CAIE Further Paper 3 2020 November — Question 7 10 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2020
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeVariable force (position x) - find velocity
DifficultyChallenging +1.8 This is a challenging Further Maths mechanics problem requiring integration of a variable force to find velocity (using v dv/dx = a), then separating variables and integrating again to find the x-t relationship. The algebraic manipulation is non-trivial, particularly showing the exponential relationship in part (b), and requires careful handling of initial conditions. However, the solution path follows standard techniques for variable acceleration problems, making it demanding but within the expected scope of Further Maths Paper 3.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)3.02f Non-uniform acceleration: using differentiation and integration
A particle \(P\) moving in a straight line has displacement \(x\) m from a fixed point \(O\) on the line at time \(t\) s. The acceleration of \(P\), in m s\(^{-2}\), is given by \(\frac{200}{x^2} - \frac{100}{x^3}\) for \(x > 0\). When \(t = 0\), \(x = 1\) and \(P\) has velocity \(10\) m s\(^{-1}\) directed towards \(O\).
  1. Show that the velocity \(v\) m s\(^{-1}\) of \(P\) is given by \(v = \frac{10(1-2x)}{x}\). [5]
  2. Show that \(x\) and \(t\) are related by the equation \(e^{-40t} = (2x-1)e^{2x-2}\) and deduce what happens to \(x\) as \(t\) becomes large. [5]
Question 7:
AnswerMarks
7(a)dv 100 200
v =− +
dx x3 x2
v2 50 200
= − +A
AnswerMarks Guidance
2 x2 xM1 A1 Correct equation and attempt to integrate
Correct
AnswerMarks Guidance
x=1, v=−10: A=200M1 Use initial condition
100 ( 2x−1 )2
v2 =
AnswerMarks Guidance
x2M1 Rearrange to find v2
10 ( 2x−1 )
v=± and take negative sign to meet initial condition,
x
10 ( 1−2x )
so v=
AnswerMarks Guidance
xA1 Convincingly shown (no mention of ± scores A0)
AG
5
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(b)xdx
=10dt
1−2x
1 1 
−1 dx=10dt
 
21−2x
1 x
− ln1−2x − =10t+B
AnswerMarks Guidance
4 2M1 A1 Rearrange and attempt to integrate
1
t=0, x=1: B=−
AnswerMarks Guidance
2M1 Use initial condition
2x−2=−40t−ln(1−2x) so e −40t =( 2x−1 ) e2x−2A1 Convincingly shown, working required
AG
1
For large values of t, x→
AnswerMarks Guidance
2B1 CAO
5
Question 7:
--- 7(a) ---
7(a) | dv 100 200
v =− +
dx x3 x2
v2 50 200
= − +A
2 x2 x | M1 A1 | Correct equation and attempt to integrate
Correct
x=1, v=−10: A=200 | M1 | Use initial condition
100 ( 2x−1 )2
v2 =
x2 | M1 | Rearrange to find v2
10 ( 2x−1 )
v=± and take negative sign to meet initial condition,
x
10 ( 1−2x )
so v=
x | A1 | Convincingly shown (no mention of ± scores A0)
AG
5
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | xdx
=10dt
1−2x
1 1 
−1 dx=10dt
 
21−2x

1 x
− ln1−2x − =10t+B
4 2 | M1 A1 | Rearrange and attempt to integrate
1
t=0, x=1: B=−
2 | M1 | Use initial condition
2x−2=−40t−ln(1−2x) so e −40t =( 2x−1 ) e2x−2 | A1 | Convincingly shown, working required
AG
1
For large values of t, x→
2 | B1 | CAO
5
A particle $P$ moving in a straight line has displacement $x$ m from a fixed point $O$ on the line at time $t$ s. The acceleration of $P$, in m s$^{-2}$, is given by $\frac{200}{x^2} - \frac{100}{x^3}$ for $x > 0$. When $t = 0$, $x = 1$ and $P$ has velocity $10$ m s$^{-1}$ directed towards $O$.

\begin{enumerate}[label=(\alph*)]
\item Show that the velocity $v$ m s$^{-1}$ of $P$ is given by $v = \frac{10(1-2x)}{x}$. [5]
\item Show that $x$ and $t$ are related by the equation $e^{-40t} = (2x-1)e^{2x-2}$ and deduce what happens to $x$ as $t$ becomes large. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q7 [10]}}
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