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 question requiring integration of a variable force with exponential decay. Part (a) involves solving a separable differential equation F=ma with the force (2v-1)²e^(-t), requiring substitution u=2v-1 and integration. Part (b) requires a second integration to find displacement. While the algebra is moderately involved and requires careful manipulation of exponentials, the solution path is relatively standard for Further Maths variable force problems—identify the differential equation, separate variables, integrate, and apply initial conditions. The exponential and quadratic terms add computational complexity but don't require novel insight beyond standard FM techniques.
Spec1.08h Integration by substitution6.06a Variable force: dv/dt or v*dv/dx methods
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 \(\frac{1}{10}(2v - 1)^2e^{-t}\) 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 $\frac{1}{10}(2v - 1)^2e^{-t}$ 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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