OCR Further Pure Core 1 2023 June — Question 7 11 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2023
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeForce depends on velocity v
DifficultyChallenging +1.2 This is a multi-part Further Maths mechanics question involving differential equations and Maclaurin series. Part (a)(i) requires understanding force directions (straightforward conceptual reasoning). Part (a)(ii) involves substituting initial conditions into a given differential equation to find acceleration—a standard technique requiring no solving. Part (b) applies routine Maclaurin series methods with given initial conditions. While it combines several topics (forces, differential equations, series), each step follows standard procedures without requiring novel insight. The Further Maths context and multi-step nature place it above average difficulty, but it remains a structured, methodical question.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.10b Model with differential equations: kinematics and other contexts6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.06a Variable force: dv/dt or v*dv/dx methods

7 An engineer is modelling the motion of a particle \(P\) of mass 0.5 kg in a wind tunnel. \(P\) is modelled as travelling in a straight line. The point \(O\) is a fixed point within the wind tunnel. The displacement of \(P\) from \(O\) at time \(t\) seconds is \(x\) metres, for \(t \geqslant 0\). You are given that \(x \geqslant 0\) for all \(t \geqslant 0\) and that \(P\) does not reach the end of the wind tunnel.
If \(t \geqslant 0\), then \(P\) is subject to three forces which are modelled in the following way.
  • The first force has a magnitude of \(5 ( t + 1 ) \cosh t \mathrm {~N}\) and acts in the positive \(x\)-direction.
  • The second force has a magnitude of \(0.5 x \mathrm {~N}\) and acts towards \(O\).
  • The third force has a magnitude of \(\left| \frac { d x } { d t } \right| \mathrm { N }\) and acts in the direction of motion of the particle.
      1. Show that the Maclaurin series for \(\mathrm { f } ( t )\) up to and including the term in \(t\) is \(6 - 5 t\).
      2. Use your answer to part (a)(ii) to show that the term in \(t ^ { 2 }\) in the Maclaurin series for \(\mathrm { f } ( t )\) is \(- 3 t ^ { 2 }\).
      3. By differentiating the differential equation in part (a) with respect to \(t\), show that the term in \(t ^ { 3 }\) in the Maclaurin series for \(\mathrm { f } ( t )\) is \(0.5 t ^ { 3 }\). You are given that the complete Maclaurin series for the function f is valid for all values of \(t \geqslant 0\).
        After 0.25 seconds \(P\) has travelled 1.43 m towards the origin.
      1. By using the Maclaurin series for \(\mathrm { f } ( t )\) up to and including the term in \(t ^ { 3 }\), evaluate the suitability of the model for determining the displacement of \(P\) from \(O\) when \(t = 0.25\).
      2. Explain why it might not be sensible to use the Maclaurin series for \(\mathrm { f } ( t )\) up to and including the term in \(t ^ { 3 }\) to evaluate the suitability of the model for determining the displacement of \(P\) from \(O\) when \(t = 10\).

Question 7:
AnswerMarks Guidance
7(a) (i)
e.g. sign of is +ve because the force is in the same
dt
AnswerMarks Guidance
direction of motionB1 Convincingly shown
[1]
AnswerMarks
(ii)1 d 2 x
= 5 c o s h 0 − 3 − 5
2 d t 2
= − 3
d 2 x
 = − 6 ( m s − 2 )
AnswerMarks
d t 2M1
A1dx
Substitutes t = 0 and 5
dt
Or 6 towards O
[2]
AnswerMarks Guidance
(b)(i) Maclaurin: x=f(0)+f '(0)t+...
dx 
When t =0,(x=6=)f(0)=6, =  =f '(0)=−5
 dt 
AnswerMarks Guidance
 x=6−5t....B1 First two terms ofMaclaurin’s formula, possibly in generalised form, must
be seen
AG Convincingly shown.
[1]
AnswerMarks
(ii)t 2
3rd term of Maclaurin is f ''( 0 )
2 !
With (from (a)(ii)) f''(0)=−6
t 2
So 3rd term is ( − 6 ) = − 3 t 2
AnswerMarks Guidance
2 !B1 AG Convincingly shown
[1]
AnswerMarks Guidance
7(b) (iii)
 5 c o s h t + 5 ( t + 1 ) s i n h t − 0 .5 + =
d t d t 2 2 d t 3
1
W h e n t = 0 , 5 + 0 + 2 .5 − 6 = f '''( 0 )
2
 f '''( 0 = ) 3
t 3 1
 4 t h t e r m = f '''( 0 ) = t 3
AnswerMarks
3 ! 2M1
A1
M1
AnswerMarks
A1d
For (cosht)=sinhtand product rule attempted
dt
Fully correct differentiation
Substitute t = 0
AG. Allow embedded answer.
[4]
AnswerMarks Guidance
(c)(i) 1
x = 6 − 5 t − 3 2 + t 3 t
2
 W h e n t = 0 . 2 5 , x = 6 − 1 . 2 5 − 0 . 1 8 7 5 − 0 . 0 0 7 8  4 . 5 7 0 . . .
 D i s t a n c e t r a v e l l e d 6 − 4 . 5 7 0  1 . 4 3 0 . . .
AnswerMarks Guidance
So suitable as value closeB1 Substitutes t = 0.25 to obtain an approximation and correct
conclusion
[1]
AnswerMarks
(ii)e.g.
more terms may be required
Higher terms may be large
The candidate calculates the term in t4 (11t4/12) and indicates
AnswerMarks Guidance
that this term is large for values of t > 1.B1 Allow any correct explanation that explains/implies that for t > 1
some of the higher power terms are large and so non-negligible.
f(10) = 156 is too large is not enough
[1]
Question 7:
7 | (a) | (i) | dy
e.g. sign of is +ve because the force is in the same
dt
direction of motion | B1 | Convincingly shown
[1]
(ii) | 1 d 2 x
= 5 c o s h 0 − 3 − 5
2 d t 2
= − 3
d 2 x
 = − 6 ( m s − 2 )
d t 2 | M1
A1 | dx
Substitutes t = 0 and 5
dt
Or 6 towards O
[2]
(b) | (i) | Maclaurin: x=f(0)+f '(0)t+...
dx 
When t =0,(x=6=)f(0)=6, =  =f '(0)=−5
 dt 
 x=6−5t.... | B1 | First two terms ofMaclaurin’s formula, possibly in generalised form, must
be seen
AG Convincingly shown.
[1]
(ii) | t 2
3rd term of Maclaurin is f ''( 0 )
2 !
With (from (a)(ii)) f''(0)=−6
t 2
So 3rd term is ( − 6 ) = − 3 t 2
2 ! | B1 | AG Convincingly shown
[1]
7 | (b) | (iii) | d x d 2 x 1 d 3 x
 5 c o s h t + 5 ( t + 1 ) s i n h t − 0 .5 + =
d t d t 2 2 d t 3
1
W h e n t = 0 , 5 + 0 + 2 .5 − 6 = f '''( 0 )
2
 f '''( 0 = ) 3
t 3 1
 4 t h t e r m = f '''( 0 ) = t 3
3 ! 2 | M1
A1
M1
A1 | d
For (cosht)=sinhtand product rule attempted
dt
Fully correct differentiation
Substitute t = 0
AG. Allow embedded answer.
[4]
(c) | (i) | 1
x = 6 − 5 t − 3 2 + t 3 t
2
 W h e n t = 0 . 2 5 , x = 6 − 1 . 2 5 − 0 . 1 8 7 5 − 0 . 0 0 7 8  4 . 5 7 0 . . .
 D i s t a n c e t r a v e l l e d 6 − 4 . 5 7 0  1 . 4 3 0 . . .
So suitable as value close | B1 | Substitutes t = 0.25 to obtain an approximation and correct
conclusion
[1]
(ii) | e.g.
more terms may be required
Higher terms may be large
The candidate calculates the term in t4 (11t4/12) and indicates
that this term is large for values of t > 1. | B1 | Allow any correct explanation that explains/implies that for t > 1
some of the higher power terms are large and so non-negligible.
f(10) = 156 is too large is not enough
[1]
7 An engineer is modelling the motion of a particle $P$ of mass 0.5 kg in a wind tunnel.\\
$P$ is modelled as travelling in a straight line. The point $O$ is a fixed point within the wind tunnel. The displacement of $P$ from $O$ at time $t$ seconds is $x$ metres, for $t \geqslant 0$.

You are given that $x \geqslant 0$ for all $t \geqslant 0$ and that $P$ does not reach the end of the wind tunnel.\\
If $t \geqslant 0$, then $P$ is subject to three forces which are modelled in the following way.

\begin{itemize}
  \item The first force has a magnitude of $5 ( t + 1 ) \cosh t \mathrm {~N}$ and acts in the positive $x$-direction.
  \item The second force has a magnitude of $0.5 x \mathrm {~N}$ and acts towards $O$.
  \item The third force has a magnitude of $\left| \frac { d x } { d t } \right| \mathrm { N }$ and acts in the direction of motion of the particle.
\begin{enumerate}[label=(\alph*)]
\item The engineer applies the equation " $F = m a$ " to the model of the motion of $P$ and derives the following differential equation.\\
$5 ( t + 1 ) \operatorname { cosht } - 0.5 x + \frac { d x } { d t } = 0.5 \frac { d ^ { 2 } x } { d t ^ { 2 } }$
\begin{enumerate}[label=(\roman*)]
\item Explain the sign of the $\frac { \mathrm { dx } } { \mathrm { dt } }$ term in the engineer's differential equation.
\end{itemize}

When $t = 0$ the displacement of $P$ is 6 m , and it is travelling towards $O$ with a speed of $5 \mathrm {~ms} ^ { - 1 }$.
\item Without attempting to solve the differential equation, find the acceleration of $P$ when $t = 0$.

Let the particular solution to the differential equation in part (a) be a function f such that $\mathrm { x } = \mathrm { f } ( \mathrm { t } )$ for $t \geqslant 0$.

The particular solution to the differential equation can be expressed as a Maclaurin series.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Show that the Maclaurin series for $\mathrm { f } ( t )$ up to and including the term in $t$ is $6 - 5 t$.
\item Use your answer to part (a)(ii) to show that the term in $t ^ { 2 }$ in the Maclaurin series for $\mathrm { f } ( t )$ is $- 3 t ^ { 2 }$.
\item By differentiating the differential equation in part (a) with respect to $t$, show that the term in $t ^ { 3 }$ in the Maclaurin series for $\mathrm { f } ( t )$ is $0.5 t ^ { 3 }$.

You are given that the complete Maclaurin series for the function f is valid for all values of $t \geqslant 0$.\\
After 0.25 seconds $P$ has travelled 1.43 m towards the origin.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item By using the Maclaurin series for $\mathrm { f } ( t )$ up to and including the term in $t ^ { 3 }$, evaluate the suitability of the model for determining the displacement of $P$ from $O$ when $t = 0.25$.
\item Explain why it might not be sensible to use the Maclaurin series for $\mathrm { f } ( t )$ up to and including the term in $t ^ { 3 }$ to evaluate the suitability of the model for determining the displacement of $P$ from $O$ when $t = 10$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2023 Q7 [11]}}