CAIE Further Paper 3 2024 June — Question 6 9 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeMotion with exponential force
DifficultyChallenging +1.8 This is a Further Maths mechanics problem requiring integration of a variable force equation with exponential terms. Part (a) involves solving a separable differential equation (F=ma with force depending on both v and t), requiring careful algebraic manipulation and substitution. Part (b) requires a second integration. While the techniques are standard for FM students (separable ODEs, integration by parts/substitution), the exponential-velocity coupling and multi-step algebraic manipulation elevate this above routine exercises, though it remains a structured question with clear methodology.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)3.02f Non-uniform acceleration: using differentiation and integration4.10c Integrating factor: first order equations
A particle \(P\) of mass \(2\) kg moving on a horizontal straight line has displacement \(x\) m from a fixed point \(O\) on the line and velocity \(v\) m s\(^{-1}\) at time \(t\) s. The only horizontal force acting on \(P\) has magnitude \(\frac{1}{10}(2v - 1)^2 e^{-t}\) N and acts towards \(O\). When \(t = 0\), \(x = 1\) and \(v = 3\).
  1. Find an expression for \(v\) in terms of \(t\). [5]
  2. Find an expression for \(x\) in terms of \(t\). [4]
Question 6:
AnswerMarks
6(a)dv 1 dv 1
2  2v12 et so  etdt
dt 10 2v12 20
p
qet  A
AnswerMarks Guidance
2v1*M1 Separate variables and attempt to integrate both
sides.
Where p and q are constants.
1 1
  et  A
AnswerMarks Guidance
22v1 20A1 AEF
 3 
t0, v3, A 
 
AnswerMarks Guidance
 20DM1 Substituting the boundary condition and obtain a
value.
1 5et
v 
AnswerMarks Guidance
2 3et 1*M1 A1 Find v in terms of t . AEF.
5
AnswerMarks Guidance
6(b)Integrate: x ptqln(ret s) B *M1
1 5
x t ln(3et 1) B
AnswerMarks Guidance
2 3A1 AEF
5
t 0, x1, B1 ln2
AnswerMarks Guidance
3DM1 Substituting the boundary condition and obtain a
value.
1 5 (3et 1)
x1 t ln
AnswerMarks Guidance
2 3 2A1 AEF
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(a) ---
6(a) | dv 1 dv 1
2  2v12 et so  etdt
dt 10 2v12 20
p
qet  A
2v1 | *M1 | Separate variables and attempt to integrate both
sides.
Where p and q are constants.
1 1
  et  A
22v1 20 | A1 | AEF
 3 
t0, v3, A 
 
 20 | DM1 | Substituting the boundary condition and obtain a
value.
1 5et
v 
2 3et 1 | *M1 A1 | Find v in terms of t . AEF.
5
--- 6(b) ---
6(b) | Integrate: x ptqln(ret s) B | *M1
1 5
x t ln(3et 1) B
2 3 | A1 | AEF
5
t 0, x1, B1 ln2
3 | DM1 | Substituting the boundary condition and obtain a
value.
1 5 (3et 1)
x1 t ln
2 3 2 | A1 | AEF
4
Question | Answer | Marks | Guidance
A particle $P$ of mass $2$ kg moving on a horizontal straight line has displacement $x$ m from a fixed point $O$ on the line and velocity $v$ m s$^{-1}$ at time $t$ s. The only horizontal force acting on $P$ has magnitude $\frac{1}{10}(2v - 1)^2 e^{-t}$ N and acts towards $O$. When $t = 0$, $x = 1$ and $v = 3$.

\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v$ in terms of $t$. [5]
\item Find an expression for $x$ in terms of $t$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q6 [9]}}
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