CAIE Further Paper 3 2023 June — Question 6 9 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeForce depends on velocity v
DifficultyChallenging +1.8 This is a Further Maths mechanics question requiring two integrations with substitution. Part (a) uses the chain rule form a = v(dv/dx) to separate variables, yielding a square root integration. Part (b) requires solving for v(t) from dv/dt, then integrating again for x(t). While the techniques are standard for Further Maths (variable acceleration, separation of variables, substitution), the multi-step nature, careful algebraic manipulation with square roots, and application of two initial conditions make this moderately challenging but within expected Further Maths scope.
Spec1.08h Integration by substitution6.06a Variable force: dv/dt or v*dv/dx methods
A particle \(P\) moving in a straight line has displacement \(x\)m from a fixed point \(O\) on the line and velocity \(v\)m s\(^{-1}\) at time \(t\)s. The acceleration of \(P\), in m s\(^{-2}\), is given by \(6\sqrt{v + 9}\). When \(t = 0\), \(x = 2\) and \(v = 72\).
  1. Find an expression for \(v\) in terms of \(x\). [4]
  2. Find an expression for \(x\) in terms of \(t\). [5]
Question 6:
AnswerMarks
6Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9231/31 Cambridge International AS & A Level – Mark Scheme May/June 2023
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2023 Page 5 of 15
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(a)dv
v 6v v9 and attempt to separate variables and integrate
AnswerMarks Guidance
dxM1
2 v9 6xAA1
x2, v72 A6M1 Use initial condition to find constant.
v9x12
AnswerMarks Guidance
9A1 Correct, AEF.
4
AnswerMarks Guidance
6(b)  dx 9  x2 2x  , dx 9dt   1  1  1  dx9dt
 dt xx2  2x x2M1 Separate variables and write in the form
a b 
  dxdt
 x xc
1  x 
ln 9tB
AnswerMarks Guidance
2 x2A1 Integrate, any correct form.
1 1
t 0, x2 B ln
AnswerMarks Guidance
2 2M1 Use initial condition to find constant.
 2x  2x
18t ln  e18t 
AnswerMarks Guidance
x2 x2M1 Take logarithms
2e18t 2
x or x
AnswerMarks Guidance
2e18t 2e18t 1A1 Any correct form
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9231/31 Cambridge International AS & A Level – Mark Scheme May/June 2023
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2023 Page 5 of 15
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | dv
v 6v v9 and attempt to separate variables and integrate
dx | M1
2 v9 6xA | A1
x2, v72 A6 | M1 | Use initial condition to find constant.
v9x12
9 | A1 | Correct, AEF.
4
--- 6(b) ---
6(b) |   dx 9  x2 2x  , dx 9dt   1  1  1  dx9dt
 dt xx2  2x x2 | M1 | Separate variables and write in the form
a b 
  dxdt
 x xc
1  x 
ln 9tB
2 x2 | A1 | Integrate, any correct form.
1 1
t 0, x2 B ln
2 2 | M1 | Use initial condition to find constant.
 2x  2x
18t ln  e18t 
x2 x2 | M1 | Take logarithms
2e18t 2
x or x
2e18t 2e18t 1 | A1 | Any correct form
5
Question | Answer | Marks | Guidance
A particle $P$ moving in a straight line has displacement $x$m from a fixed point $O$ on the line and velocity $v$m s$^{-1}$ at time $t$s. The acceleration of $P$, in m s$^{-2}$, is given by $6\sqrt{v + 9}$. When $t = 0$, $x = 2$ and $v = 72$.

\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v$ in terms of $x$. [4]

\item Find an expression for $x$ in terms of $t$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q6 [9]}}
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