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1995 Paper 2 Q10
D: 1600.0 B: 1484.0
momentum collisions coefficient of restitution three particles sequential collisions inequality conservation of momentum change of reference frame

Three small spheres of masses \(m_{1},m_{2}\) and \(m_{3},\) move in a straight line on a smooth horizontal table. (Their order on the straight line is the order given.) The coefficient of restitution between any two spheres is \(e\). The first moves with velocity \(u\) towards the second whilst the second and third are at rest. After the first collision the second sphere hits the third after which the velocity of the second sphere is \(u.\) Find \(m_{1}\) in terms of \(m_{2},m_{3}\) and \(e\). deduce that \[ m_{2}e>m_{3}(1+e+e^{2}). \] Suppose that the relation between \(m_{1},m_{2}\) and \(m_{3}\) is that in the formula you found above, but that now the first sphere initially moves with velocity \(u\) and the other two spheres with velocity \(v\), all in the same direction along the line. If \(u>v>0\) use the first part to find the velocity of the second sphere after two collisions have taken place. (You should not need to make any substantial computations but you should state your argument clearly.)

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1992 Paper 3 Q12
D: 1700.0 B: 1500.0
circular motion conservation of momentum energy conservation hemispherical bowl particle on hemisphere normal reaction moving frame constraint equation Lagrangian-style mechanics

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TikZ diagram
A smooth hemispherical bowl of mass \(2m\) is rigidly mounted on a light carriage which slides freely on a horizontal table as shown in the diagram. The rim of the bowl is horizontal and has centre \(O\). A particle \(P\) of mass \(m\) is free to slide on the inner surface of the bowl. Initially, \(P\) is in contact with the rim of the bowl and the system is at rest. The system is released and when \(OP\) makes an angle \(\theta\) with the horizontal the velocity of the bowl is \(v\)? Show that \[3v=a\dot{\theta}\sin\theta \] and that \[ v^{2}=\frac{2ga\sin^{3}\theta}{3(3-\sin^{2}\theta)}, \] where \(a\) is the interior radius of the bowl. Find, in terms of \(m,g\) and \(\theta,\) the reaction between the bowl and the particle.

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1991 Paper 1 Q13
D: 1516.0 B: 1484.0
circular motion conservation of momentum semicircular groove lamina impulse smooth constraint particle in groove average force

\(\ \)\vspace{-1.5cm} \noindent

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A heavy smooth lamina of mass \(M\) is free to slide without rotation along a straight line on a fixed smooth horizontal table. A smooth groove \(ABC\) is inscribed in the lamina, as indicated in the above diagram. The tangents to the groove at \(A\) and at \(B\) are parallel to the line. When the lamina is stationary, a particle of mass \(m\) (where \(m < M\)) enters the groove at \(A\). The particle is travelling, with speed \(V\), parallel to the line and in the plane of the lamina and table. Calculate the speeds of the particle and of the lamina, when the particle leaves the groove at \(C\). Suppose now that the lamina is held fixed by a peg attached to the line. Supposing that the groove \(ABC\) is a semicircle of radius \(r\), obtain the value of the average force per unit time exerted on the peg by the lamina between the instant that the particle enters the groove and the instant that it leaves it.

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1990 Paper 1 Q11
D: 1500.0 B: 1484.0
projectiles conservation of momentum collision time of flight trajectory angle two-dimensional motion particle mechanics

A shell of mass \(m\) is fired at elevation \(\pi/3\) and speed \(v\). Superman, of mass \(2m\), catches the shell at the top of its flight, by gliding up behind it in the same horizontal direction with speed \(3v\). As soon as Superman catches the shell, he instantaneously clasps it in his cloak, and immediately pushes it vertically downwards, without further changing its horizontal component of velocity, but giving it a downward vertical component of velocity of magnitude \(3v/2\). Calculate the total time of flight of the shell in terms of \(v\) and \(g\). Calculate also, to the nearest degree, the angle Superman's flight trajectory initially makes with the horizontal after releasing the shell, as he soars upwards like a bird. {[}Superman and the shell may be regarded as particles.{]}

Solution: The particle has initial velocity \(\displaystyle \binom{v \cos \frac{\pi}{3}}{v \sin \frac{\pi}{3}}\) and acceleration \(\displaystyle \binom{0}{-g}\). It will have zero vertical speed (ie be at the top of its trajectory) when \(t = \frac{\sqrt{3}v}{2g}\). Since \(0 = v^2-u^2 + 2as\) the height achieved will be \(\frac{3v^2}{8g}\) At this point it will need to travel the same distance again, but this time the initial speed is \(\frac{3v}{2}\) so: \begin{align*} && \frac{3v^2}{8g} &= \frac{3v}{2} t + \frac12 g t^2 \\ \Rightarrow && 0 &= 4g^2t^2+12vgt - 3v^2 \\ \Rightarrow && t &= \l \frac{-3+2\sqrt{3}}{2} \r \frac{v}{g} \end{align*} Therefore the total time is: \begin{align*} \l \frac{\sqrt{3}}{2} - \frac32 + \sqrt{3} \r \frac{v}{g} &= \frac{3\sqrt{3}-3}{2}\frac{v}{g} \end{align*} \begin{align*} COM(\uparrow): && 0 &= 2m v_y - m \frac{3}{2}v \\ \Rightarrow &&v_y &= \frac34 v \\ COM(\rightarrow): && 3mV &= 2m (3v) +m \frac{v}{2} \\ \Rightarrow && V &= \frac{13}6\\ \end{align*} Therefore superman is now travelling at a vector of \(\displaystyle \binom{\frac{13}6}{\frac34}v\) ie an angle of \(\tan^{-1} \frac 9{26}\) to the horizontal, approximately \(19^\circ\)

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1990 Paper 2 Q14
D: 1600.0 B: 1484.0
momentum collisions elastic collision Newton's experimental law chain of collisions conservation of energy conservation of momentum inequality

The identical uniform smooth spherical marbles \(A_{1},A_{2},\ldots,A_{n},\) where \(n\geqslant3,\) each of mass \(m,\) lie in that order in a smooth straight trough, with each marble touching the next. The marble \(A_{n+1},\) which is similar to \(A_{n}\) but has mass \(\lambda m,\) is placed in the trough so that it touches \(A_{n}.\) Another marble \(A_{0},\) identical to \(A_{n},\) slides along the trough with speed \(u\) and hits \(A_{1}.\) It is given that kinetic energy is conserved throughout.

  1. Show that if \(\lambda<1,\) there is a possible subsequent motion in which only \(A_{n}\) and \(A_{n+1}\) move (and \(A_{0}\) is reduced to rest), but that if \(\lambda>1,\) such a motion is not possible.
  2. If \(\lambda>1,\) show that a subsequent motion in which only \(A_{n-1},A_{n}\) and \(A_{n+1}\) move is not possible.
  3. If \(\lambda>1,\) find a possible subsequent motion in which only two marbles move.

Solution: Without loss of generality, let \(m = u = 1\).

  1. \begin{align*} \text{COM}: && 1&= v_n + \lambda v_{n+1} \\ && &= v_n + \lambda v_{n+1}\\ \text{COE}: && \frac12 &= \frac12 v_n^2 + \frac12 \lambda v_{n+1}^2 \\ && 1 &= v_n^2 +\lambda v_{n+1}^2 \\ \\ \Rightarrow && v_n^2 + 2\lambda v_n v_{n+1} + \lambda^2 v_{n+1}^2 &= v_n^2 + \lambda v_{n+1}^2 \\ && \lambda v_{n+1}^2 &= v_{n+1}^2 - 2 v_n v_{n+1} \\ && \lambda v_{n+1} &= (v_{n+1} - 2v_n) \\ && (1-\lambda)v_{n+1} &= 2v_n \end{align*} Since \(v_{n+1} > v_n > 0\) this is only possible if \(\lambda < 1\)
  2. \begin{align*} \text{COM}: && 1&= v_{n-1}+v_n+\lambda v_{n+1} \\ && 1&= v_{n-1} + v_n + \lambda v_{n+1} \\ \text{COE}: && \frac12 &= \frac12 v_{n-1}^2+\frac12v_n^2+\frac12\lambda v_{n+1}^2 \\ && 1&= v_{n-1}^2 + v_n^2 + \lambda v_{n+1}^2 \\ \\ \Rightarrow && 1 &= v_{n-1}^2 + v_n^2 + \lambda \frac{(1-v_{n-1}-v_n)^2}{\lambda^2} \\ &&&= v_{n-1}^2 + v_n^2 + \frac{(1-v_{n-1}-v_n)^2}{\lambda} \\ \Rightarrow && 1 &< v_{n-1}^2 + v_n^2 + (1-v_{n-1}-v_n)^2 \\ &&&= 2v_{n-1}^2+2v_n^2 + 1-2v_{n-1}-2v_{n-2} +2v_{n-1}v_n\\ \Rightarrow && v_{n-1}+v_n & <(v_{n-1}+v_n)^2 - v_{n-1}v_n \end{align*} but this cannot be true since \(0 < v_{n-1}+v_n < 1\) and \(v_n v_{n-1} > 0\)
  3. The only way this is possible is if the first and last marble are moving. \begin{align*} \text{COM}: && 1 &= v_0 +\lambda v_{n+1} \\ \text{COE}: && \frac12 &= \frac12 v_0^2 + \frac12 \lambda^2 v_{n+1} \\ && 1 &= v_0^2 + \lambda v_{n+1}^2 \\ \Rightarrow && 2v_0 + \lambda v_{n+1} &= v_{n+1} \\ \Rightarrow && v_{n+1} &=\frac{2}{1-\lambda} v_0 \\ \Rightarrow && v_0 &= \frac{1-\lambda}{1+\lambda} \\ && v_{n+1} &= \frac{2}{1+\lambda} \end{align*} which will work since \(v_0\) can travel backwards.

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1990 Paper 3 Q12
D: 1700.0 B: 1500.0
momentum collisions wedge elastic collision conservation of momentum energy conservation smooth surface projectile particle on wedge coefficient of restitution

A uniform smooth wedge of mass \(m\) has congruent triangular end faces \(A_{1}B_{1}C_{1}\) and \(A_{2}B_{2}C_{2},\) and \(A_{1}A_{2},B_{1}B_{2}\) and \(C_{1}C_{2}\) are perpendicular to these faces. The points \(A,B\) and \(C\) are the midpoints of \(A_{1}A_{2},B_{1}B_{2}\) and \(C_{1}C_{2}\) respectively. The sides of the triangle \(ABC\) have lengths \(AB=AC=5a\) and \(BC=6a.\) The wedge is placed with \(BC\) on a smooth horizontal table, a particle of mass \(2m\) is placed at \(A\) on \(AC,\) and the system is released from rest. The particle slides down \(AC,\) strikes the table, bounces perfectly elastically and lands again on the table at \(D\). At this time the point \(C\) of the wedge has reached the point \(E\). Show that \(DE=\frac{192}{19}a.\)

Solution: Conservation of energy, tells us that \(2m \cdot g \cdot 4a = 8amg\) is equal to \(\frac12 m v_{wedge}^2 + \frac12(2m)v_{particle}^2\). Conservation of momentum (horizontally) tells us that \(m v_{wedge}+2mv_{particle, \rightarrow} = 0 \Rightarrow v_{particle, \rightarrow} = -\frac12 v_{wedge}\).

TikZ diagram
We know that the particle must remain on the slope, and so \(v_{particle,\downarrow} = \frac{4}{3} \frac{3}{2} v_{wedge} = 2v_{wedge}\). In conclusion, we have: \begin{align*} && 8amg &= \frac12 m v_{wedge}^2 + \frac12 (2m)\left ((-\tfrac12 v_{wedge})^2 + (2v_{wedge})^2 \right ) \\ &&&= \frac{19}{4}mv_{wedge}^2 \\ \Rightarrow && v_{wedge}^2 &= \frac{32}{19}ag \end{align*}. To calculate the time the ball bounces for, note that: \(s = ut + \frac12 at^2 \Rightarrow 0 = 2v_{wedge} - \frac12 gt \Rightarrow t = \frac{4v_{wedge}}{g}\). During this time, the wedge (and ball) who horizontally are moving apart with speed \(\frac32 v_{wedge}\) we have they move apart by: \begin{align*} && DE &= \underbrace{\frac32 v_{wedge}}_{\text{speed they move apart}} \cdot \underbrace{\frac{4v_{wedge}}{g}}_{\text{time they are moving apart for}} \\ &&&= \frac{6}{g} v_{wedge}^2 \\ &&&= \frac{6}{g}\frac{32}{19}ag \\ &&&= \frac{192}{19}a \end{align*}

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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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