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2018 Paper 1 Q10
D: 1500.0 B: 1471.6
Newton's laws connected particles train tension inequality compression system of equations acceleration

A train is made up of two engines, each of mass \(M\), and \(n\) carriages, each of mass \(m\). One of the engines is at the front of the train, and the other is coupled between the \(k\)th and \((k+1)\)th carriages. When the train is accelerating along a straight, horizontal track, the resistance to the motion of each carriage is \(R\) and the driving force on each engine is \(D\), where \(2D >nR\,\). The tension in the coupling between the engine at the front and the first carriage is \(T\).

  1. Show that \[ T = \frac{n(mD+MR)}{nm+2M}\,. \]
  2. Show that \(T\) is greater than the tension in any other coupling provided that \(k> \frac12n\,\).
  3. Show also that, if \(k> \frac12n\,\), then at least one of the couplings is in compression (that is, there is a negative tension in the coupling).

Solution:

  1. \(\,\)
    TikZ diagram
    \begin{align*} \text{N2}(\leftarrow, \text{first engine}): && D-T &= Ma \\ \text{N2}(\leftarrow, \text{rest of train}): && T-nR+D &= (M+nm)a \\ \Rightarrow && \frac{D-T}{M} &= \frac{T+D-nR}{M+nm} \\ \Rightarrow && T \left ( \frac{1}{M+nm}+\frac{1}{M} \right) &= \frac{D}{M} + \frac{nR-D}{M+nm} \\ \Rightarrow && T \left ( 2M+nm\right) &= DM +Dnm + nRM - DM \\ &&&= n(mD+MR) \\ \Rightarrow && T &= \frac{n(mD+MR)}{2M+nm} \end{align*}
  2. The greatest coupling must occur behind an engine, because each carriage behind an engine acts as a drag. Therefore we need only consider the couple between the second engine and the rest of the carriages:
    TikZ diagram
    \begin{align*} \text{N2}(\leftarrow, \text{up to second engine}): && 2D - T_2 - kR &= (2M+km)a \\ \text{N2}(\leftarrow, \text{everything else}): && T_2 - (n-k)R &= (n-k)ma \\ \Rightarrow && \frac{2D-T_2-kR}{2M+km} &= \frac{T_2-(n-k)R}{(n-k)m} \\ \Rightarrow && T_2 \left (\frac{1}{(n-k)m} + \frac{1}{2M+km} \right) &= \frac{2D-kR}{2M+km} + \frac{R}{m} \\ \Rightarrow && T_2 \left (2M+ nm \right) &= (2D-kR)m(n-k) + R(2M+km)(n-k) \\ \Rightarrow && T_2 &= \frac{(n-k)\left (2Dm+2RM \right)}{2M+nm} \\ &&&= \frac{2(n-k)(mD + MR)}{2M+nm} \end{align*} Therefore \(T > T_2\) provided \(2(n-k) < n \Leftrightarrow k > \frac12n\)
  3. If there is a coupling which is in negative tension, it must be between the two engines. In particular, if there is one, there must be one directly in front of the first engine.
    TikZ diagram
    \begin{align*} \text{N2}(\leftarrow, \text{before second engine}): && D - T_3 - kR &= (M+km)a \\ \text{N2}(\leftarrow, \text{everything else}): && T_3 +D- (n-k)R &= (M+(n-k)m)a \\ \Rightarrow && \frac{D-T_3-kR}{M+km} &= \frac{T_3+D-(n-k)R}{M+(n-k)m} \\ \Rightarrow && T_3 \left ( \frac{1}{M+(n-k)m} + \frac{1}{M+km} \right) &= \frac{D-(n-k)R}{M+(n-k)m}+\frac{kR-D}{M+km} \\ \Rightarrow && T_3 (2M+nm) &= (D-(n-k)R)(M+km)+(kR-D)(M+(n-k)m) \\ &&&= D(M+km-M-(n-k)m) + R(kM+k(n-k)m-(n-k)M-k(n-k)m) \\ &&&= D(n-2k)m+RM(2k-n)m \\ &&&= (n-2k)m(D-RM) \end{align*} Therefore \(T_3\) is negative if \(k > \frac12n\) so there are some connections in compression.

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2009 Paper 2 Q11
D: 1600.0 B: 1516.0
work-energy-power Newton's second law power resistance train logarithmic approximation work-energy theorem kinematics integration optimisation

A train consists of an engine and \(n\) trucks. It is travelling along a straight horizontal section of track. The mass of the engine and of each truck is \(M\). The resistance to motion of the engine and of each truck is \(R\), which is constant. The maximum power at which the engine can work is \(P\). Obtain an expression for the acceleration of the train when its speed is \(v\) and the engine is working at maximum power. The train starts from rest with the engine working at maximum power. Obtain an expression for the time \(T\) taken to reach a given speed \(V\), and show that this speed is only achievable if \[ P>(n+1)RV\,. \]

  1. In the case when \((n+1) RV/P\) is small, use the approximation \(\ln (1-x) \approx -x -\frac12 x^2\) (valid for small \( x \)) to obtain the approximation \[ PT\approx \tfrac 12 (n+1) MV^2\, \] and interpret this result.
  2. In the general case, the distance moved from rest in time \(T\) is \(X\). {\em Write down}, with explanation, an equation relating \(P\), \(T\), \(X\), \(M\), \(V\), \(R\) and \(n\) and hence show that \[ X= \frac{2PT - (n+1)MV^2}{2(n+1)R} \,. \]

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2004 Paper 2 Q11
D: 1600.0 B: 1500.0
variable force power resistance Newton's second law differential equation separable ODE exponential train terminal velocity

The maximum power that can be developed by the engine of train \(A\), of mass \(m\), when travelling at speed \(v\) is \(Pv^{3/2}\,\), where \(P\) is a constant. The maximum power that can be developed by the engine of train \(B\), of mass \(2m\), when travelling at speed \(v\) is \(2Pv^{3/2}.\) For both \(A\) and \(B\) resistance to motion is equal to \(kv\), where \(k\) is a constant. For \(t\le0\), the engines are crawling along at very low equal speeds. At \(t = 0\,\), both drivers switch on full power and at time \(t\) the speeds of \(A\) and \(B\) are \(v_{\vphantom{\dot A}\!A}\) and \(v_{\vphantom{\dot B}\hspace{-1pt}B},\) respectively.

  1. Show that \[ v_{\vphantom{\dot A}\!A} = \frac{P^2 \left(1-\e^{-kt/2m}\right)^2}{k^2} \] and write down the corresponding result for \(v_{\vphantom{\dot B}B}\).
  2. Find \(v_{\vphantom{\dot B}A}\) and \(v_{\vphantom{\dot B}B}\) when \(9 v_{\vphantom{\dot B}A} =4v_{\vphantom{\dot B}B}\;\). [Not on original paper] Show that \(1 < v_{\vphantom{\dot B}\hspace{-1pt}B} /v_{\vphantom{\dot A}\!A} < 4\) for \(t > 0\,\).
  3. Both engines are switched off when \(9 v_{\vphantom{\dot B}A} =4v_{\vphantom{\dot B}B}\,\). Show that thereafter \(k^2 v_{\vphantom{\dot B}B}^2 = 4 P^2 v_{\vphantom{\dot B}A}\,\).

Solution:

  1. \(\,\) \begin{align*} && P &= Fv \\ \text{N2}(\rightarrow): && Pv^{1/2} - kv &= ma \\ \Rightarrow && \dot{v} &= \frac{P}{m} \sqrt{v}-\frac{k}{m}v \\ \Rightarrow && \int \d t & = \int \frac{m}{\sqrt{v}\left (P-k\sqrt{v} \right)} \d v \\ &&t &= -\frac{2m}k\ln\left (P - k\sqrt{v_A}\right) + C \\ t = 0, v_A \approx 0: && 0 & =-\frac{2m}{k} \ln P+ C \\ \Rightarrow && C &= \frac{2m}{k} \ln P \\ \Rightarrow && e^{-kt/2m} &= \frac{P- k \sqrt{v_A}}{P} \\ \Rightarrow && v_A &= \frac{P^2(1-e^{-kt/2m})^2}{k^2} \end{align*} The equation of motion for \(B\) is \(\dot{v_B} = \frac{P}{m}\sqrt{v} - \frac{k}{2m} v\), ie \(k \to \frac{k}{2}\), so \[ v_B = \frac{4P^2(1-e^{-kt/4m})^2}{k^2} \]
  2. Suppose \(9v_A = 4v_B\), then and let \(e^{-kt/4m} = X\) \begin{align*} && 9 \frac{P^2(1-e^{-kt/2m})^2}{k^2} &= 4 \frac{4P^2(1-e^{-kt/4m})^2}{k^2} \\ \Rightarrow && \frac{3}{4} &= \frac{1-X}{1-X^2} \\ \Rightarrow && 0 &= 3X^2-4X+1 \\ &&&= (3X-1)(X-1) \\ \Rightarrow && X &= 1, \frac13 \\ X = 1: && t &= 0 \\ X = \frac13: && e^{-kt/4m} &= \frac13\\ \Rightarrow && t &= \frac{4m}{k}\ln 3 \\ && v_A &= \frac{P^2(1-\frac19)^2}{k^2} \\ &&&= \frac{64P^2}{81k^2} \\ && v_B &= \frac{P^2(1-\frac13)^2}{k^2} \\ &&&= \frac{4P^2}{9k^2} \end{align*} Notice also that \begin{align*} && \frac{v_B}{v_A} &= 4\left ( \frac{1-X}{1-X^2} \right)^2 \\ &&&= 4 \frac{1}{(1+X)^2} \end{align*} Since \(X \in (0,1)\) \(\frac{v_B}{v_A} \in (1, 4)\)
  3. Once the engines are switched off, the equation of motion for \(A\) is (where \(t\) is measured from that point) \begin{align*} && \dot{v} &= -\frac{k}{m}v \\ \Rightarrow && v &= Ae^{-kt/m} \\ \Rightarrow && v_A &= \frac{64P^2}{81k^2}e^{-kt/m} \end{align*} Similarly, \(v_B = \frac{4P^2}{9k^2}e^{-kt/2m}\) so \begin{align*} && \frac{v_A}{v_B^2} &= \frac{64P^2}{81k^2} \cdot \frac{81k^4}{16P^4} = \frac{4k^2}{P^2} \end{align*} as required.

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