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2018 Paper 3 Q9
D: 1700.0 B: 1484.0
momentum collisions coefficient of restitution recurrence relation elastic wall two-body collision second order recurrence inequality

A particle \(P\) of mass \(m\) is projected with speed \(u_0\) along a smooth horizontal floor directly towards a wall. It collides with a particle \(Q\) of mass \(km\) which is moving directly away from the wall with speed \(v_0\). In the subsequent motion, \(Q\) collides alternately with the wall and with \(P\). The coefficient of restitution between \(Q\) and \(P\) is \(e\), and the coefficient of restitution between \(Q\) and the wall is 1. Let \(u_n\) and \(v_n\) be the velocities of \(P\) and \(Q\), respectively, towards the wall after the \(n\)th collision between \(P\) and \(Q\).

  1. Show that, for \(n\ge2\), \[ (1+k)u_{n} - (1-k)(1+e)u_{n-1} + e(1+k)u_{n-2} =0\,. \tag{\(*\)} \]
  2. You are now given that \(e=\frac12\) and \(k = \frac1{34}\), and that the solution of \((*)\) is of the form \[ \phantom{(n\ge0)} u_n= A\left( \tfrac 7{10}\right)^n + B\left( \tfrac 5{7 }\right)^n \ \ \ \ \ \ (n\ge0) \,, \] where \(A\) and \(B\) are independent of \(n\). Find expressions for \(A\) and \(B\) in terms of \(u_0\) and \(v_0\). Show that, if \(0 < 6u_0 < v_0\), then \(u_n\) will be negative for large \(n\).

Solution:

  1. Just before collision \(n-1\): Velocity of \(P\) is \(u_{n-2}\) Velocity of \(Q\) is \(-v_{n-2}\) \begin{align*} COM: && mu_{n-2}+km(-v_{n-2}) &= mu_{n-1}+kmv_{n-1} \\ \Rightarrow && u_{n-2}-kv_{n-2} &= u_{n-1}+kv_{n-1} \\ NEL: && v_{n-1}-u_{n-1} &= -e((-v_{n-2})-u_{n-2}) \\ \Rightarrow && v_{n-1}-u_{n-1} &= e(v_{n-2}+u_{n-2}) \end{align*} \begin{align*} &&kv_{n-1} &= u_{n-2} - kv_{n-2}-u_{n-1} \\ &&kv_{n-1}&= ku_{n-1}+kev_{n-2}+keu_{n-2} \\ \Rightarrow && kv_{n-2}(1+e) &= u_{n-2}(1-ke)-u_{n-1}(1+k) \\ \Rightarrow && kv_{n-1}(1+e) &= u_{n-1}(1-ke)-u_{n}(1+k) \\ && k(1+e)v_{n-1}-k(1+e)u_{n-1} &= k(1+e)e(v_{n-2}+u_{n-2}) \\ \Rightarrow && u_{n-1}(1-ke)-u_{n}(1+k)-k(1+e)u_{n-1} &= e(u_{n-2}(1-ke)-u_{n-1}(1+k))+k(1+e)eu_{n-2} \\ \Rightarrow && 0 &= (1+k)u_n + ((ke-1)+k(1+e)-e(1+k))u_{n-1} \\ &&& \quad \quad + (e(1-ke)+k(1+e)e)u_{n-2} \\ \Rightarrow && 0 &= (1+k)u_n- (1-k)(1+e)u_{n-1} +e(1+k)u_{n-2} \end{align*}
  2. \(u_0 = A + B\) \begin{align*} &&& \begin{cases}u_0 - kv_0 &= kv_1 + u_1 \\ \frac12 (u_0+v_0) &= v_1 - u_1 \\ \end{cases} \\ \Rightarrow && (1+k)u_1 &= u_0 - kv_0 - \frac{k}{2}(u_0 + v_0) \\ \Rightarrow && u_1 &= \frac{1}{k+1} \l u_0 (1-\frac{k}{2}) - \frac32 k v_0 \r \\ &&&= \frac{67}{70} u_0 - \frac{3}{70} v_0 \end{align*} Therefore \(A+B = u_0, \frac{49A+50B}{70} = \frac{67}{70} u_0 - \frac{3}{70} v_0\) \begin{align*} && A+B &= u_0 \\ && 49A+50B &= 67u_0 - 3v_0 \\ \Rightarrow && 50u_0 - A &= 67u_0 - 3v_0 \\ \Rightarrow && A &= -17u_0 + 3v_0 \\ && B &= 18u_0 - 3v_0 \end{align*} If \(0 < 6u_0 < v_0\), then \(B < 0\) and as \(n \to \infty\) we will find that \(\l \frac57 \r^n\) dominates \(\l \frac7{10} \r^n\) and so our velocity will be negative and the particle will change direction

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2012 Paper 2 Q8
D: 1600.0 B: 1485.7
sequences recurrence relation inequality strictly increasing sequence algebraic manipulation second order recurrence quadratic recurrence

The positive numbers \(\alpha\), \(\beta\) and \(q\) satisfy \(\beta-\alpha >q\). Show that \[ \frac{\alpha^2+\beta^2 -q^2}{\alpha\beta}-2> 0\,. \] The sequence \(u_0\), \(u_1\), \(\ldots\) is defined by \(u_0=\alpha\), \(u_1=\beta\) and \[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u_{n+1} = \frac {u_{n}^2 -q^2}{u_{n-1}} \ \ \ \ \ \ \ \ \ \ \ (n\ge1), \] where \(\alpha\), \(\beta\) and \(q\) are given positive numbers (and \(\alpha\) and \(\beta\) are such that no term in the sequence is zero). Prove that \(u_n(u_n+u_{n+2}) = u_{n+1}(u_{n-1}+u_{n+1})\,\). Prove also that \[ u_{n+1} -pu_n + u_{n-1}=0 \] for some number \(p\) which you should express in terms of \(\alpha\), \(\beta\) and \(q\). Hence, or otherwise, show that if \(\beta> \alpha+q\), the sequence is strictly increasing (that is, \(u_{n+1}-u_n > 0\) for all \(n\)). Comment on the case \(\beta =\alpha +q\).

Solution: \begin{align*} && \beta - \alpha &> q \\ \Rightarrow &&(\beta - \alpha)^2 &> q^2 \\ \Rightarrow && \beta^2 +\alpha^2 - 2\beta \alpha &> q^2 \\ \Rightarrow && \alpha^2+\beta^2-q^2 -2 \beta \alpha &> 0 \\ \Rightarrow && \frac{\alpha^2+\beta^2-q^2}{\alpha\beta} - 2 &> 0 \end{align*} \begin{align*} && u_n(u_n+u_{n+2}) &= u_n \cdot \left (u_n + \frac {u_{n+1}^2 -q^2}{u_{n}}\right) \\ &&&= u_n^2 + u_{n+1}^2-q^2 \\ &&&= u_n^2 + u_{n+1}^2 - (u_n^2-u_{n-1}u_{n+1}) \\ &&&= u_{n+1}^2 + u_{n+1}u_{n-1} \\ &&&= u_{n+1}(u_{n-1}+u_{n+1}) \\ \\ && u_{n+1}-pu_n+u_{n-1} &= -pu_n+\frac{u_{n}(u_{n-2}+u_n)}{u_{n-1}} \\ &&&= \frac{u_n(u_{n}-pu_{n-1}+u_{n-2})}{u_{n-1}} \end{align*} Therefore if \(u_2 -pu_1 + u_0 = 0\) it is always zero, ie if \begin{align*} && u_2 &= p\beta - \alpha \\ && u_2 &= \frac{\beta^2-q^2}{\alpha} \\ \Rightarrow && \frac{\beta^2-q^2}{\alpha} &= p\beta - \alpha \\ \Rightarrow && p &= \frac{\alpha^2+\beta^2-q^2}{\alpha\beta} \end{align*} If \(\beta > \alpha + q\) we must have that \(p > 2\), and so \(u_{n+1}-u_n = (p-1)u_n - u_{n-1} > u_n-u_{n-1} > 0\), therefore the sequence is strictly increasing. If \(\beta = \alpha + q\) the sequence follows \(u_{n+1} - 2u_n + u_{n-1} =0\) and so \(u_{n+1}-u_n = u_n - u_{n-1}\) for all \(n\) (which is still increasing - it's an arithmetic progression with common difference \(\beta - \alpha\)).

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