OCR Further Pure Core 1 2020 November — Question 10 13 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2020
SessionNovember
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeApplied/modelling contexts
DifficultyStandard +0.8 This is a substantial Further Maths question requiring Newton's second law formulation, integrating factor method, asymptotic analysis, and qualitative DE analysis. While the techniques are standard for FM students, the multi-part structure, applied context, and need to interpret long-term behavior elevate it above routine exercises. The integrating factor itself is straightforward, but the complete solution path requires sustained reasoning across 5 parts.
Spec3.03d Newton's second law: 2D vectors4.10a General/particular solutions: of differential equations4.10c Integrating factor: first order equations4.10d Second order homogeneous: auxiliary equation method

10 A particle of mass 0.5 kg is initially at point \(O\). It moves from rest along the \(x\)-axis under the influence of two forces \(F _ { 1 } \mathrm {~N}\) and \(F _ { 2 } \mathrm {~N}\) which act parallel to the \(x\)-axis. At time \(t\) seconds the velocity of the particle is \(v \mathrm {~ms} ^ { - 1 }\). \(F _ { 1 }\) is acting in the direction of motion of the particle and \(F _ { 2 }\) is resisting motion.
In an initial model
  • \(F _ { 1 }\) is proportional to \(t\) with constant of proportionality \(\lambda > 0\),
  • \(F _ { 2 }\) is proportional to \(v\) with constant of proportionality \(\mu > 0\).
    1. Show that the motion of the particle can be modelled by the following differential equation.
$$\frac { 1 } { 2 } \frac { d v } { d t } = \lambda t - \mu v$$
  • Solve the differential equation in part (a), giving the particular solution for \(v\) in terms of \(t\), \(\lambda\) and \(\mu\). You are now given that \(\lambda = 2\) and \(\mu = 1\).
  • Find a formula for an approximation for \(v\) in terms of \(t\) when \(t\) is large. In a refined model

  • Question 10:
    AnswerMarks Guidance
    10(a) F =ma
    (F is in the direction of motion and F is resisting motion)
    1 2
    F =λt−µv
    1 dv
    ⇒ =λt−µv AG
    AnswerMarks
    2 dtM1
    A13.3
    2.1Use of Newton II with constants of
    proportionality. F = ma must be seen
    [2]
    Alternative method
    dv
    +2µv=2λt
    AnswerMarks
    dtE
    M1
    AE:m+2µ=0
    AnswerMarks Guidance
    ⇒m=−2µ⇒CF = Ae−2µtM1
    A1M1 E
    E
    A1
    AnswerMarks Guidance
    PI :v=at+bPI :v=at+b M1
    dv
    ⇒ =a⇒a+2µat+2µb=2λt
    dt
    λ a λ
    ∴a = , b=− =−
    µ 2µ 2µ2
    λ λ
    ∴GS:v= t− + Ae−2µt
    AnswerMarks
    µ 2µ2M1
    A1
    AnswerMarks Guidance
    A1M1 C
    C
    A
    A1
    A1
    [7]
    AnswerMarks
    (c)λ λ λ
    v= t− + e−2µt
    µ 2µ2 2µ2
    λ=2,µ=1⇒v=2t−1+e−2t
    When t is large, e−2t is very small
    AnswerMarks
    so v≈2t−1M1
    A13.4
    3.3Consider the behaviour of the exponential
    function in their equation from (b) soi
    or v≈2t
    [2]
    AnswerMarks
    (d)1 dv
    =2−v oe
    AnswerMarks Guidance
    2 dtB1 3.5c
    [1]
    AnswerMarks
    (e)dv
    As v approaches 2, →0
    dt
    AnswerMarks Guidance
    i.e. v approaches a constant value.B1 3.4
    [1]
    Alternative method
    dv
    +2µv=2λt
    dt
    E
    E
    E
    C
    C
    A
    Question 10:
    10 | (a) | F =ma
    (F is in the direction of motion and F is resisting motion)
    1 2
    F =λt−µv
    1 dv
    ⇒ =λt−µv AG
    2 dt | M1
    A1 | 3.3
    2.1 | Use of Newton II with constants of
    proportionality. F = ma must be seen
    [2]
    Alternative method
    dv
    +2µv=2λt
    dt | E
    M1
    AE:m+2µ=0
    ⇒m=−2µ⇒CF = Ae−2µt | M1
    A1 | M1 | E
    E
    A1
    PI :v=at+b | PI :v=at+b | M1 | C
    dv
    ⇒ =a⇒a+2µat+2µb=2λt
    dt
    λ a λ
    ∴a = , b=− =−
    µ 2µ 2µ2
    λ λ
    ∴GS:v= t− + Ae−2µt
    µ 2µ2 | M1
    A1
    A1 | M1 | C
    C
    A
    A1
    A1
    [7]
    (c) | λ λ λ
    v= t− + e−2µt
    µ 2µ2 2µ2
    λ=2,µ=1⇒v=2t−1+e−2t
    When t is large, e−2t is very small
    so v≈2t−1 | M1
    A1 | 3.4
    3.3 | Consider the behaviour of the exponential
    function in their equation from (b) soi
    or v≈2t
    [2]
    (d) | 1 dv
    =2−v oe
    2 dt | B1 | 3.5c
    [1]
    (e) | dv
    As v approaches 2, →0
    dt
    i.e. v approaches a constant value. | B1 | 3.4
    [1]
    Alternative method
    dv
    +2µv=2λt
    dt
    E
    E
    E
    C
    C
    A
    10 A particle of mass 0.5 kg is initially at point $O$. It moves from rest along the $x$-axis under the influence of two forces $F _ { 1 } \mathrm {~N}$ and $F _ { 2 } \mathrm {~N}$ which act parallel to the $x$-axis. At time $t$ seconds the velocity of the particle is $v \mathrm {~ms} ^ { - 1 }$.\\
    $F _ { 1 }$ is acting in the direction of motion of the particle and $F _ { 2 }$ is resisting motion.\\
    In an initial model
    
    \begin{itemize}
      \item $F _ { 1 }$ is proportional to $t$ with constant of proportionality $\lambda > 0$,
      \item $F _ { 2 }$ is proportional to $v$ with constant of proportionality $\mu > 0$.
    \begin{enumerate}[label=(\alph*)]
    \item Show that the motion of the particle can be modelled by the following differential equation.
    \end{itemize}
    
    $$\frac { 1 } { 2 } \frac { d v } { d t } = \lambda t - \mu v$$
    \item Solve the differential equation in part (a), giving the particular solution for $v$ in terms of $t$, $\lambda$ and $\mu$.
    
    You are now given that $\lambda = 2$ and $\mu = 1$.
    \item Find a formula for an approximation for $v$ in terms of $t$ when $t$ is large.
    
    In a refined model
    
    \begin{itemize}
      \item $F _ { 1 }$ is constant, acting in the direction of motion with magnitude 2 N ,
      \item $F _ { 2 }$ is as before with $\mu = 1$.
    \item Write down a differential equation for the refined model.
    \item Without solving the differential equation in part (d), write down what will happen to the velocity in the long term according to this refined model.
    \end{itemize}
    \end{enumerate}
    
    \hfill \mbox{\textit{OCR Further Pure Core 1 2020 Q10 [13]}}