OCR Further Mechanics 2019 June — Question 5 14 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2019
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeApplied/modelling contexts
DifficultyStandard +0.3 This is a standard Further Maths mechanics question combining Newton's second law with a first-order linear differential equation. Part (a) requires setting up the equation from given forces, part (b) involves solving using integrating factor (a routine technique at this level), and part (c) requires integration of velocity. While it's multi-step and requires careful algebra, all techniques are standard and the question provides significant scaffolding through 'show that' parts.
Spec4.10c Integrating factor: first order equations6.06a Variable force: dv/dt or v*dv/dx methods

5 A particle \(P\) of mass 4.5 kg is free to move along the \(x\)-axis. In a model of the motion it is assumed that \(P\) is acted on by two forces:
  • a constant force of magnitude \(f \mathrm {~N}\) in the positive \(x\) direction;
  • a resistance to motion, \(R \mathrm {~N}\), whose magnitude is proportional to the speed of \(P\).
At time \(t\) seconds the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\). When \(t = 0 , P\) is at the origin \(O\) and is moving in the positive direction with speed \(u \mathrm {~ms} ^ { - 1 }\), and when \(v = 5 , R = 2\). \begin{enumerate}[label=(\alph*)] \item Show that, according to the model, \(\frac { d v } { d t } = \frac { 10 f - 4 v } { 45 }\). \item
  1. By solving the differential equation in part (a), show that \(\mathrm { v } = \frac { 1 } { 2 } \left( 5 \mathrm { f } - ( 5 \mathrm { f } - 2 \mathrm { u } ) \mathrm { e } ^ { - \frac { 4 } { 45 } \mathrm { t } } \right)\).
  2. Describe briefly how, according to the model, the speed of \(P\) varies over time in each of the following cases.

Question 5:
AnswerMarks Guidance
5(a) R k v 25kk 0.4
dv dv 10f 4v
F ma f 0.4v4.5  
AnswerMarks
dt dt 45M1
A1
AnswerMarks Guidance
[2]3.3
3.3AG Must be clear equation has
come from NII
Alternate Method
Must include e𝑘𝑡
4 4
𝐼𝐹 = e∫ 45 d𝑡 = e45 𝑡
B1
d 4 𝑡 2 4 𝑡
(𝑣e45 )= 𝑓e45 oe
d𝑡 9
M1
4 5 4
𝑡 𝑡
𝑣e45 = 𝑓e45 +𝑐
2
5
𝑐 = 𝑢− 𝑓
AnswerMarks
2M1
M1
5 5 −4
𝑡
𝑣 = 𝑓+(𝑢− 𝑓e45 )→
2 2
1 −4
𝑣 = (5𝑓−(5𝑓−2𝑢)e45 𝑡 )
AnswerMarks
2A1
[5]
Must include e𝑘𝑡
AnswerMarks Guidance
5(b) (ii)
as t
u2.5f v2.5f, constant
u2.5f  v decreases (from u) and approaches 2.5f
AnswerMarks
as tB1
B1
B1
AnswerMarks
[3]3.4
3.4
AnswerMarks
3.4Allow the idea that v = 2.5f for large
t, and allow technically inaccurate
statements (eg “v speeds up”)
provided that intent is clear
AnswerMarks
See aboveSC: If B0B1B0 or B0B0B0
awarded. If mentions v
approaches 2.5f for cases 1
and 3 award B1 or if
mentions v increases in case
1 and v decreases in case 3
award B1
AnswerMarks Guidance
5(c) dx 1 4t 
 5e 45 f oe
dt 2
x2.5ft 45 fe  4 4 5 t c
8
45
c f and use of t = 9 to find x
8
45 3e0.8
x f
AnswerMarks
8B1
*M1
Dep*
M1
A1
AnswerMarks
[4]1.1
3.4
1.1
AnswerMarks
1.1Could be in a definite integral
Substituting t = 0, x = 0 to obtain c,
 45 4t 9
or x 2.5ft fe 45 with
 
8
0
correct use of limits
Question 5:
5 | (a) | R k v 25kk 0.4
dv dv 10f 4v
F ma f 0.4v4.5  
dt dt 45 | M1
A1
[2] | 3.3
3.3 | AG | Must be clear equation has
come from NII
Alternate Method
Must include e𝑘𝑡
4 4
𝐼𝐹 = e∫ 45 d𝑡 = e45 𝑡
B1
d 4 𝑡 2 4 𝑡
(𝑣e45 )= 𝑓e45 oe
d𝑡 9
M1
4 5 4
𝑡 𝑡
𝑣e45 = 𝑓e45 +𝑐
2
5
𝑐 = 𝑢− 𝑓
2 | M1
M1
5 5 −4
𝑡
𝑣 = 𝑓+(𝑢− 𝑓e45 )→
2 2
1 −4
𝑣 = (5𝑓−(5𝑓−2𝑢)e45 𝑡 )
2 | A1
[5]
Must include e𝑘𝑡
5 | (b) | (ii) | u2.5f  v increases (from u) and approaches 2.5f
as t
u2.5f v2.5f, constant
u2.5f  v decreases (from u) and approaches 2.5f
as t | B1
B1
B1
[3] | 3.4
3.4
3.4 | Allow the idea that v = 2.5f for large
t, and allow technically inaccurate
statements (eg “v speeds up”)
provided that intent is clear
See above | SC: If B0B1B0 or B0B0B0
awarded. If mentions v
approaches 2.5f for cases 1
and 3 award B1 or if
mentions v increases in case
1 and v decreases in case 3
award B1
5 | (c) | dx 1 4t 
 5e 45 f oe
dt 2
x2.5ft 45 fe  4 4 5 t c
8
45
c f and use of t = 9 to find x
8
45 3e0.8
x f
8 | B1
*M1
Dep*
M1
A1
[4] | 1.1
3.4
1.1
1.1 | Could be in a definite integral
Substituting t = 0, x = 0 to obtain c,
 45 4t 9
or x 2.5ft fe 45 with
 
8
0
correct use of limits
5 A particle $P$ of mass 4.5 kg is free to move along the $x$-axis. In a model of the motion it is assumed that $P$ is acted on by two forces:

\begin{itemize}
  \item a constant force of magnitude $f \mathrm {~N}$ in the positive $x$ direction;
  \item a resistance to motion, $R \mathrm {~N}$, whose magnitude is proportional to the speed of $P$.
\end{itemize}

At time $t$ seconds the velocity of $P$ is $v \mathrm {~ms} ^ { - 1 }$. When $t = 0 , P$ is at the origin $O$ and is moving in the positive direction with speed $u \mathrm {~ms} ^ { - 1 }$, and when $v = 5 , R = 2$.
\begin{enumerate}[label=(\alph*)]
\item Show that, according to the model, $\frac { d v } { d t } = \frac { 10 f - 4 v } { 45 }$.
\item \begin{enumerate}[label=(\roman*)]
\item By solving the differential equation in part (a), show that $\mathrm { v } = \frac { 1 } { 2 } \left( 5 \mathrm { f } - ( 5 \mathrm { f } - 2 \mathrm { u } ) \mathrm { e } ^ { - \frac { 4 } { 45 } \mathrm { t } } \right)$.
\item Describe briefly how, according to the model, the speed of $P$ varies over time in each of the following cases.

\begin{itemize}
\end{enumerate}\item $\mathrm { u } < 2.5 \mathrm { f }$
  \item $\mathrm { u } = 2.5 \mathrm { f }$
  \item $u > 2.5 f$
\item In the case where $\mathrm { u } = 2 \mathrm { f }$, find in terms of $f$ the exact displacement of $P$ from $O$ when $t = 9$.
\end{itemize}
\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2019 Q5 [14]}}