Problems

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Solution:

1. Assuming the particles have mass \(m\), and speed \(v\) just before collision, then \begin{align*} \text{COE}: && \underbrace{\frac12 m u^2}_{\text{initial kinetic energy}} + \underbrace{0}_{\text{initial GPE}} &= \underbrace{\frac12m v^2}_{\text{kinetic energy just before collision}} + \underbrace{mgr\left(1-\frac1{\sqrt{2}}\right)}_{\text{GPE just before collision}} \\ \Rightarrow && v^2 &= u^2 - gr(2-\sqrt{2}) \end{align*} Therefore the particles has velocity \(\frac{\sqrt{2}}2v \binom{\pm 1}{1}\) before the collision. By symmetry, the impulse between the particles will be horizontal, so the vertical velocities will be unchanged at \(\frac{\sqrt{2}}{2}v\). By conservation of momentum (or symmetry) the particles will have equal but opposite velocities after the collision (say \(w\)) satisfying: \[ e = \frac{2w}{2\frac{\sqrt{2}}{2}v} \] ie \(w = \frac{\sqrt{2}}2 e v\) as required.
2. Once the particles have rebounded, they will be projectiles whilst the strings are slack. If we consider the left-most point \(A = (0,0)\) then the particles colide at \(\left ( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)\) and the position at time \(t\) after the collision (before the string goes slack) will be: \begin{align*} \mathbf{x}_t &= \frac{\sqrt{2}}{2}r\binom{1}{-1} + \frac{\sqrt{2}}{2} vt \binom{-e}{1} + \frac12 gt^2 \binom{0}{-1} \end{align*} The string will go taught when \(|\mathbf{x}_t|^2 = r^2\), ie \begin{align*} && r^2 &= \left ( \frac{\sqrt{2}}{2} r - \frac{\sqrt{2}}{2}evt \right)^2 + \left (-\frac{\sqrt{2}}{2} r + \frac{\sqrt{2}}{2}vt -\frac12 gt^2 \right)^2 \\ \Rightarrow && r^2 &= \frac12 \left (r - evt \right)^2 + \frac12 \left (-r+vt - \frac{\sqrt{2}}{2}gt^2 \right)^2 \\ \Rightarrow && 2r^2 &= \left (r - evt \right)^2 + \left (-r+vt - \frac{\sqrt{2}}{2}gt^2 \right)^2 \\ \end{align*} as required.
3. Suppose \(z = \frac{vt}{r}\), \(c = \frac{\sqrt{2}v^2}{rg}\), then \begin{align*} && 2r^2 &= \left (r - evt \right)^2 + \left (-r+vt - \frac{\sqrt{2}}{2}gt^2 \right)^2 \\ \Leftrightarrow && 2 &= \left (1 - e\frac{vt}{r} \right)^2 + \left (-1 + \frac{vt}{r}- \frac{\sqrt{2}}{2} \frac{gt^2}{r} \right)^2 \\ \Leftrightarrow && 2 &= \left (1 - ez \right)^2 + \left (-1 +z- \frac{v^2t^2}{r^2} \frac{gr}{\sqrt{2}v^2}\right)^2 \\ \Leftrightarrow && 2 &= \left (1 - ez \right)^2 + \left (-1 +z- \frac{z^2}{c} \right)^2 \\ \Leftrightarrow && 2 &= 1-2ez + e^2z^2 + 1 + z^2 +\frac{z^4}{c^2} - 2z-2\frac{z^3}{c}+2\frac{z^2}{c} \\ \Leftrightarrow && 0 &= z(-2e-2) + z^2(e^2+1 + \frac{2}{c}) + z^3(-\frac{2}{c}) + z^4 \frac{1}{c^2} \\ \underbrace{\Leftrightarrow}_{z \neq 0} && 0 &= \left ( \frac{z}{c} \right)^3 - 2\left ( \frac{z}{c} \right)^2 + \left ( \frac{z}{c} \right) (1 + e^2 + \frac{2}{c} ) - \frac{2}{c}(1+e) \end{align*} as required, (where on the last step we divide by \(z/c\)).
4. If a cubic has \(3\) equal, non-zero roots then it must have the form \((z-a)^3 = z^3 -3az^2 + 3a^2 z -a^3 = 0\), so \(3a = 2\), and so the expansion must be \(\left ( \frac{z}{c} \right)^3 - 2\left ( \frac{z}{c} \right)^2 + \frac{4}{3}\left ( \frac{z}{c} \right) - \frac{8}{27} = 0\) \begin{align*} && \frac{2}{c}(1+e) &= \frac{8}{27} \\ \Rightarrow && \frac{2}{c} &= \frac{8}{27} \frac{1}{1+e} \\ && 1 + e^2 + \frac{2}{c} &= \frac43 \\ \Rightarrow && e^2 + \frac{8}{27(1+e)} &= \frac{1}{3} \\ \Rightarrow && 27(1+e)e^2+8 &= 9(1+e) \\ \Rightarrow && 27e^3 + 27e^2-9e-1 &= 0 \\ \Rightarrow && (3e-1)(9e^2+12e+1) &= 0 \end{align*} The only (positive) root is \(e = \frac13\), therefore \(e = \frac13\). We must also have \begin{align*} && \frac{2}{c} \frac43 &= \frac{8}{27} \\ \Rightarrow && c &= 9 \\ \Rightarrow && \frac{\sqrt{2}v^2}{rg} &= 9 \\ \Rightarrow && v^2 &= \frac{9\sqrt{2}rg}{2} \end{align*} as required. If we consider the path of the particle acting as a projectile, iff the path is tangent to the circle then there will be exactly one solution for \(z/c\) and (importantly) it will be a repeated root. Therefore the particle rejoins the circle at a tangent and the tension is acting perpendicularly to the direction of motion (ie no energy loss).
5. Since the only energy lost is lost in the collision, we can apply conservation of energy again: \begin{align*} \text{COE:} && \frac12 m U^2 &= \frac12 m \frac12v^2(1+e^2) + mgr\left (1 - \frac1{\sqrt{2}} \right) \\ \Rightarrow && U^2 &= \frac12 \frac{9 \sqrt{2}}{2}gr(1+\frac19) + gr(2 - \sqrt{2}) \\ &&&= \left (\frac{5\sqrt{2}}{2}+2 - \sqrt{2} \right)gr \\ &&&= \left (\frac{4+3\sqrt{2}}{2} \right)gr \end{align*}

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Solution:

1. The line to the circle is tangent, and the point it meets the circle is \((a \cos \theta, a \sin \theta)\) and it will be a distance \(b - a \theta\) away, therefore it is at \((a \cos \theta - (b-a \theta) \sin \theta, a \sin \theta + (b-a \theta) \cos \theta)\)
2. The velocity will be \(\displaystyle \binom{-a \dot{\theta}\sin \theta-b \dot{\theta}\cos \theta + a \dot{\theta} \sin \theta + a \theta \dot{\theta} \cos \theta}{ a \dot{\theta} \cos \theta - b \dot{\theta} \sin \theta -a \dot{\theta} \cos \theta + a \theta \dot{\theta} \sin \theta}= \binom{-b \dot{\theta}\cos \theta + a \theta \dot{\theta} \cos \theta}{ - b \dot{\theta} \sin \theta + a \theta \dot{\theta} \sin \theta}\) Therefore the speed will be \(\dot{\theta}(b-a\theta)\)
3. Conservation of energy and the fact that the tension is perpendicular to the velocity means no work is being done on the particle and hence it's speed is unchanged. So \(u = \dot{\theta}(b-a\theta)\).
4. Note that the acceleration is \begin{align*} && \mathbf{a} &= \frac{\d}{\d t} \left (-\dot{\theta}(b-a\theta) \binom{\cos \theta}{\sin \theta} \right) \\ &&&=-u \dot{\theta}\binom{-\sin \theta}{\cos \theta} \\ \Rightarrow && T &= ma \\ &&&= \frac{mu^2}{b - a \theta} \end{align*} It would be valuable to have \(\theta\) in terms of \(t\), so we want to solve \begin{align*} &&\frac{\d \theta}{\d t} (b-a\theta) &= u \\ \Rightarrow && b \theta - a\frac{\theta^2}{2} + C &= ut \\ t = 0, \theta = 0: && C &= 0 \\ \Rightarrow && b\theta - \frac{a}{2} \theta^2 &= ut \\ \Rightarrow && \theta &= \frac{b \pm \sqrt{b^2-2aut}}{a} \end{align*} At \(t\) increases, \(\theta\) increases so \(a\theta = b -\sqrt{b^2-2aut}\) or \(b-a \theta = \sqrt{b^2-2aut}\) and the result follows

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Solution:

1. \(\mathbf{r} = (a \sin \theta - s) \mathbf{i}+a\cos \theta\mathbf{j}\), so \begin{align*} && \dot{\mathbf{r}} &=(a \dot{\theta} \cos \theta - \dot{s}) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j}\\ \\ \text{COM}(\rightarrow): && 0 &= M(-\dot{s}) + m(a \dot{\theta} \cos \theta - \dot{s}) \\ \Rightarrow && \dot{s} &= \frac{ma \dot{\theta} \cos \theta}{m+M} \\ &&&= \left ( 1- \frac{M}{m+M} \right) a\dot{\theta} \cos \theta \\ &&&= (1 - k) a\dot{\theta} \cos \theta \\ \\ \Rightarrow && \dot{\mathbf{r}} &=(a \dot{\theta} \cos \theta - \dot{s}) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j} \\ &&&= (a \dot{\theta} \cos \theta - (1 - k) a\dot{\theta} \cos \theta) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j} \\ &&&= a\dot{\theta} \left ( k \cos \theta \mathbf{i} - \sin \theta \mathbf{j} \right) \end{align*}
2. \(\,\) \begin{align*} COE: &&\underbrace{0}_{\text{k.e.}}+ \underbrace{mga}_{\text{GPE}} &= \underbrace{\frac12 m \mathbf{\dot{r}}\cdot\mathbf{\dot{r}}}_{\text{k.e. }P} + \underbrace{mg a\cos \theta}_{\text{GPE}} + \underbrace{\frac12 M \dot{s}^2}_{\text{k.e. hemisphere}} \\ \Rightarrow && 2amg(1-\cos \theta) &= a^2m \dot{\theta}^2(k^2 \cos^2 \theta + \sin^2 \theta)+ M(1 - k)^2 a^2\dot{\theta}^2 \cos^2 \theta \\ \Rightarrow && 2mg(1-\cos \theta) &= a \dot{\theta}^2 \left (m\sin^2 \theta + (mk^2 + M(1-k)^2)\cos^2 \theta \right) \\ &&&= a \dot{\theta}^2 \left (m\sin^2 \theta + mk\cos^2 \theta \right) \\ \Rightarrow && 2g(1-\cos \theta) &= a \dot{\theta}^2 \left (\sin^2 \theta + k\cos^2 \theta \right) \\ \end{align*}
3. The equation of motion is \(m \ddot{\mathbf{r}} = \mathbf{R} - mg\mathbf{j}\) and the particle will leave the surface when \(\mathbf{R} = 0\). If we take the component in the directions suggested: \begin{align*} && \ddot{\mathbf{r}} &= a\ddot{\theta}(k \cos \theta \mathbf{i}- \sin \theta \mathbf{j}) + a \dot{\theta}(-k\dot{\theta} \sin \theta \mathbf{i}- \dot{\theta} \cos \theta \mathbf{j}) \\ &&&= ak (\ddot{\theta} \cos \theta - \dot{\theta}^2 \sin \theta) \mathbf{i} -a(\ddot{\theta} \sin \theta + \dot{\theta}^2 \cos \theta) \mathbf{j} \\ \Rightarrow && \mathbf{\ddot{r}} \cdot (\sin \theta \mathbf{i} + k \cos \theta \mathbf{j}) &= ak (\ddot{\theta} \cos \theta - \dot{\theta}^2 \sin \theta) \sin \theta -ak(\ddot{\theta} \sin \theta + \dot{\theta}^2 \cos \theta)\cos \theta \\ &&&= - ak \dot{\theta}^2 \\ && (-g\mathbf{j}) \cdot (\sin \theta \mathbf{i} + k \cos \theta \mathbf{j}) &= -gk \cos \theta \\ \mathbf{R} = 0: && gk \cos \theta &= ak \dot{\theta}^2 \\ \Rightarrow && g \cos \theta &= a \dot{\theta}^2 \end{align*}
4. \(\,\) \begin{align*} && 2g(1-\cos \theta) &= a \dot{\theta}^2(k \cos^2 \theta + \sin^2 \theta) \\ && a \dot{\theta}^2 &= g \cos \alpha \\ \Rightarrow && 2g(1-\cos \alpha) &= g \cos \alpha(k \cos^2 \alpha + (1-\cos^2 \alpha)) \\ \Rightarrow && 0 &= g(k-1)c^3+3gc-2g \\ \Rightarrow && 0 &= (k-1)c^3+3c - 2 \end{align*} When \(c =1, f(c) = k > 0\) when \(c = \frac23, f(c) = k-1 < 0\). Therefore there is a root with \(\cos \alpha > \frac23\)

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Solution: Applying conservation of energy, since there are no external forces (other than gravity) the condition to reach the house (with any speed) is the initial GPE is larger than the final GPE, ie: \begin{align*} && m g x \sin \alpha &\geq m g d \sin \beta \\ \Rightarrow && x \sin \alpha &\geq d \sin \beta \end{align*} Let \(T_1\) be the time taken on the downward section, and \(T_2\) the time taken on the upward section, then: \begin{align*} && s &= ut + \frac12 a t^2 \\ \Rightarrow && x &= \frac12 g \sin \alpha T_1^2 \\ \Rightarrow && T_1^2 &= \frac{2x}{g \sin \alpha} \\ && v &= u + at \\ \Rightarrow && v &= T_1 g \sin \alpha \\ && mg x \sin \alpha &= mg d \sin \beta + \frac12 m w^2 \\ \Rightarrow && w &= \sqrt{2(x \sin \alpha - d \sin \beta)} \\ && w &= v - g \sin \beta T_2 \\ \Rightarrow && T_2 &= \frac{v - w}{g \sin \beta} \\ \Rightarrow && T &= T_1 + T_2 \\ &&&= \sqrt{\frac{2x}{g \sin \alpha}} + \frac{\sqrt{\frac{2x}{g \sin \alpha}} g \sin \alpha- \sqrt{2(x \sin \alpha - d \sin \beta)}}{g \sin \beta} \\ &&&= \left ( \frac{2}{g \sin \alpha} \right)^{\tfrac12} \left ( \sqrt{x} + \sqrt{x}k - \sqrt{k^2x-kd}\right) \end{align*} Differentiating wrt to \(x\), we obtain: \begin{align*} && \frac{\d T}{\d x} &= C(-(1+k)x^{-1/2}+k^2(k^2 x - kd)^{-1/2}) \\ \text{set to }0: && 0 &= k^2(k^2 x - kd)^{-1/2} - (1+k)x^{-1/2} \\ \Rightarrow && \sqrt{x} k^2 &= \sqrt{k^2x - kd} (1+k) \\ \Rightarrow && x k^4 &= (k^2x-kd)(1+k)^2 \\ \Rightarrow && x(k^4-k^2(1+k)^2) &= -kd(1+k)^2 \\ \Rightarrow && x(2k^2+k) &= d \\ \Rightarrow && x &= \frac{d}{(2k^2+k)} \end{align*}

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Solution: \begin{align*} && v^2 &= u^2 +2as \\ \Rightarrow && V^2 &= 2 g \cdot (8h)\\ \Rightarrow && V &=4\sqrt{hg}\\ \end{align*} When the first particle collides with the ground, the second particle is at \(9h\) traveling with speed \(V\), the first particle is at \(0\) traveling (upwards) with speed \(\tfrac12 V\). For a collision we need: \begin{align*} && \underbrace{\frac12 V t- \frac12 g t^2}_{\text{position of A}} &= \underbrace{9h - Vt - \frac12 gt^2}_{\text{position of B}} \\ \Rightarrow && \frac32Vt &= 9h \\ \Rightarrow && t &= \frac{6h}{V} \\ \\ && \underbrace{\frac12 V t- \frac12 g t^2}_{\text{position of A}} &= \frac12 V \frac{6h}{V} - \frac12 g t^2 \\ &&&= 3h - \frac12 g\frac{36h^2}{16hg} \\ &&&= 3h - \frac{9}{8}h \\ &&&= \frac{15}{8}h \end{align*} Just before the collision, \(A\) will be moving with velocity (taking upwards as positive) \begin{align*} && u_A &= \frac12 V-gt \\ &&&= 2\sqrt{hg}-g \frac{6h}{V} \\ &&&= 2\sqrt{hg} - g \frac{6h}{4\sqrt{hg}} \\ &&&= 2\sqrt{hg}-\frac32\sqrt{hg} \\ &&&= \frac12 \sqrt{hg} \end{align*} Similarly, for \(B\). \begin{align*} && u_B &= -V -gt \\ &&&= -4\sqrt{hg} - \frac32\sqrt{hg} \\ &&&= -\frac{11}{2}\sqrt{hg} \end{align*} Considering \(A\), to figure out \(v_A\). \begin{align*} && v^2 &= u^2 + 2as \\ && V^2 &= v_A^2 + 2g\frac{15}{8}h \\ && 16hg &= v_A^2 + \frac{15}{4}gh \\ \Rightarrow && v_A^2 &= \frac{49}{4}gh \\ \Rightarrow && v_A &= -\frac{7}{2}\sqrt{gh} \end{align*}

To keep things clean, lets use units of \(\sqrt{hg}\) so we don't need to focus on that for now: \begin{align*} \text{COM}: && \frac12 - \frac{11}{2} &= -\frac{7}{2}+v_B \\ \Rightarrow && v_B& =-\frac{3}{2} \\ \text{NEL}: && e &= \frac{2}{6} = \frac13 \end{align*}

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Solution:

First, notice that the centre of mass will lie directly below \(A\) and will be \(\frac{m}{M+m}\) of the way between \(O\) and \(P\). Therefore \(\sin \beta = \frac{m}{M+m}\). The cosine rule states that: \begin{align*} && AP^2 &= a^2 + a^2 - 2a^2 \cos \angle AOP \\ \Rightarrow && \frac{AP^2}{a^2} &= 2 - 2 \sin \beta \\ &&&= \frac{2M+2m-2m}{M+m} \\ &&&= \frac{2M}{M+m} \\ \Rightarrow && \frac{AP}{a} &= \sqrt{\frac{2M}{M+m}} \end{align*} Considering conservation of energy, we have: Rotational kinetic energy for the disc: \(\frac12 I \dot{\theta}^2\) Kinetic energy for the particle: \(\frac12 m (\dot{\theta} \sqrt{2-2\sin \beta} a)^2 = (1- \sin \beta)ma^2 \dot{\theta}^2\) GPE: The important thing is the vertical location of \(G\). The triangle \(OAG\) will still have angle \(\beta\) at \(A\). The vertical height below is: \(\cos \theta \cdot AG = \cos \theta a \cos \beta\). The distance from when \(\theta = 0\) will be \(a \cos \beta (1- \cos \theta)\) and so the GPE will be \((M+m)ga \cos \beta ( 1- \cos \theta)\) we can therefore say by conservation of energy: \[ \frac12 I \dot{\theta}^2 + (1- \sin \beta)ma^2 \dot{\theta}^2+(M+m)ga \cos \beta ( 1- \cos \theta) \] is constant. Suppose \(m = \frac32 M\) and \(I = \frac32 Ma^2\) then differentiating the constant wrt to \(\theta\) gives \(\sin \beta = \frac{m}{M+m} = \frac{3}{5}, \cos \beta = \frac45\) \begin{align*} && 0 &= \frac12 \frac32 M a^2 \cdot 2 \dot{\theta}\ddot{\theta} + (1- \sin \beta)\frac32M a^2 2 \dot{\theta}\ddot{\theta} + (M+\frac32M) ga \cos \beta \sin \theta \cdot \dot{\theta} \\ \Rightarrow && 0 &= \frac32 \ddot{\theta} + 3(1-\sin \beta) \ddot{\theta} + \frac{5}{2}\frac{g}{a} \cos \beta \sin \theta \\ &&&= (\frac32 + \frac65) \ddot{\theta} + \frac{2g}{a} \sin \theta \\ &&&= \frac{27}{10} \ddot{\theta} + \frac{2g}{a} \sin \theta \end{align*} If \(\theta\) is small, we can approximate this by: \(\frac{27}{10} \ddot{\theta} + \frac{2g}{a} \theta = 0\) which will have period \(\displaystyle 2 \pi \sqrt{\frac{27a}{10\cdot2 g}} = 2 \pi \sqrt{\frac{3a}{5g}}\) as required.

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Solution:

\begin{align*} \text{N2}(\swarrow): &&T +mg \cos \alpha &= m \frac{V^2}{b} \\ \end{align*} So the string goes slack when \(bg\cos \alpha = V^2 \Rightarrow V = \sqrt{bg \cos \alpha}\). Once the string goes slack, the particle moves as a projectile. It's initial speed is \(V\binom{-\cos \alpha}{\sin \alpha}\) and it's position is \(\binom{b\sin \alpha}{b\cos \alpha}\): \begin{align*} && \mathbf{s} &= \binom{b\sin \alpha}{b\cos \alpha}+Vt \binom{-\cos \alpha}{\sin \alpha} + \frac12 gt^2 \binom{0}{-1} \\ &&&= \binom{b\sin \alpha - Vt \cos \alpha}{b\cos \alpha + Vt \sin \alpha - \frac12 gt^2} \\ |\mathbf{s}|^2 = b^2 \Rightarrow && b^2 &= \left ( \binom{b\sin \alpha}{b\cos \alpha}+Vt \binom{-\cos \alpha}{\sin \alpha} + \frac12 gt^2 \binom{0}{-1} \right)^2 \\ &&&= b^2 + V^2t^2+\frac14 g^2 t^4 -gb\cos \alpha t^2-V\sin \alpha gt^3 \\ \Rightarrow && 0 &= V^2t^2 + \frac14 g^2 t^4 - V^2 t^2- V \sin \alpha g t^3 \\ &&&= \frac14 g^2 t^4 - V \sin \alpha gt^3 \\ \Rightarrow && gT &= 4V \sin \alpha \end{align*} The particle will have velocity \(\displaystyle \binom{-V \cos \alpha}{V \sin \alpha - 4V \sin \alpha} = \binom{-V \cos \alpha}{-3V \sin \alpha}\) so the angle \(\beta\) will satisfy \(\tan \beta = 3 \tan \alpha\). The particle will come to an instantaneous rest if all the momentum is destroyed, ie if the particle is travelling parallel to the string. \begin{align*} && 3 \tan \alpha &= \frac{b\cos \alpha + Vt \sin \alpha - \frac12 gt^2}{b\sin \alpha - Vt \cos \alpha} \\ &&&= \frac{\frac{V^2}{g}+\frac{4V^2\sin^2\alpha}{g} - \frac{8V^2\sin^2 \alpha}{g}}{\frac{V^2\sin \alpha}{g \cos \alpha} - \frac{4V^2 \sin \alpha \cos \alpha}{g}} \\ &&&= \frac{1 -4\sin^2 \alpha}{\tan \alpha(1 - 4\cos^2 \alpha)} \\ \Leftrightarrow&& 3 \frac{\sin^2 \alpha}{1-\sin^2 \alpha} &= \frac{1- 4 \sin^2 \alpha}{-3+4\sin^2 \alpha} \\ \Leftrightarrow && -9 \sin^2 \alpha + 12 \sin^4 \alpha &= 1 - 5 \sin^2 \alpha + 4 \sin^4 \alpha \\ \Leftrightarrow && 0 &= 1+4 \sin^2 \alpha - 8\sin^4 \alpha \\ \Leftrightarrow && \sin^2 \alpha &= \frac{1 + \sqrt{3}}4 \end{align*} (taking the only positive root)

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Solution:

1. \(\,\) \begin{align*} \bf{COE}: && \frac12 m \cdot 5ag &= mg\cdot 2a + \frac12 m v^2 \\ \Rightarrow && v^2 &= ag \\ && v &= \sqrt{ag} \end{align*} If it just clears the second wall, we must have: \begin{align*} && 0 &= \sqrt{ag} \sin \theta t - \frac12 gt^2 \\ \Rightarrow && t &= \frac{2\sqrt{ag}\sin \theta}{g} \\ && a &= \sqrt{ag} \cos \theta t \\ &&&=\sqrt{ag} \cos \theta \frac{2\sqrt{ag}\sin \theta}{g} \\ &&&= a \sin 2 \theta \\ \Rightarrow && \theta &= 45^{\circ} \end{align*} Imagine firing the particle backwards from the top of the wall at \(45^\circ\) then \begin{align*} && -2a &= \sqrt{ag}\cdot \left ( -\frac1{\sqrt{2}} \right) t - \frac12 g t^2 \\ \Rightarrow && 0 &= gt^2+\sqrt{2ag} t -4a \\ &&&= (\sqrt{g}t -\sqrt{2} \sqrt{a})(\sqrt{g}t +2\sqrt{2} \sqrt{a}) \\ \Rightarrow && t &= \sqrt{\frac{2a}{g}} \\ \Rightarrow && s &= \left ( -\frac1{\sqrt{2}} \right) \sqrt{ag} \sqrt{\frac{2a}{g}} \\ &&&= -a \end{align*} Therefore the \(A\) is \(a\) from the wall.
2. When it passes over the first wall, \begin{align*} \bf{COE}: && \frac52amg &= (2a+h)mg + \frac12 m v^2 \\ \Rightarrow && v^2 &= (a-2h)g \end{align*} Now imagine firing a particle with this speed in any direction. The question is asking whether we can ever travel \(2a\) without descending more than \(h\). \begin{align*} && a &= \sqrt{(a-2h)g} \cos \beta t \\ \Rightarrow && t &= \frac{a}{\sqrt{(a-2h)g} \cos \beta}\\ && -h &= \sqrt{(a-2h)g} \sin \beta t - \frac12 g t^2 \\ &&&= a \tan \beta - \frac12 \frac{a^2}{(a-2h)} \sec^2 \beta \\ &&&= a \tan \beta - \frac{a^2}{2(a-2h)}(1+ \tan^2 \beta )\\ \Rightarrow && 0 &= \frac{a^2}{2(a-2h)} \tan^2 \beta-a \tan \beta + \frac{a^2-2ah+4h^2}{2(a-2h)} \\ && \Delta &= a^2 - \frac{a^2}{a-2h} \frac{a^2-2ah+4h^2}{a-2h} \\ &&&= \frac{a^2}{(a-2h)^2}\left ( a^2-4ah+4h^2-a^2+2ah-4h^2\right) \\ &&&= \frac{a^2}{(a-2h)^2}\left ( -2ah\right) < 0 \\ \end{align*} So there are no solutions if \(h > 0\)

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Solution: \begin{align*} \text{N2}(\downarrow): && mg -T &= m\ddot{y} \\ \text{N2}(\rightarrow): && T &= 2m\ddot{x} \\ \Rightarrow && g &= \ddot{y}+2\ddot{x} \\ \Rightarrow && \tfrac12gt^2 &= y + 2x \end{align*} At time \(T = \sqrt{6a/g}\), we have \(y + 2x = 3a\), note also that \(\dot{y}+2\dot{x} = gt\) \begin{array}{ccc} & \text{KE} & \text{GPE} & \text{EPE} \\ \text{Initial} & 0 & 0 & 0 \\ \text{Final} & \frac12m\dot{y}^2 + \frac12(2m)\dot{x}^2 & -mgy & \frac{\lambda (y-x)^2}{2a} \end{array} Also note when we head over the table, \(x = a\) and \(y = a\) \begin{align*} \text{COE}: && 0 &= \frac12m(gT-2\dot{x})^2+m\dot{x}^2-mga+\frac{\lambda(0)^2}{2a} \\ \Rightarrow && 0 &= (gT-2\dot{x})^2+2\dot{x}^2-2ga \\ &&&= (\sqrt{6ag}-2\dot{x})^2+2\dot{x}^2-2ga \\ &&&= 6\dot{x}^2-4\sqrt{6ag}+4ag \\ \Rightarrow &&&= (\sqrt{6}\dot{x} - 2\sqrt{ag})^2 \\ \Rightarrow && \dot{x} &= \sqrt{2ag/3} \end{align*} as required.

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Solution:

By energy considerations, the initial energy is \(0\). Therefore: \begin{align*} && \l 3a^2 + \ell^2\r \dot{\theta}^2 &= \l 3a + 2\ell \r g \sin \theta \\ \Rightarrow && \l 3a^2 + \ell^2\r 2\dot{\theta} \ddot{\theta} &= \l 3a + 2\ell \r g \cos\theta \dot{\theta} \tag{\(\frac{\d}{\d t}\)} \\ \Rightarrow && 2\l 3a^2 + \ell^2\r \ddot{\theta} &= \l 3a + 2\ell \r g \cos\theta \\ \Rightarrow && \ddot{\theta} &= \boxed{\frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta} \\ \end{align*} \begin{align*} \text{N}2(\perp PQ): && mg \cos \theta - R &= m \ell \ddot{\theta} \\ && R &= mg \cos \theta - m \ell \l \frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta \r \\ && &= mg\cos \theta \l 1 - \ell \frac{3a + 2\ell }{2(3a^2 + \ell^2)} \r \\ && &= mg \cos \theta \l \frac{6a^2 + 2\ell^2 - 3a\ell - 2\ell^2}{2(3a^2 + \ell^2)} \r \\ && &= mg \cos \theta \l \frac{3a(2a - \ell)}{2(3a^2 + \ell^2)} \r > 0 \tag{since \(2a > \ell\)} \end{align*} At limiting equilibrium, \(F = \mu R\). \begin{align*} \text{N}2(\parallel PQ): && \mu R - mg \sin \theta &= m \ell \dot{\theta}^2 \\ \Rightarrow && \mu mg \cos \theta \l \frac{3a(2a - \ell)}{2(3a^2 + \ell^2)} \r - mg \sin \theta &= m \ell \frac{(3a+2\ell)}{(3a^2+\ell^2)} g \sin \theta \\ \Rightarrow && \mu \l 3a(2a - \ell) \r - \l 2(3a^2 + \ell^2) \r \tan \theta &= 2\ell (3a+2\ell) \tan \theta \\ \Rightarrow && \mu \l 3a(2a - \ell) \r &= \l 6a\ell + 6a^2 + 6\ell^2 \r \tan \theta \\ \Rightarrow && \tan \theta &= \boxed{\frac{\mu a(2a-\ell)}{2(a^2 + a\ell + \ell^2)}} \end{align*} If \(\ell > 2a\), then the initial reaction force will be \(0\), ie the particle will have no contact with the rod. In other words, the rod will rotate faster than the particle will free-fall and the particle immediately loses contact with the rod.

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Solution:

1. Since \(R\) is constant, \(F=ma \Rightarrow \text{acc} = \frac{R}{m}\) and \(v^2 = u^2 + 2as\) so \(0 = u^2 - 2 \frac{R}{m}a\), ie \(a = \frac{m u^2}{2R}\)
2. By conservation of momentum, the bullet/block combination will eventually be travelling at \(v = \frac{m}{m+M}u\). The bullet will slow down to this speed in a time of \(\frac{m}{m+M}u = u - \frac{R}{m} T \Rightarrow T = \frac{Mm}{R(m+M)}u\) and will travel \(b+c = \frac{\left ( 1 - \left ( \frac{m}{m+M} \right)^2\right)u^2m}{2R}= \left ( 1 - \left ( \frac{m}{m+M} \right)^2\right)a\). The block will travell \(c = \frac12 \frac{R}{M} \frac{M^2m^2u^2}{R^2(m+M)^2} = \frac{Mm}{(m+M)^2}a\) Therefore the \(b = \left ( 1 - \left ( \frac{m}{m+M} \right)^2\right)a - \frac{Mm}{(m+M)^2}a = \frac{M}{M+m}a\)

Work done by friction is the relative gain in KE for the block. ie \(R \cdot c = \frac12 M \left ( \frac{m}{m+M}u\right)^2 \Rightarrow c = \frac{Mm}{(M+m)^2}a\).

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Solution:

1. The sine rule tells us: \begin{align*} && \frac{\frac58 a}{\sin(30^\circ + \theta)} &= \frac{a}{\sin(90^{\circ}-\theta)} \\ \Rightarrow &&\frac58 \cos \theta &= \frac12 \cos \theta+ \frac{\sqrt{3}}2 \sin \theta \\ \Rightarrow && \frac{1}{4\sqrt{3}} &= \tan \theta \\ \Rightarrow && \sin \theta &= \sqrt{\frac{1}{48+1}} = \frac17 \end{align*}
2. \(\,\) \begin{align*} && \text{initial energy} &= \frac12(5m)v^2 + \frac12 (3m)v^2 - 3m \cdot g \cdot a \cos(30^{\circ}+\theta) -5m \cdot g \cdot a\cos(30^\circ - \theta) \\ &&&= 4m v^2 - amg(4\sqrt{3} \cos \theta + \sin \theta) \\ &&&= 4mv^2 - 7amg \\ && \text{energy at top} &= \frac12 m v_{top}^2 + 7amg \end{align*} We need this equation to be positive for all values of \(v_{top} \geq 0\), so \(4mv^2 \geq 14amg \Rightarrow v_0 = \sqrt{\frac{7ag}2}\)