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2017 Paper 1 Q10
Particles \(P_1\), \(P_2\), \(\ldots\) are at rest on the \(x\)-axis, and the \(x\)-coordinate of \(P_n\) is \(n\). The mass of \(P_n\) is \(\lambda^nm\). Particle \(P\), of mass \(m\), is projected from the origin at speed \(u\) towards \(P_1\). A series of collisions takes place, and the coefficient of restitution at each collision is \(e\), where \(0 < e <1\). The speed of \(P_n\) immediately after its first collision is \(u_n\) and the speed of \(P_n\) immediately after its second collision is \(v_n\). No external forces act on the particles.
- Show that \(u_1=\dfrac{1+e}{1+\lambda}\, u\) and find expressions for \(u_n\) and \(v_n\) in terms of \(e\), \(\lambda\), \(u\) and \(n\).
- Show that, if \(e > \lambda\), then each particle (except \(P\)) is involved in exactly two collisions.
- Describe what happens if \(e=\lambda\) and show that, in this case, the fraction of the initial kinetic energy lost approaches \(e\) as the number of collisions increases.
- Describe what happens if \(\lambda e=1\). What fraction of the initial kinetic energy is \mbox{eventually} lost in this case?
Solution:
- \begin{align*} \text{COM}: && mu &= mv + \lambda m u_1 \\ \Rightarrow && u &= v + \lambda u_1 \tag{1} \\ \text{NEL}: && e &= \frac{u_1-v}{u} \\ \Rightarrow && eu &= u_1 - v \tag{2} \\ (1)+(2) && (1+e)u &= (1+\lambda) u_1 \\ \Rightarrow && u_1 &= \frac{1+e}{1+\lambda}u \\ && v &= u_1 - eu \\ &&&= \frac{1+e - (1+\lambda)e}{1+\lambda} u \\ &&&= \frac{1-\lambda e}{1+\lambda}u \end{align*} Note that subsequent (first (and second)) are the same as these, therefore: \begin{align*} u_n &= \left ( \frac{1+e}{1+\lambda} \right)^n u \\ v_n &= \frac{1-\lambda e}{1+\lambda } u_n \\ &= \frac{1-\lambda e}{1+\lambda } \left ( \frac{1+e}{1+\lambda} \right)^n u \end{align*}
- If \(e > \lambda\) then \((1-\lambda e) > 1-e^2 > 0\) and \begin{align*} \frac{v_{n+1}}{v_n} &= \frac{1+e}{1+\lambda} > 1 \end{align*} So the particles are moving away from each other - hence no more collisions.
- If \(e = \lambda\) then \(u_n = u\) and \(v_n = (1-\lambda)u\) so all the particles end up moving at the same speed. \begin{align*} \text{initial k.e.} &= \frac12 m u^2 \\ \text{final k.e.} &= \frac12 m((1-e)u)^2 + \sum_{n = 1}^{\infty} \frac12 \lambda^n m ((1-e)u)^2 \\ &= \frac12mu^2(1-e)^2 \left ( \sum_{n=0}^{\infty} e^n \right) \tag{\(e = \lambda\)} \\ &= \frac12 mu^2(1-e)^2 \frac{1}{1-e} \\ &= \frac12m u^2 (1-e) \\ \text{change in k.e.} &= \frac12 m u^2 - \frac12m u^2 (1-e) \\ &= e\frac12m u^2 \end{align*} Ie the total energy lost approaches a fraction of \(e\).
- If \(\lambda e = 1\), after the second collision the particle will be stationary. ie \begin{align*} \text{initial k.e.} &= \frac12 m u^2 \\ \text{k.e. after }n\text{ collisions} &= \frac12 \lambda^n m \left (\left ( \frac{1+e}{1+\lambda} \right)^n u \right)^2\\ &= \frac12 \lambda^n m \left ( \frac{1+\frac1{\lambda}}{1+\lambda} \right)^{2n} u&2\\ &= \frac12 \lambda^n m \left ( \frac{1+\frac1{\lambda}}{1+\lambda} \right)^{2n} u\\ &= \frac12 \lambda^n m \left ( \frac{1}{\lambda} \right)^{2n} u\\ &= \frac12 m \lambda^{-n} u\\ &\to 0 \end{align*} Eventually we lose all the kinetic energy.
2005 Paper 2 Q11
A plane is inclined at an angle \(\arctan \frac34\) to the horizontal and a small, smooth, light pulley~\(P\) is fixed to the top of the plane. A string, \(APB\), passes over the pulley. A particle of mass~\(m_1\) is attached to the string at \(A\) and rests on the inclined plane with \(AP\) parallel to a line of greatest slope in the plane. A particle of mass \(m_2\), where \(m_2>m_1\), is attached to the string at \(B\) and hangs freely with \(BP\) vertical. The coefficient of friction between the particle at \(A\) and the plane is \( \frac{1}{2}\). The system is released from rest with the string taut. Show that the acceleration of the particles is \(\ds \frac{m_2-m_1}{m_2+m_1}g\). At a time \(T\) after release, the string breaks. Given that the particle at \(A\) does not reach the pulley at any point in its motion, find an expression in terms of \(T\) for the time after release at which the particle at \(A\) reaches its maximum height. It is found that, regardless of when the string broke, this time is equal to the time taken by the particle at \(A\) to descend from its point of maximum height to the point at which it was released. Find the ratio \(m_1 : m_2\). \noindent [Note that \(\arctan \frac34\) is another notation for \(\tan^{-1} \frac34\,\).]