Problems

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Solution: \begin{align*} \text{N2}: && 2\ddot{x}_1 &= -\frac{2}{(x_2-x_1)^3}\\ \text{N2}: && \ddot{x}_2 &= \frac{2}{(x_2-x_1)^3}\\ \Rightarrow && \ddot{x}_2 - \ddot{x}_1 &= \frac{3}{(x_1-x_2)^3} \\ \Rightarrow && \frac{\d^2 z}{\d t^2} &= \frac{3}{z^3} \\ \Rightarrow && v \frac{\d v}{\d z} &= \frac{3}{z^3} \\ \Rightarrow && \int v \d v &= \int \frac{3}{z^3} \d z \\ \Rightarrow && \frac{v^2}{2} &= -\frac{3}{2}z^{-2} + C \\ \Rightarrow && v^2 &= -3 z^{-2} + C' \\ t=0,z=1,v=-1: && 1 &= -3+C \Rightarrow C = 4 \\ \Rightarrow && \frac{\d z}{\d t} &= -\sqrt{4-3z^{-2}} \\ \Rightarrow && \int \d t &= -\int \frac{1}{\sqrt{4-3z^{-2}}} \d z \\ \Rightarrow && t &= \int \frac{z}{\sqrt{4z^2-3}} \d z \\ \Rightarrow && t &= -\frac14\sqrt{4z^2-3} + C \\ t=0, z = 1: && 0 &= -\frac14+C \\ \Rightarrow && C &= \frac14\\ \Rightarrow && 4t -1 &= -\sqrt{4z^2-3} \\ \Rightarrow && 16t^2+1-8t &= 4z^2-3 \\ \Rightarrow && z &= \sqrt{4t^2-2t+1} \end{align*} \begin{align*} && 2\ddot{x}_1 + \ddot{x}_2 &= 0 \\ \Rightarrow && 2x_1+x_2 &= At + B \\ t = 0, v = -1: && 2x_1+x_2 &= -t+1 \\ \\ \Rightarrow && x_2-x_1 &= \sqrt{4t^2-2t+1}\\ && 2x_1+x_2 &= 1-t \\ \Rightarrow && x_1 &= \frac13 \left (1-t-\sqrt{4t^2-2t+1} \right) \\ && x_2 &= \frac13(1-t + \sqrt{4t^2-2t+1}) \end{align*} This method of considering the relative position and considering the motion of the centre of mass is extremely common for solving systems of particles problems.

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Solution:

1. \(\,\) \begin{align*} && \ddot{x} &= kx \dot{x} \\ \Rightarrow && \frac{\d v}{\d x} \dot{x} &= k x \dot{x} \\ \Rightarrow && \int \d v &= \int k x \d x \\ \Rightarrow && v &= \frac12kx^2 + C \\ t=0, x = d, \dot{x} = kd^2: && kd^2 &= \frac12kd^2 + C \\ \Rightarrow && \dot{x} &= \frac12k(x^2+d^2) \\ \Rightarrow && \frac{\d x}{\d t} &= \frac12k(x^2+d^2) \\ \Rightarrow && \int \d t &= \int \frac{1}{\frac12k(x^2+d^2)} \d x \\ &&&= \frac{2}{kd}\tan^{-1} \frac{x}{d} \\ \Rightarrow && t &= \frac{2}{kd}\tan^{-1} \frac{x}{d} + C' \\ t = 0, x = d: && 0 &= \frac{\pi}{2kd} + C' \\ \Rightarrow && t &= \frac{2}{kd}\tan^{-1} \frac{x}{d}-\frac{\pi}{2kd} \end{align*} As \(x \to \infty\), \(t \to \frac{2}{kd} \frac{\pi}{2} - \frac{\pi}{2kd} = \frac{\pi}{2kd} \)
2. \(\,\) \begin{align*} && v &= \frac12kx^2 + C \\ t=0, x = d, \dot{x} = U && U &= \frac12kd^2 + C \\ \Rightarrow && \dot{x} &= \frac12k(x^2-d^2)+U \\ \Rightarrow && \frac{\d x}{\d t} &=\frac12k(x^2-d^2)+U \\ \Rightarrow && \int \d t &= \int \frac{1}{\frac12k(x^2-d^2)+U} \d x \\ && &=\frac{2}{k} \int \frac{1}{x^2-d^2+\frac{2U}k} \d x \\ &&&= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} \\ \Rightarrow && t &= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} + C'' \\ t = 0, \dot{x} = d: && 0 &= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-d}{d+\sqrt{d^2-\frac{2U}k}} + C'' \\ \Rightarrow && t &= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \left ( \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} \frac{d+\sqrt{d^2-\frac{2U}k}}{ \sqrt{d^2-\frac{2U}k}-d} \right ) \end{align*} as \(t \to \infty\) the denominator needs to head to \(0\), ie \(x \to -\sqrt{d^2-\frac{2U}k}\)

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Solution: Looking at the forces on Jane, \(kv < g \Rightarrow v < \frac{g}{k}\). For Karen we have \begin{align*} kv + (2k^2/g)v^2 &< g\\ -g^2 + gkv + (2k^2)v^2 &< 0 \\ (2kv-g)(kv+g) &< 0\\ \Rightarrow v &< \frac{g}{2k} \end{align*} \begin{align*} && \dot{v} &= g - kv \\ \Rightarrow && \frac{\dot{v}}{g - kv} &= 1 \\ \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv\\ && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ && &= \left [-\frac{1}{k} \ln \l g - kv \r \right ]_0^{g/(3k)} \\ && &= \frac{1}{k} \ln \l g \r - \frac{1}{k} \ln \l \frac{2}{3}g \r\\ &&&= \frac{1}{k} \ln \l \frac{3}{2} \r \end{align*} \begin{align*} && \dot{v} &= g - kv - (2k^2/g)v^2 \\ \Rightarrow && \frac{\dot{v}}{g - kv - (2k^2/g)v^2} &= 1 \\ \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv - (2k^2/g)v^2} dv \\ && &= \int_0^{g/(3k)} \frac{g}{(g-2kv)(kv+g)} dv\\ && &= \int_0^{g/(3k)} \l \frac{2}{3(g-2kv)} + \frac{1}{3(kv+g)} \r dv\\ && &= \left [ \l -\frac{1}{3k} \ln (g-2kv) + \frac{1}{3k}\ln(kv+g) \r \right ]_0^{g/(3k)} \\ && &= \left [ \l -\frac{1}{3k}\ln \l \frac{g}{3} \r + \frac{1}{3k}\ln \l \frac{4g}{3} \r \r \right ] - \left [- \frac1{3k} \ln(g) + \frac{1}{3k} \ln (g) \right ] \\ && &= \frac{1}{3k} \ln \l 4 \r \end{align*} NB: \(\sqrt[3]{4} \approx 1.58 > \frac{3}{2}\) so Karen has been in free-fall for longer, but not \emph{much} longer than Jane.

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Solution: While the force is constant, the train is accelerating at \(\frac{F}{M}\), and since \(u = \frac{F}{M} t_1 \Rightarrow t_1 = \frac{Mu^2}{Fu} = \frac{Mu^2}{P}\). Once the train is being driven at a constant rate, we can observe that change in energy will be power times time, ie \(Pt_2 = \frac{1}{2}M(v^2 - u^2) \Rightarrow t_2 = \frac{M}{2P} ( v^2 - u^2)\). Therefore the total time will be \(t_1 + t_2 = \frac{M}{2P} ( u^2 + v^2)\). During the first period, the distance will be: \(s_1 = \frac12 \frac{F}{M} t_1^2 = \frac12 \frac{F}{M} \frac{M^2u^2}{F^2} = \frac{Mu^3}{2P}\) In the second period, \(P = Fu\) and so \(\text{Force} = \frac{P}{v} \Rightarrow M v \frac{\d v}{\d x} = \frac{P}{v} \Rightarrow M \l \frac{v^3}{3} - \frac{u^3}{3}\r = Ps_2\) and therefore total distance will be: \(\frac{M}{6P}(2v^{3}+u^{3})\)

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Solution: While descending, the body experiences the force \(-\frac15mg - 2mv^2\). \begin{align*} \text{N2:} && m \dot{v} &= -\frac15 mg - 2mv^2 \\ \Rightarrow && \frac{\dot{v}}{\frac15g + 2v^2} &= -1 \\ \Rightarrow && \frac{1}{2}\tan^{-1} v_1 - \frac{1}{2}\tan^{-1} {v_0} &= -T \end{align*} We care about when \(v_1 = 0\), ie \(\displaystyle T = \frac{1}{2}\tan^{-1} {v_0} < \frac12 \frac{\pi}2 = \frac{\pi}4\) seconds. If the body enters at speed \(1\ \mathrm{ms}^{-1}.\) then for the first part of it's journey it will experience forces \(-\frac15mg - 2mv^2\) and so: \begin{align*} \text{N2:} && m v \frac{\d v}{\d x} &= -\frac15 mg - 2mv^2 \\ \Rightarrow && \int \frac{v}{2(1 + v^2)} \d v &= \int -1 \d x \\ \Rightarrow && \frac14 \ln (1 + v^2) &= -x \end{align*} Therefore the depth is \(\frac14 \ln 2\) metres. When the body is rising, it experiences forces of: \(\frac15mg - 2mv^2\) and so: \begin{align*} \text{N2:} && m v \frac{\d v}{\d x} &= \frac15mg - 2mv^2 \\ \Rightarrow && \int \frac{v}{2(1 - v^2)} \d v &= \int -1 \d x \\ \Rightarrow && -\frac14 \ln (1 - v^2) &= \frac14 \ln 2 \\ \Rightarrow && 1-v^2 &= \frac12 \\ \Rightarrow && v &= \frac{1}{\sqrt{2}} \ \mathrm{ms}^{-1} \end{align*} This will take \begin{align*} \text{N2:} && m \dot{v} &= \frac15mg - 2mv^2 \\ \Rightarrow && \frac{\dot{v}}{2(1-v^2)} &= -1 \\ \Rightarrow && \dot{v} \frac{1}{4}\l \frac{1}{1 - v} + \frac{1}{1+v} \r &= -1 \\ \Rightarrow && \frac{1}{4} \l -\ln(1 - v) + \ln(1 + v)\r &= -T \end{align*} Since \(v = \frac{1}{\sqrt{2}}\) \begin{align*} T &= \frac{1}{4} \ln \l \frac{1+ \frac1{\sqrt{2}}}{1 - \frac1{\sqrt{2}}}\r \\ &= \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \end{align*} and therefore the total time will be: \begin{align*} & \frac12 \tan^{-1} 1 + \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \\ =& \frac{\pi}{8} + \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \end{align*}

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Solution: Using Newton's second law, we have: \begin{align*} && -M\omega(v^2+V^2)/v &= M v \frac{\d v}{\d x} \\ \Rightarrow && \frac{v^2}{v^2+V^2} \frac{\d v}{\d x} &= -\omega \\ \Rightarrow && \omega X &= \int_{V/2}^V \frac{v^2}{v^2+V^2} \d v \\ &&&= \int_{V/2}^V \l 1 - \frac{V^2}{v^2+V^2} \r \d v \\ &&&= \left [v - V\tan^{-1} \frac{v}{V} \right]_{V/2}^V \\ &&&= V \l \frac12 - \tan^{-1} 1 + \tan^{-1} \frac12 \r \\ \Rightarrow X &= \frac{V}{\omega} \l \tan^{-1} \frac12 + \frac12 - \frac{\pi}{4} \r \end{align*}. The resistive force just before the field changes is \(M \omega (\frac{V^2}{4} + V^2)/\frac{V}{2} = \frac52MV\omega\). Therefor the constant resistive force is between \(\frac{11}4MV\omega\) and \(\frac{9}{4}MV \omega\) and acceleration is \(\frac{11}{4}V\omega, \frac{9}{4}V\omega\). Since \(v^2 = u^2 + 2as \Rightarrow s = \frac{v^2-u^2}{2a} = \frac{\frac{V^2}{4}}{2kV\omega} = \frac{V}{8k\omega}\) therefore \(z \in \left [ \frac{V}{22\omega},\frac{V}{18 \omega} \right]\)

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Solution:

Considering the pulley system with the lift (now of mass \(M-m\)) and the counterweight of mass \(M\). Once they start moving, since they are connected by a light inextensible string they must move with the same speed (and by extension the same acceleration). (Up to sign) \begin{align*} \text{N2(lift,}\uparrow):&&(M-m)a &= T-(M-m)g \\ \text{N2(couterweight,}\downarrow):&&Ma &= Mg - T \\ \Rightarrow && (2M-m)a &= mg \\ \Rightarrow && a &= \frac{mg}{2M-m} \end{align*} We could treat the situation as the tile travelling a distance of \(h\) with acceleration \(\displaystyle g \left ( 1 + \frac{m}{2M-m} \right) = g \frac{2M}{2M-m}\). \begin{align*} t &= \sqrt{\frac{2h}{g \frac{2M}{2M-m}}} \\ &= \sqrt{\frac{(2M-m)h}{Mg}} \\ \end{align*} Since the collision between the lift and tile is perfectly inelastic, they immediately coalesce. There is also an impulse in the pulley system, which goes over the pulley, which means there is an impulse acting vertically on the lift and the counterweight. Assume afterwards the lift (and tile) is travelling upwards with speed \(V\) and the counterweight is travelling downwards with speed \(V\) (ie velocity \(-V\)). \begin{align*} \text{for the lift/tile}: && I_{LC} &= (\text{momentum after}) - (\text{momentum before})\\ &&&= MV - ((M-m)at +m(-g)t) \\ &&&= MV - Mat + m(a-g)t \\ \text{for the counterweight}: && I_{LC} &= (\text{momentum after}) - (\text{momentum before})\\ &&&= M(-V) - (M(-a)t) \\ &&&= -MV +Mat \\ \Rightarrow && 2MV &= m(g-a)t + 2Mat \\ &&&= t (2Ma -ma+mg) \\ &&&= 0 \\ \Rightarrow && V &= 0 \end{align*} Therefore, the lift ends up stationary. The energy lost in the collision is: \begin{align*} && E &= KE_{before} - KE_{after} \\ &&&= \underbrace{\frac12 (M-m)a^2 t^2}_{lift} + \underbrace{\frac12 mg^2 t^2}_{tile} + \underbrace{\frac12 Ma^2 t^2}_{counterweight} - \underbrace{0}_{\text{everything is at rest after}} \\ &&&= \frac12 \l (M-m)a^2 + mg^2 + Ma^2 \r t^2 \\ &&&= \frac12 \l 2Ma^2-ma^2 + mg^2 \r t^2 \\ &&&= \frac12 \l (2M-m)a^2 + mg^2 \r t^2 \\ &&&= \frac12 \l mga + mg^2 \r t^2 \\ &&&= \frac12 mg (a + g)t^2 \\ &&&= \frac12 mg \left ( \frac{mg}{2M-m} + g\right ) \frac{(2M-m)h}{Mg} \\ &&&= \frac12 mg \left ( \frac{mg +2Mg - mg}{2M-m} \right) \frac{(2M-m)h}{Mg} \\ &&&= mgh \end{align*} as required.

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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: \[ x(t) = \begin{cases} \frac{v_0}{\mu} \sin \mu t & 0 \leq t \leq \frac{\pi}{\mu} \\ -\frac{v_0}{\kappa} \l 1-e^{-\kappa(t-\frac{\pi}{\mu})}\r & t > \frac{\pi}{\mu}\end{cases}\]