CAIE Further Paper 3 2022 November — Question 4 7 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2022
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeLimiting or terminal velocity
DifficultyChallenging +1.8 This is a Further Maths variable force mechanics problem requiring setting up F=ma with velocity-dependent forces, separating variables, and integrating to find v(x). While the integration involves a non-trivial substitution and the setup requires careful sign consideration, it follows a standard Further Mechanics template. The 6-mark allocation and straightforward part (b) confirm it's challenging but not exceptional for Further Maths students who have practiced this topic.
Spec1.08k 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 \(5\) kg moves along a horizontal straight line. At time \(t\) s, the velocity of \(P\) is \(v\) m s\(^{-1}\) and its displacement from a fixed point \(O\) on the line is \(x\) m. The forces acting on \(P\) are a force of magnitude \(\frac{500}{v}\) N in the direction \(OP\) and a resistive force of magnitude \(\frac{1}{2}v^2\) N. When \(t = 0\), \(x = 0\) and \(v = 5\).
  1. Find an expression for \(v\) in terms of \(x\). [6]
  2. State the value that the speed approaches for large values of \(x\). [1]
Question 4:
AnswerMarks
4Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
AnswerMarks
4(a)dv 500 1
5v = − v2
AnswerMarks Guidance
dx v 2B1 Sight of m or 5 is required.
10v2dv
=dx
AnswerMarks Guidance
1000−v3M1 Separate variables and attempt to integrate into a log term.
− 10 ln ( 1000−v3) =x(+A)
AnswerMarks
3A1
10
x=0, v=5, A=− ln875
AnswerMarks Guidance
3M1 Evaluate constant: correct initial condition used.
10 875
x= ln
AnswerMarks Guidance
3 1000−v3M1 Make v the subject: correct use of logs.
1
v=(1000−875e−0.3x3
AnswerMarks Guidance
 A1 1
v=5(8−7e−0.3x3 : A0 if eln terms.
 
6
AnswerMarks Guidance
4(b)Maximum value of v is 10 B1
1
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | dv 500 1
5v = − v2
dx v 2 | B1 | Sight of m or 5 is required.
10v2dv
=dx
1000−v3 | M1 | Separate variables and attempt to integrate into a log term.
− 10 ln ( 1000−v3) =x(+A)
3 | A1
10
x=0, v=5, A=− ln875
3 | M1 | Evaluate constant: correct initial condition used.
10 875
x= ln
3 1000−v3 | M1 | Make v the subject: correct use of logs.
1
v=(1000−875e−0.3x3
  | A1 | 1
v=5(8−7e−0.3x3 : A0 if eln terms.
 
6
--- 4(b) ---
4(b) | Maximum value of v is 10 | B1 | No FT: result can be found from initial equation.
1
Question | Answer | Marks | Guidance
A particle $P$ of mass $5$ kg moves along a horizontal straight line. At time $t$ s, the velocity of $P$ is $v$ m s$^{-1}$ and its displacement from a fixed point $O$ on the line is $x$ m. The forces acting on $P$ are a force of magnitude $\frac{500}{v}$ N in the direction $OP$ and a resistive force of magnitude $\frac{1}{2}v^2$ N. When $t = 0$, $x = 0$ and $v = 5$.

\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v$ in terms of $x$. [6]
\item State the value that the speed approaches for large values of $x$. [1]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q4 [7]}}
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Q1 4 Q2 6 Q3 6 Q4 7 Q5 9 Q6 9 Q7 9