OCR MEI M4 2013 June — Question 3 24 marks

Exam BoardOCR MEI
ModuleM4 (Mechanics 4)
Year2013
SessionJune
Marks24
PaperDownload PDF ↗
TopicAdvanced work-energy problems
TypeVariable resistance or force differential equation
DifficultyChallenging +1.2 This is a multi-part mechanics question requiring power-force relationships (P=Fv), differential equations with partial fractions, and integration. While it involves several steps and techniques (showing a given result, solving separable DEs, finding limiting values), each individual step follows standard M4 procedures without requiring novel insight. The 'show that' parts provide scaffolding, and the techniques are all textbook applications for this module level.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods
3 A model car of mass 2 kg moves from rest along a horizontal straight path. After time \(t \mathrm {~s}\), the velocity of the car is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The power, \(P \mathrm {~W}\), developed by the engine is initially modelled by \(P = 2 v ^ { 3 } + 4 v\). The car is subject to a resistance force of magnitude \(6 v \mathrm {~N}\).
  1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = ( 1 - v ) ( 2 - v )\) and hence show that \(t = \ln \frac { 2 - v } { 2 ( 1 - v ) }\).
  2. Hence express \(v\) in terms of \(t\). Once the power reaches 4.224 W it remains at this constant value with the resistance force still acting.
  3. Verify that the power of 4.224 W is reached when \(v = 0.8\) and calculate the value of \(t\) at this instant.
  4. Find \(v\) in terms of \(t\) for the motion at constant power. Deduce the limiting value of \(v\) as \(t \rightarrow \infty\).
3 A model car of mass 2 kg moves from rest along a horizontal straight path. After time $t \mathrm {~s}$, the velocity of the car is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The power, $P \mathrm {~W}$, developed by the engine is initially modelled by $P = 2 v ^ { 3 } + 4 v$. The car is subject to a resistance force of magnitude $6 v \mathrm {~N}$.\\
(i) Show that $\frac { \mathrm { d } v } { \mathrm {~d} t } = ( 1 - v ) ( 2 - v )$ and hence show that $t = \ln \frac { 2 - v } { 2 ( 1 - v ) }$.\\
(ii) Hence express $v$ in terms of $t$.

Once the power reaches 4.224 W it remains at this constant value with the resistance force still acting.\\
(iii) Verify that the power of 4.224 W is reached when $v = 0.8$ and calculate the value of $t$ at this instant.\\
(iv) Find $v$ in terms of $t$ for the motion at constant power. Deduce the limiting value of $v$ as $t \rightarrow \infty$.

\hfill \mbox{\textit{OCR MEI M4 2013 Q3 [24]}}
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