Year: 2004
Paper: 3
Question Number: 11
Course: UFM Mechanics
Section: Variable Force
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
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 $t \,$.
\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.