Question 6 - A-Level Maths

CAIE Further Paper 3 2021 November — Question 6 9 marks

Exam Board	CAIE
Module	Further Paper 3 (Further Paper 3)
Year	2021
Session	November
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Variable Force
Type	Variable force (position x) - find velocity
Difficulty	Challenging +1.8 This is a Further Maths mechanics problem requiring integration of a variable force using F=ma and v dv/dx = a, followed by separation of variables. While the algebra is moderately complex (involving x^(-3) terms and careful manipulation), the solution path is standard for FM students who recognize the technique. The 9 marks and multi-step nature elevate it above average, but it doesn't require novel insight beyond applying known methods systematically.
Spec	1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods

A particle \(P\) of mass \(2\) kg moves along a horizontal straight line. The point \(O\) is a fixed point on this line. At time \(t\) s the velocity of \(P\) is \(v\) m s\(^{-1}\) and the displacement of \(P\) from \(O\) is \(x\) m. A force of magnitude \(\left(8x - \frac{128}{x^3}\right)\) N acts on \(P\) in the direction \(OP\). When \(t = 0\), \(x = 8\) and \(v = -15\).

Show that \(v = -\frac{2}{3}(x^2 - 4)\). [5]
Find an expression for \(x\) in terms of \(t\). [4]

Show mark scheme Show mark scheme source

Question 6:

Answer	Marks
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

9231/32 Cambridge International AS & A Level – Mark Scheme October/November 2021

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 2021 Page 5 of 12

Answer	Marks	Guidance
Question	Answer	Marks

Answer	Marks
6(a)	dv 128

2v =8x−

Answer	Marks	Guidance
dx x3	M1	Separate variables and integrate, + c not required for M1.
v2 =4x2 +64x −2 +c	A1	OE.
x = 8, v = ‒15 and c = ‒32	M1	Use initial condition.

4 ( ) 64

v2 = x4 −8x2 +16 or 4x2 −32+

Answer	Marks	Guidance
x2 x2	A1	Correct expression for v2, AEF.

4 ( )2 2( )

v2 = x2 −4 giving v=− x2 −4

Answer	Marks	Guidance
x2 x	A1	Convincingly shown, e.g. v is negative initially, AG.

5

Answer	Marks	Guidance
6(b)	1 ln(x2 −4)=−2t (+A )
2	M1	dx

Use v= and integrate.

dt

1 t=0, x=8, A= ln60

Answer	Marks	Guidance
2	DM1	Use initial condition.

1 x2 −4 x2 −4

ln  =−2t giving =e −4t

Answer	Marks	Guidance
2  60  60	M1	Remove log.
x= 4+60e −4t	A1	CAO

4

Answer	Marks	Guidance
Question	Answer	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
9231/32 Cambridge International AS & A Level – Mark Scheme October/November 2021
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 2021 Page 5 of 12
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | dv 128
2v =8x−
dx x3 | M1 | Separate variables and integrate, + c not required for M1.
v2 =4x2 +64x −2 +c | A1 | OE.
x = 8, v = ‒15 and c = ‒32 | M1 | Use initial condition.
4 ( ) 64
v2 = x4 −8x2 +16 or 4x2 −32+
x2 x2 | A1 | Correct expression for v2, AEF.
4 ( )2 2( )
v2 = x2 −4 giving v=− x2 −4
x2 x | A1 | Convincingly shown, e.g. v is negative initially, AG.
5
--- 6(b) ---
6(b) | 1 ln(x2 −4)=−2t (+A )
2 | M1 | dx
Use v= and integrate.
dt
1
t=0, x=8, A= ln60
2 | DM1 | Use initial condition.
1 x2 −4 x2 −4
ln  =−2t giving =e −4t
2  60  60 | M1 | Remove log.
x= 4+60e −4t | A1 | CAO
4
Question | Answer | Marks | Guidance

Show LaTeX source

A particle $P$ of mass $2$ kg moves along a horizontal straight line. The point $O$ is a fixed point on this line. At time $t$ s the velocity of $P$ is $v$ m s$^{-1}$ and the displacement of $P$ from $O$ is $x$ m.

A force of magnitude $\left(8x - \frac{128}{x^3}\right)$ N acts on $P$ in the direction $OP$. When $t = 0$, $x = 8$ and $v = -15$.

\begin{enumerate}[label=(\alph*)]
\item Show that $v = -\frac{2}{3}(x^2 - 4)$. [5]

\item Find an expression for $x$ in terms of $t$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q6 [9]}}

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