CAIE Further Paper 3 2020 June — Question 2 6 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2020
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeAcceleration as function of velocity (separation of variables)
DifficultyStandard +0.8 This is a standard Further Maths mechanics question requiring formation and solution of a first-order differential equation (F=ma with resistance). While it involves multiple steps (setting up equation, separating variables, integrating, applying initial conditions), it follows a well-established method taught in Further Maths courses. The 6 marks reflect the working required rather than exceptional difficulty. It's moderately harder than average A-level due to being Further Maths content, but remains a textbook-style exercise.
Spec6.06a Variable force: dv/dt or v*dv/dx methods

A particle \(Q\) of mass \(m\) kg falls from rest under gravity. The motion of \(Q\) is resisted by a force of magnitude \(mkv\) N, where \(v\) ms\(^{-1}\) is the speed of \(Q\) at time \(t\) s and \(k\) is a positive constant. Find an expression for \(v\) in terms of \(g\), \(k\) and \(t\). [6]

Question 2:
AnswerMarks
2dv
=dt
AnswerMarks
g−kvM1
− 1 ln ( g−kv )=t+ A
AnswerMarks
kM1
1
t =0, v=0: A=− lng
AnswerMarks
kA1
1 g−kv
t =− ln
AnswerMarks
k gM1
g( )
v= 1−e −kt
AnswerMarks
kM1A1
6
AnswerMarks Guidance
QuestionAnswer Marks
Question 2:
2 | dv
=dt
g−kv | M1
− 1 ln ( g−kv )=t+ A
k | M1
1
t =0, v=0: A=− lng
k | A1
1 g−kv
t =− ln
k g | M1
g( )
v= 1−e −kt
k | M1A1
6
Question | Answer | Marks
A particle $Q$ of mass $m$ kg falls from rest under gravity. The motion of $Q$ is resisted by a force of magnitude $mkv$ N, where $v$ ms$^{-1}$ is the speed of $Q$ at time $t$ s and $k$ is a positive constant.

Find an expression for $v$ in terms of $g$, $k$ and $t$. [6]

\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q2 [6]}}