2004 Paper 3 Q11

Year: 2004
Paper: 3
Question Number: 11

Course: UFM Mechanics
Section: Variable Force

Difficulty: 1700.0 Banger: 1500.0

Problem

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.
Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

Banger Comparisons: 0

Show LaTeX source
Problem source
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 source
\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.