CAIE Further Paper 3 2021 June — Question 5 4 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2021
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeLimiting or terminal velocity
DifficultyStandard +0.8 This is part of a multi-part variable force mechanics question requiring integration of equations of motion and finding maximum displacement. While the individual steps (integrating to find velocity, then position, and setting v=0 for maximum height) are standard Further Maths techniques, the context of variable force and the need to correctly handle initial conditions and interpret 'upwards motion' adds moderate complexity beyond routine exercises.
Spec6.06a Variable force: dv/dt or v*dv/dx methods
The displacement of \(P\) from \(O\) is \(x\) m at time \(t\) s.
  1. Find an expression for \(x\) in terms of \(t\), while \(P\) is moving upwards. [2]
  2. Find, correct to 3 significant figures, the greatest height above \(O\) reached by \(P\). [2]
Question 5:
AnswerMarks
5Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
AnswerMarks
5(a)dv
m =−mg−2mv
AnswerMarks Guidance
dtB1 Use of SUVAT implies 0 marks.
N2L, must include .
m
( ) ( )
AnswerMarks Guidance
ln 5+v =−2t +AM1 Separate variables and integrate 3-term N2L, condone omission
of constant.
( )
AnswerMarks Guidance
ln 5+v =−2t +AA1 FT FT only sign error in N2L.
t=0, v=20, A=ln25M1 Use correct initial condition.
2t =ln   5 2 + 5 v    , e2t = 5 2 + 5 vM1 Remove all logs.
v = 25e−2t −5A1
Alternative method for question 5(a)
dv
m =−mg−2mv
AnswerMarks Guidance
dtB1 N2L, must include .
m
dv
+2v=−g : Integrating factor = e2t
AnswerMarks
dtM1
( )
d ve2t
=−ge2t, ve2t =− g e2t (+A )
AnswerMarks Guidance
dt 2M1 Integrate both sides, condone omission of constant.
g
ve2t =− e2t +A
AnswerMarks Guidance
2A1 FT FT only sign error in N2L.
QuestionAnswer Marks
5(a)t =0, v=20, A=25 M1
g
ve2t =− e2t +25, v=25e −2t −5
AnswerMarks
2A1
6
AnswerMarks Guidance
5(b)x=− 25 e −2t −5t (+B )
2M1 Use of SUVAT implies 0 marks.
Integrate their expression from part (a).
25
t=0, x=0, B=
2
25( )
x= 1−e −2t −5t
AnswerMarks Guidance
2A1 FT v=Pekt +Q
FT only expressions of the form for P, Q non-zero.
2
AnswerMarks
5(c)1
Greatest height when v = 0, so t = 0.8047…or ln5
AnswerMarks Guidance
2M1 Use of SUVAT in part (a) or part (b) implies 0 marks.
Find value of t, may be embedded.
AnswerMarks Guidance
x = 5.98 mA1 CWO
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | dv
m =−mg−2mv
dt | B1 | Use of SUVAT implies 0 marks.
N2L, must include .
m
( ) ( )
ln 5+v =−2t +A | M1 | Separate variables and integrate 3-term N2L, condone omission
of constant.
( )
ln 5+v =−2t +A | A1 FT | FT only sign error in N2L.
t=0, v=20, A=ln25 | M1 | Use correct initial condition.
2t =ln   5 2 + 5 v    , e2t = 5 2 + 5 v | M1 | Remove all logs.
v = 25e−2t −5 | A1
Alternative method for question 5(a)
dv
m =−mg−2mv
dt | B1 | N2L, must include .
m
dv
+2v=−g : Integrating factor = e2t
dt | M1
( )
d ve2t
=−ge2t, ve2t =− g e2t (+A )
dt 2 | M1 | Integrate both sides, condone omission of constant.
g
ve2t =− e2t +A
2 | A1 FT | FT only sign error in N2L.
Question | Answer | Marks | Guidance
5(a) | t =0, v=20, A=25 | M1 | Use correct initial condition.
g
ve2t =− e2t +25, v=25e −2t −5
2 | A1
6
--- 5(b) ---
5(b) | x=− 25 e −2t −5t (+B )
2 | M1 | Use of SUVAT implies 0 marks.
Integrate their expression from part (a).
25
t=0, x=0, B=
2
25( )
x= 1−e −2t −5t
2 | A1 FT | v=Pekt +Q
FT only expressions of the form for P, Q non-zero.
2
--- 5(c) ---
5(c) | 1
Greatest height when v = 0, so t = 0.8047…or ln5
2 | M1 | Use of SUVAT in part (a) or part (b) implies 0 marks.
Find value of t, may be embedded.
x = 5.98 m | A1 | CWO
2
Question | Answer | Marks | Guidance
The displacement of $P$ from $O$ is $x$ m at time $t$ s.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find an expression for $x$ in terms of $t$, while $P$ is moving upwards. [2]
\item Find, correct to 3 significant figures, the greatest height above $O$ reached by $P$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q5 [4]}}
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