Edexcel M3 2008 January — Question 2 8 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2008
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeInverse power force - non-gravitational context
DifficultyStandard +0.8 This M3 variable force question requires setting up and solving a differential equation using F=ma with v(dv/dx), integrating with an inverse square force law, applying initial conditions, and finding when v=0. It involves multiple sophisticated steps beyond standard mechanics, but follows a recognizable M3 pattern for this topic.
Spec6.02c Work by variable force: using integration6.06a Variable force: dv/dt or v*dv/dx methods
2. A particle \(P\) of mass 0.1 kg moves in a straight line on a smooth horizontal table. When \(P\) is a distance \(x\) metres from a fixed point \(O\) on the line, it experiences a force of magnitude \(\frac { 16 } { 5 x ^ { 2 } } \mathrm {~N}\) away from \(O\) in the direction \(O P\). Initially \(P\) is at a point 2 m from \(O\) and is moving towards \(O\) with speed \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find the distance of \(P\) from \(O\) when \(P\) first comes to rest.
Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(m\ 'a'\ = \pm\frac{16}{5x^2}\), with acceleration in any form e.g. \(\frac{d^2x}{dt^2}\), \(v\frac{dv}{dx}\), \(\frac{dv}{dt}\)B1 part (a)
Uses \(a = v\frac{dv}{dx}\) to obtain \(kv\frac{dv}{dx} = \pm k\cdot\frac{32}{x^2}\)M1
Separates variables: \(k\int v\,dv = k'\int\frac{32}{x^2}\,dx\)dM1
Obtains \(\frac{1}{2}v^2 = \mp\frac{32}{x}\ (+C)\) or equivalent e.g. \(\frac{0.1}{2}v^2 = -\frac{16}{5x}\ (+C)\)A1
Substituting \(x=2\), \(v=\pm 8 \Rightarrow 32 = -16 + C \Rightarrow C = 48\)M1 A1
\(v=0 \Rightarrow \frac{32}{x} = 48 \Rightarrow x = \frac{2}{3}\text{ m}\)M1 A1 cao N.B. \(-\frac{2}{3}\) is not acceptable for final answer
Total: 8 marks
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $m\ 'a'\ = \pm\frac{16}{5x^2}$, with acceleration in any form e.g. $\frac{d^2x}{dt^2}$, $v\frac{dv}{dx}$, $\frac{dv}{dt}$ | B1 | part (a) |
| Uses $a = v\frac{dv}{dx}$ to obtain $kv\frac{dv}{dx} = \pm k\cdot\frac{32}{x^2}$ | M1 | |
| Separates variables: $k\int v\,dv = k'\int\frac{32}{x^2}\,dx$ | dM1 | |
| Obtains $\frac{1}{2}v^2 = \mp\frac{32}{x}\ (+C)$ or equivalent e.g. $\frac{0.1}{2}v^2 = -\frac{16}{5x}\ (+C)$ | A1 | |
| Substituting $x=2$, $v=\pm 8 \Rightarrow 32 = -16 + C \Rightarrow C = 48$ | M1 A1 | |
| $v=0 \Rightarrow \frac{32}{x} = 48 \Rightarrow x = \frac{2}{3}\text{ m}$ | M1 A1 cao | N.B. $-\frac{2}{3}$ is not acceptable for final answer |

**Total: 8 marks**

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2. A particle $P$ of mass 0.1 kg moves in a straight line on a smooth horizontal table. When $P$ is a distance $x$ metres from a fixed point $O$ on the line, it experiences a force of magnitude $\frac { 16 } { 5 x ^ { 2 } } \mathrm {~N}$ away from $O$ in the direction $O P$. Initially $P$ is at a point 2 m from $O$ and is moving towards $O$ with speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

Find the distance of $P$ from $O$ when $P$ first comes to rest.\\

\hfill \mbox{\textit{Edexcel M3 2008 Q2 [8]}}
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