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2009 Paper 3 Q11
D: 1700.0 B: 1500.1
variable force momentum variable mass Newton's second law differential equation conservation of momentum dust cloud comet separable ODE limiting behaviour

A comet in deep space picks up mass as it travels through a large stationary dust cloud. It is subject to a gravitational force of magnitude \(M\!f\) acting in the direction of its motion. When it entered the cloud, the comet had mass \(M\) and speed \(V\). After a time \(t\), it has travelled a distance \(x\) through the cloud, its mass is \(M(1+bx)\), where~\(b\) is a positive constant, and its speed is \(v\).

  1. In the case when \(f=0\), write down an equation relating \(V\), \(x\), \(v\) and \(b\). Hence find an expression for \(x\) in terms of \(b\), \(V\) and \(t\).
  2. In the case when \(f\) is a non-zero constant, use Newton's second law in the form \[ \text{force} = \text{rate of change of momentum} \] to show that \[ v = \frac{ft+V}{1+bx}\,. \] Hence find an expression for \(x\) in terms of \(b\), \(V\), \(f\) and \(t\). Show that it is possible, if \(b\), \(V\) and \(f\) are suitably chosen, for the comet to move with constant speed. Show also that, if the comet does not move with constant speed, its speed tends to a constant as \(t\to\infty\).

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1993 Paper 1 Q2
D: 1500.0 B: 1516.0
geometric series sum of geometric series convergence infinite series pattern recognition partial fractions limiting behaviour

If \(\left|r\right|\neq1,\) show that \[ 1+r^{2}+r^{4}+\cdots+r^{2n}=\frac{1-r^{2n+2}}{1-r^{2}}\,. \] If \(r\neq1,\) find an expression for \(\mathrm{S}_{n}(r),\) where \[ \mathrm{S}_{n}(r)=r+r^{2}+r^{4}+r^{5}+r^{7}+r^{8}+r^{10}+\cdots+r^{3n-1}. \] Show that, if \(\left|r\right|<1,\) then, as \(n\rightarrow\infty,\) \[ \mathrm{S}_{n}(r)\rightarrow\frac{1}{1-r}-\frac{1}{1-r^{3}}\,. \] If \(\left|r\right|\neq1,\) find an expression for \(\mathrm{T}_{n}(r),\) where \[ \mathrm{T}_{n}(r)=1+r^{2}+r^{3}+r^{4}+r^{6}+r^{8}+r^{9}+r^{10}+r^{12}+r^{14}+r^{15}+r^{16}+\cdots+r^{6n}. \] If \(\left|r\right|<1,\) find the limit of \(\mathrm{T}_{n}(r)\) as \(n\rightarrow\infty.\) What happens to \(\mathrm{T}_{n}(r)\) as \(n\rightarrow\infty\) in the three cases \(r>1,r=1\) and \(r=-1\)? In each case give reasons for your answer.

Solution: \begin{align*} && S &= 1 + r^2 + r^4 + \cdots + r^{2n} \\ && r^2S &= \quad \,\,\,\, r^2 + r^4 + \cdots+r^{2n}+r^{2n+2} \\ \Rightarrow && (1-r^2)S &= 1 - r^{2n+2} \\ \Rightarrow && S &= \frac{1-r^{2n+2}}{1-r^2} \end{align*} \begin{align*} && S_n(r) &= r + r^2 + r^4 + r^5 + r^7 + \cdots + r^{3n-1} \\ &&&= 1 + r + r^2 + \cdots + r^{3n} - (1 + r^3 + r^6 + r^{3n}) \\ &&&= \frac{1-r^{3n+1}}{1-r} - \frac{1-r^{3n+3}}{1-r^3} \\ \\ \Rightarrow && \lim_{n \to \infty} S_n(r) &= \frac{1-0}{1-r} - \frac{1-0}{1-r^3} = \frac{1}{1-r} - \frac{1}{1-r^3} \end{align*} \begin{align*} && T_n(r) &= 1 + r^2 + r^3 + r^4 + r^6 + \cdots + r^{6n} \\ &&&= \frac{1-r^{6n+6}}{1-r^6} + \frac{r^2-r^{6n+2}}{1-r^6} + \frac{r^3-r^{6n+3}}{1-r^6} + \frac{r^4-r^{6n+4}}{1-r^6} \\ &&&= \frac{1+r^2+r^3+r^4-r^{6n}(r^2+r^3+r^4+r^6))}{1-r^6} \\ \\ &&\lim_{n \to \infty} T_n(r) &= \frac{1+r^2+r^3+r^4}{1-r^6} \end{align*} If \(r > 1\) clear it diverges. if \(r = 1\) same story. if \(r = -1\) the sums in blocks of \(4\) are all \(1+1-1+1 = 2 > 0\) and so it also diverges.

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