CAIE Further Paper 3 2021 June — Question 5 6 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2021
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeAir resistance kv - vertical motion
DifficultyStandard +0.8 This is a standard Further Maths mechanics problem requiring formation and solution of a differential equation with variable resistance. While it involves multiple steps (applying Newton's second law, separating variables, integrating, applying initial conditions), the technique is well-practiced in FM syllabi and follows a predictable method. The 6 marks reflect computational work rather than conceptual difficulty, placing it moderately above average but not requiring novel insight.
Spec6.06a Variable force: dv/dt or v*dv/dx methods
A particle \(P\) of mass \(m\) kg is projected vertically upwards from a point \(O\), with speed \(20\) m s\(^{-1}\), and moves under gravity. There is a resistive force of magnitude \(2mv\) N, where \(v\) m s\(^{-1}\) is the speed of \(P\) at time \(t\) s after projection.
  1. Find an expression for \(v\) in terms of \(t\), while \(P\) is moving upwards. [6]
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
A particle $P$ of mass $m$ kg is projected vertically upwards from a point $O$, with speed $20$ m s$^{-1}$, and moves under gravity. There is a resistive force of magnitude $2mv$ N, where $v$ m s$^{-1}$ is the speed of $P$ at time $t$ s after projection.

\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v$ in terms of $t$, while $P$ is moving upwards. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q5 [6]}}
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Q1 3 Q2 5 Q3 3 Q3 3 Q4 8 Q5 6 Q5 4 Q6 3 Q6 6 Q7 4 Q7 5