[ \frac{2nu}{2k +n(n+1)}\,. \] In the case \(2k = N(N+1)\), where \(N\) is a positive integer, determine the number of collisions that occur. Show that the total kinetic energy lost in all the collisions is \[ \tfrac12 mu^2\bigg( \sum_{n=2}^{N+1} \frac 1 n \bigg)\,. \]
Solution: \begin{align*} \text{COM}: && \sum_{i=1}^n im \cdot \frac{u}{i} &= \left ( km + \sum_{i=1}^n im \right) v \\ \Rightarrow && nu &= \left ( k + \frac{n(n+1)}{2} \right) v \\ \Rightarrow && v &= \frac{2nu}{2k + n(n+1)} \end{align*} If \(2k = N(N+1)\), there will be no more collisions when \(v_n > \frac{u}{n+1}\), ie \begin{align*} && \frac{u}{n+1} &<\frac{2nu}{2k + n(n+1)} \\ \Leftrightarrow && N(N+1) + n(n+1) &< 2n(n+1) \\ \Leftrightarrow && N(N+1) &< n(n+1) \\ \end{align*} Therefore \(n = N+1\) and there will be \(N+1\) collisions. The loss of kinetic energy is: \begin{align*} && \text{initial k.e.} &= \sum_{k=1}^{N+1} \frac12 im \cdot \frac{u^2}{i^2} \\ &&&= \frac12 m u^2 \left ( \sum_{k=1}^{N+1} \frac{1}{i}\right) \\ && \text{final k.e.} &= \frac12 \left ( k + \frac{(N+1)(N+2)}{2}\right)m \left ( \frac{2(N+1)u}{N(N+1)+(N+1)(N+2)} \right)^2 \\ &&&= \frac12 m u^2 \frac{2(N+1)^2}{(N+1)(2N+2)} \\ &&&= \frac12 mu^2 \\ \Rightarrow && \Delta \text{ k.e.} &= \frac12 m u^2 \left ( \sum_{k=2}^{N+1} \frac{1}{i}\right) \end{align*}
A modern villa has complicated lighting controls. In order for the light in the swimming pool to be on, a particular switch in the hallway must be on and a particular switch in the kitchen must be on. There are four identical switches in the hallway and four identical switches in the kitchen. Guests cannot tell whether the switches are on or off, or what they control. Each Monday morning a guest arrives, and the switches in the hallway are either all on or all off. The probability that they are all on is \(p\) and the probability that they are all off is \(1-p\). The switches in the kitchen are each on or off, independently, with probability \(\frac12\).
Solution:
In this question, you may assume that \(\displaystyle \int_0^\infty \!\!\! \e^{-x^2/2} \d x = \sqrt{\tfrac12 \pi}\,\). The number of supermarkets situated in any given region can be modelled by a Poisson random variable, where the mean is \(k\) times the area of the given region. Find the probability that there are no supermarkets within a circle of radius \(y\). The random variable \(Y\) denotes the distance between a randomly chosen point in the region and the nearest supermarket. Write down \(\P(Y < y)\) and hence show that the probability density function of \(Y\) is \(\displaystyle 2\pi y k \e^{-\pi k y^2}\) for \(y\ge0\). Find \(\E(Y)\) and show that \(\var(Y) = \dfrac{4-\pi}{4\pi k}\).
Solution: A circle radius \(y\) has a number of supermarkets \(X\) where \(X \sim Po(k \pi y^2)\). \[ \mathbb{P}(X = 0) = e^{-k\pi y^2} \frac{1}{0!} = e^{-k\pi y^2} \] The probability \(\mathbb{P}(Y < y) = 1-\mathbb{P}(Y \geq y) = 1-e^{-k\pi y^2}\), and in particular \(f_Y(y) = 2k\pi y e^{-k\pi y^2}\) (by differentiating). \begin{align*} && \mathbb{E}(Y) &= \int_0^\infty yf_Y(y) \d y \\ &&&= \int_0^\infty 2\pi y^2 k e^{-\pi k y^2} \d y \\ \sigma^2 = \frac{1}{2k\pi}:&&&= \pi k \sqrt{2 \pi}\sigma \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi} \sigma }y^2 e^{-\frac12 \cdot 2\pi k y^2} \d y \\ &&&=\pi k \sqrt{2 \pi}\sigma \mathbb{E}\left (N(0, \sigma^2)^2 \right) \\ &&&= \pi k \sqrt{2 \pi}\sigma\sigma^2 \\ &&&= \pi k \sqrt{2 \pi} \frac{1}{(2k\pi)^{3/2}} \\ &&&= \frac{1}{2\sqrt{k}} \end{align*} \begin{align*} && \mathbb{E}(Y^2) &= \int_0^\infty y^2f_Y(y) \d y \\ &&&= \int_0^\infty 2\pi y^3 k e^{-\pi k y^2} \d y \\ &&&= \int_0^{\infty}y^2 2y \pi k e^{-\pi k y^2} \d y \\ \\ &&&= \left [-y^2 e^{-\pi k y^2}\right]_0^{\infty}+\int_0^\infty 2ye^{-\pi k y^2} \d y \\ &&&= \left [-\frac{1}{\pi k}e^{-\pi k y^2} \right]_0^{\infty} \\ &&&= \frac{1}{\pi k} \\ \Rightarrow && \textrm{Var}(Y) &= \mathbb{E}(Y^2) - \left [ \mathbb{E}(Y)\right]^2 \\ &&&= \frac{1}{\pi k} - \frac{1}{4k} \\ &&&= \frac{4 - \pi}{4\pi k} \end{align*}