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2023 Paper 3 Q9
D: 1500.0 B: 1500.0
newton's laws and connected particles string constraint conservation of momentum circular motion two-particle system differential equations parametric equations

Two particles, \(A\) of mass \(m\) and \(B\) of mass \(M\), are fixed to the ends of a light inextensible string \(AB\) of length \(r\) and lie on a smooth horizontal plane. The origin of coordinates and the \(x\)- and \(y\)-axes are in the plane. Initially, \(A\) is at \((0,\,0)\) and \(B\) is at \((r,\,0)\). \(B\) is at rest and \(A\) is given an instantaneous velocity of magnitude \(u\) in the positive \(y\) direction. At a time \(t\) after this, \(A\) has position \((x,\,y)\) and \(B\) has position \((X,\,Y)\). You may assume that, in the subsequent motion, the string remains taut.

  1. Explain by means of a diagram why \[X = x + r\cos\theta\] \[Y = y - r\sin\theta\] where \(\theta\) is the angle clockwise from the positive \(x\)-axis of the vector \(\overrightarrow{AB}\).
  2. Find expressions for \(\dot{X}\), \(\dot{Y}\), \(\ddot{X}\) and \(\ddot{Y}\) in terms of \(\ddot{x}\), \(\ddot{y}\), \(\dot{x}\), \(\dot{y}\), \(r\), \(\ddot{\theta}\), \(\dot{\theta}\) and \(\theta\), as appropriate. Assume that the tension \(T\) in the string is the only force acting on either particle.
  3. Show that \[\ddot{x}\sin\theta + \ddot{y}\cos\theta = 0\] \[\ddot{X}\sin\theta + \ddot{Y}\cos\theta = 0\] and hence that \(\theta = \dfrac{ut}{r}\).
  4. Show that \[m\ddot{x} + M\ddot{X} = 0\] \[m\ddot{y} + M\ddot{Y} = 0\] and find \(my + MY\) in terms of \(t\) and \(m, M, u, r\) as appropriate.
  5. Show that \[y = \frac{1}{m+M}\left(mut + Mr\sin\!\left(\frac{ut}{r}\right)\right).\]
  6. Show that, if \(M > m\), then the \(y\) component of the velocity of particle \(A\) will be negative at some time in the subsequent motion.

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2004 Paper 3 Q11
D: 1700.0 B: 1500.0
variable force inverse cube law repulsive force two-particle system second order ODE chain rule substitution Newton's second law centre of mass separation of variables integration

Particles \(P\), of mass \(2\), and \(Q\), of mass \(1\), move along a line. Their distances from a fixed point are \(x_1\) and \(x_2\), respectively where \(x_2>x_1\,\). Each particle is subject to a repulsive force from the other of magnitude \(\displaystyle {2 \over z^3}\), where \(z = x_2-x_1 \,\). Initially, \(x_1=0\), \(x_2 = 1\), \(Q\) is at rest and \(P\) moves towards \(Q\) with speed 1. Show that \(z\) obeys the equation \(\displaystyle {\mathrm{d}^2 z \over \mathrm{d}t^2} = {3 \over z^3}\). By first writing \(\displaystyle {\mathrm{d}^2 z \over \mathrm{d}t^2} = v {\mathrm{d}v \over \mathrm{d}z} \,\), where \(\displaystyle v={\mathrm{d}z \over \mathrm{d}t}\,\), show that \(z=\sqrt{4t^2-2t+1}\,\). By considering the equation satisfied by \(2x_1+x_2\,\), find \(x_1\) and \(x_2\) in terms of \(t \,\).

Solution: \begin{align*} \text{N2}: && 2\ddot{x}_1 &= -\frac{2}{(x_2-x_1)^3}\\ \text{N2}: && \ddot{x}_2 &= \frac{2}{(x_2-x_1)^3}\\ \Rightarrow && \ddot{x}_2 - \ddot{x}_1 &= \frac{3}{(x_1-x_2)^3} \\ \Rightarrow && \frac{\d^2 z}{\d t^2} &= \frac{3}{z^3} \\ \Rightarrow && v \frac{\d v}{\d z} &= \frac{3}{z^3} \\ \Rightarrow && \int v \d v &= \int \frac{3}{z^3} \d z \\ \Rightarrow && \frac{v^2}{2} &= -\frac{3}{2}z^{-2} + C \\ \Rightarrow && v^2 &= -3 z^{-2} + C' \\ t=0,z=1,v=-1: && 1 &= -3+C \Rightarrow C = 4 \\ \Rightarrow && \frac{\d z}{\d t} &= -\sqrt{4-3z^{-2}} \\ \Rightarrow && \int \d t &= -\int \frac{1}{\sqrt{4-3z^{-2}}} \d z \\ \Rightarrow && t &= \int \frac{z}{\sqrt{4z^2-3}} \d z \\ \Rightarrow && t &= -\frac14\sqrt{4z^2-3} + C \\ t=0, z = 1: && 0 &= -\frac14+C \\ \Rightarrow && C &= \frac14\\ \Rightarrow && 4t -1 &= -\sqrt{4z^2-3} \\ \Rightarrow && 16t^2+1-8t &= 4z^2-3 \\ \Rightarrow && z &= \sqrt{4t^2-2t+1} \end{align*} \begin{align*} && 2\ddot{x}_1 + \ddot{x}_2 &= 0 \\ \Rightarrow && 2x_1+x_2 &= At + B \\ t = 0, v = -1: && 2x_1+x_2 &= -t+1 \\ \\ \Rightarrow && x_2-x_1 &= \sqrt{4t^2-2t+1}\\ && 2x_1+x_2 &= 1-t \\ \Rightarrow && x_1 &= \frac13 \left (1-t-\sqrt{4t^2-2t+1} \right) \\ && x_2 &= \frac13(1-t + \sqrt{4t^2-2t+1}) \end{align*} This method of considering the relative position and considering the motion of the centre of mass is extremely common for solving systems of particles problems.

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