Work, energy and Power 2

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.

In the initial position, since the system is in equilibrium the tension in the two ropes must be \(500g\). Therefore since \(T = \frac{\lambda x}{l} \Rightarrow x = \frac{10 \cdot 500 g}{490\, 000} = \frac1{10}\) so the initial extension is \(\frac1{10}\) By conservation of momentum, if the initial speed of the plank + Nellie is \(V\), we must have \(4000 \cdot 5 = 5000 V \Rightarrow V = 4\) \begin{array}{ccc} & \text{GPE} & \text{EPE} & \text{KE} \\ \hline \text{Initially} & 5000gh & 2 \cdot \frac12 \frac{\lambda}{l} \frac{1}{100} & \frac12 \cdot 5000 \cdot 4^2 \\ & 49\,000h & 490 & 40\,000 \\ \text{Finally} & 0 & 2 \cdot \frac12 \frac{\lambda}{l} (h + \frac1{10})^2 & 0 \\ & 0 & 49\,000 (h+\frac1{10})^2 & 0 \end{array} By conservation of energy, we can set up a quadratic: \begin{align*} && 49\,000 (h+\frac1{10})^2 &= 49\,000h + 40\,490 \\ \Rightarrow && 49\,000(h + \frac1{10})^2 &= 49\,000(h + \frac1{10})+35\, 590 \\ \Rightarrow&& h + \frac1{10} &= 1.488092\cdots \\ \Rightarrow && h &= 1.49 \,\, (3\text{ s.f.}) \end{align*} When she gets off the plank, it will move according to: \begin{align*} \text{N2}(\uparrow): && \frac{\lambda x}{l} -1000g &= -1000 \ddot{x} \\ && 49 x-g &= -\ddot{x} \\ \Rightarrow && x &= A \sin 7t + B \cos 7t + 0.2 \\ && x(0) = 1.49, &x'(0) = 0 \\ \Rightarrow && B = -1.69, & A=0 \end{align*} If we continued under this motion the string would definitely reach a point \(0.1\) above \(0\), and therefore the ropes would go slack.

If \(\alpha\) is as labelled then \(\cos \alpha = \frac{1}{x}, \sin \alpha = \frac{\sqrt{x^2-1}}{x}, \tan \alpha = \sqrt{x^2-1}\). We also have the full length of the rubber band is \(2\pi - 2\alpha +2\tan \alpha\) so the extension is \(2 \l \sqrt{x^2-1} - \cos^{-1} \l \frac{1}{x}\r \r\) Therefore \(T = \frac{\l \sqrt{x^2-1} - \cos^{-1} \l \frac{1}{x}\r \r\lambda}{\pi}\). If \(x = 2\), \(T = \frac{\sqrt{3} - \frac{\pi}{3}}{\pi} \lambda, \sin \alpha = \frac{\sqrt{3}}{2}\) \begin{align*} \text{N2}(\uparrow): && 2T\sin \alpha - mg &= 0 \\ \Rightarrow && \frac{\sqrt{3} - \frac{\pi}{3}}{\pi} \lambda \sqrt{3} &= mg \\ \Rightarrow && \lambda &= \frac{\sqrt{3}\pi}{(3\sqrt{3}-\pi)}mg \end{align*}