OCR M3 2015 June — Question 4 11 marks

Exam BoardOCR
ModuleM3 (Mechanics 3)
Year2015
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeAir resistance kv - horizontal motion
DifficultyStandard +0.8 This is a variable force mechanics problem requiring differential equations (separating variables for velocity-dependent force), followed by integration to find displacement. Part (i) is standard M3 material, but part (ii) requires finding displacement via v dv/dx = a, which demands careful technique selection and multi-step integration—above average difficulty but within expected M3 scope.
Spec3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods

A particle of mass \(0.4\) kg, moving on a smooth horizontal surface, passes through a point \(O\) with velocity \(10\text{ ms}^{-1}\). At time \(t\) s after the particle passes through \(O\), the particle has a displacement \(x\) m from \(O\), has a velocity \(v\text{ ms}^{-1}\) away from \(O\), and is acted on by a force of magnitude \(\frac{1}{5}v\) N acting towards \(O\). Find
  1. the time taken for the velocity of the particle to reduce from \(10\text{ ms}^{-1}\) to \(5\text{ ms}^{-1}\), [5]
  2. the average velocity of the particle over this time. [6]

Part (ii)
Answer: \(\frac{v}{8} = 0.4v\frac{dv}{dx}\)
\(x = -3.2 \int dv\)
\(x = -3.2 v + 32\)
ave speed = \(x/(t)\)
ave speed = 7.21
OR
\(\frac{dx}{dt} = 10e^{-\frac{t}{3.2}}\)
\(x = 10 \int e^{-\frac{t}{3.2}} dt\)
\(x = 32\left(1 - e^{-\frac{t}{3.2}}\right)\)
ave speed = \(x/(t)\)
ave speed = 7.21
AnswerMarks
M1*allow sign error
A1
*M1attempt to separate variables and integrate
A1\(x = 16\) when \(v = 5\).
*M1
A1 [6]their \(x\) evaluated; accept 5/ln2
M1*for M1, fl from (i), must contain ln term
A1
*M1attempt to separate variables and integrate
A1must show constant or use limits correctly
*M1dep all 5 previous marks
A1accept 5/ln2
Guidance: \(x = 16\) when \(t = 3.2\ln(2)\)
### Part (ii)
**Answer:** $\frac{v}{8} = 0.4v\frac{dv}{dx}$

$x = -3.2 \int dv$

$x = -3.2 v + 32$

ave speed = $x/(t)$
ave speed = 7.21

OR
$\frac{dx}{dt} = 10e^{-\frac{t}{3.2}}$

$x = 10 \int e^{-\frac{t}{3.2}} dt$

$x = 32\left(1 - e^{-\frac{t}{3.2}}\right)$

ave speed = $x/(t)$
ave speed = 7.21

| M1* | allow sign error |
| A1 | |
| *M1 | attempt to separate variables and integrate |
| A1 | $x = 16$ when $v = 5$. |
| *M1 | |
| A1 [6] | their $x$ evaluated; accept 5/ln2 |
| | |
| M1* | for M1, fl from (i), must contain ln term |
| A1 | |
| *M1 | attempt to separate variables and integrate |
| A1 | must show constant or use limits correctly |
| *M1 | dep all 5 previous marks |
| A1 | accept 5/ln2 |

**Guidance:** $x = 16$ when $t = 3.2\ln(2)$
A particle of mass $0.4$ kg, moving on a smooth horizontal surface, passes through a point $O$ with velocity $10\text{ ms}^{-1}$. At time $t$ s after the particle passes through $O$, the particle has a displacement $x$ m from $O$, has a velocity $v\text{ ms}^{-1}$ away from $O$, and is acted on by a force of magnitude $\frac{1}{5}v$ N acting towards $O$. Find

\begin{enumerate}[label=(\roman*)]
\item the time taken for the velocity of the particle to reduce from $10\text{ ms}^{-1}$ to $5\text{ ms}^{-1}$, [5]
\item the average velocity of the particle over this time. [6]
\end{enumerate}

\hfill \mbox{\textit{OCR M3 2015 Q4 [11]}}