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2007 Paper 2 Q9
D: 1600.0 B: 1500.0
projectiles collision coefficient of restitution Newton's experimental law conservation of momentum cone normal impact optimisation

A solid right circular cone, of mass \(M\), has semi-vertical angle \(\alpha\) and smooth surfaces. It stands with its base on a smooth horizontal table. A particle of mass \(m\) is projected so that it strikes the curved surface of the cone at speed \(u\). The coefficient of restitution between the particle and the cone is \(e\). The impact has no rotational effect on the cone and the cone has no vertical velocity after the impact.

  1. The particle strikes the cone in the direction of the normal at the point of impact. Explain why the trajectory of the particle immediately after the impact is parallel to the normal to the surface of the cone. Find an expression, in terms of \(M\), \(m\), \(\alpha\), \(e\) and \(u\), for the speed at which the cone slides along the table immediately after impact.
  2. If instead the particle falls vertically onto the cone, show that the speed \(w\) at which the cone slides along the table immediately after impact is given by \[ w= \frac{mu(1+e)\sin\alpha\cos\alpha}{M+m\cos^2\alpha}\,. \] Show also that the value of \(\alpha\) for which \(w\) is greatest is given by \[ \cos \alpha = \sqrt{ \frac{M}{2M+m}}\ . \]

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1998 Paper 1 Q9
D: 1500.0 B: 1484.0
SHM simple pendulum Newton's experimental law coefficient of restitution collision conservation of momentum successive collisions equal pendulums period

Two small spheres \(A\) and \(B\) of equal mass \(m\) are suspended in contact by two light inextensible strings of equal length so that the strings are vertical and the line of centres is horizontal. The coefficient of restitution between the spheres is \(e\). The sphere \(A\) is drawn aside through a very small distance in the plane of the strings and allowed to fall back and collide with the other sphere \(B\), its speed on impact being \(u\). Explain briefly why the succeeding collisions will all occur at the lowest point. (Hint: Consider the periods of the two pendulums involved.) Show that the speed of sphere \(A\) immediately after the second impact is \(\frac{1}{2}u(1+e^{2})\) and find the speed, then, of sphere \(B\).

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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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1989 Paper 3 Q11
D: 1700.0 B: 1499.3
momentum and collisions coefficient of restitution oblique impact spheres string constraint rigid rod constraint Newton's experimental law impulse along line of centres geometry suspended sphere

A smooth uniform sphere, with centre \(A\), radius \(2a\) and mass \(3m,\) is suspended from a fixed point \(O\) by means of a light inextensible string, of length \(3a,\) attached to its surface at \(C\). A second smooth unifom sphere, with centre \(B,\) radius \(3a\) and mass \(25m,\) is held with its surface touching \(O\) and with \(OB\) horizontal. The second sphere is released from rest, falls and strikes the first sphere. The coefficient of restitution between the spheres is \(3/4.\) Find the speed \(U\) of \(A\) immediately after the impact in terms of the speed \(V\) of \(B\) immediately before impact. The same system is now set up with a light rigid rod replacing the string and rigidly attached to the sphere so that \(OCA\) is a straight line. The rod can turn freely about \(O\). The sphere with centre \(B\) is dropped as before. Show that the speed of \(A\) immediately after impact is \(125U/127.\)

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