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2006 Paper 3 Q11
D: 1700.0 B: 1516.0
momentum collisions coefficient of restitution energy loss lift and counterweight pulley impulsive force connected particles conservation of momentum free fall

A lift of mass \(M\) and its counterweight of mass \(M\) are connected by a light inextensible cable which passes over a fixed frictionless pulley. The lift is constrained to move vertically between smooth guides. The distance between the floor and the ceiling of the lift is \(h\). Initially, the lift is at rest, and the distance between the top of the lift and the pulley is greater than \(h\). A small tile of mass \(m\) becomes detached from the ceiling of the lift and falls to the floor of the lift. Show that the speed of the tile just before the impact is \[ \sqrt{\frac{(2M-m)gh \;}{M}}\;. \] The coefficient of restitution between the tile and the floor of the lift is \(e\). Given that the magnitude of the impulsive force on the lift due to tension in the cable is equal to the magnitude of the impulsive force on the counterweight due to tension in the cable, show that the loss of energy of the system due to the impact is \(mgh(1-e^2)\). Comment on this result.

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1989 Paper 2 Q11
D: 1600.0 B: 1484.0
work-energy-power lift and counterweight pulley perfectly inelastic collision energy loss Newton's second law relative acceleration connected particles conservation of momentum

A lift of mass \(M\) and its counterweight of mass \(M\) are connected by a light inextensible cable which passes over a light frictionless pulley. The lift is constrained to move vertically between smooth guides. The distance between the floor and the ceiling of the lift is \(h\). Initially, the lift is at rest, and the distance between the top of the lift and the pulley is greater than \(h\). A small tile of mass \(m\) becomes detached from the ceiling of the lift. Show that the time taken for it to fall to the floor is \[ t=\sqrt{\frac{\left(2M-m\right)h}{Mg}}. \] The collision between the tile and the lift floor is perfectly inelastic. Show that the lift is reduced to rest by the collision, and that the loss of energy of the system is \(mgh\). Note: the question on the STEP database is \[ t=\sqrt{\frac{2\left(M-m\right)h}{Mg}}. \]

Solution:

TikZ diagram
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*}
TikZ diagram
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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