| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Chemical reaction kinetics |
| Difficulty | Standard +0.3 This is a structured, multi-part question that guides students through verification and solving of differential equations with separable variables. Parts (i), (ii), and (iv) are routine verification/substitution tasks. Part (iii) involves standard separation of variables with partial fractions (a core C4 technique), but the algebraic manipulation is straightforward. Part (v) is trivial comparison. The question requires multiple techniques but provides significant scaffolding and asks for no novel insights—slightly easier than average for A-level. |
| Spec | 1.02y Partial fractions: decompose rational functions1.07j Differentiate exponentials: e^(kx) and a^(kx)1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(t = 0\), \(v = 5(1 - e^0) = 0\) | E1 | |
| As \(t \to \infty\), \(e^{-2t} \to 0 \Rightarrow v \to 5\) | E1 | |
| When \(t = 0.5\), \(v = 3.16 \text{ ms}^{-1}\) | B1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dv}{dt} = 5 \times (-2)e^{-2t} = 10e^{-2t}\) | B1 | |
| \(10 - 2v = 10 - 10(1 - e^{-2t}) = 10e^{-2t}\) | M1 | |
| \(\Rightarrow \frac{dv}{dt} = 10 - 2v\) | E1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dv}{dt} = 10 - 0.4v^2\) | ||
| \(\frac{10}{100 - 4v^2}\frac{dv}{dt} = 1\) | M1 | |
| \(\Rightarrow \frac{10}{25 - v^2}\frac{dv}{dt} = 4\) | ||
| \(\Rightarrow \frac{10}{(5-v)(5+v)}\frac{dv}{dt} = 4\) | E1 | |
| \(\frac{10}{(5-v)(5+v)} = \frac{A}{5-v} + \frac{B}{5+v}\) | ||
| \(10 = A(5+v) + B(5-v)\) | M1 | |
| \(v = 5 \Rightarrow 10 = 10A \Rightarrow A = 1\) | A1 | for both \(A=1, B=1\) |
| \(v = -5 \Rightarrow 10 = 10B \Rightarrow B = 1\) | ||
| \(\Rightarrow \frac{10}{(5-v)(5+v)} = \frac{1}{5-v} + \frac{1}{5+v}\) | ||
| \(\Rightarrow \int\left(\frac{1}{5-v} + \frac{1}{5+v}\right)dv = 4\int dt\) | M1 | separating variables correctly and indicating integration |
| \(\Rightarrow \ln(5+v) - \ln(5-v) = 4t + c\) | A1 | ft their \(A,B\); condone absence of \(c\) |
| when \(t=0\), \(v=0 \Rightarrow 0 = 4\times 0 + c \Rightarrow c = 0\) | A1 | ft finding \(c\) from expression of correct form |
| \(\Rightarrow \ln\left(\frac{5+v}{5-v}\right) = 4t\) | ||
| \(\Rightarrow t = \frac{1}{4}\ln\left(\frac{5+v}{5-v}\right)\) | E1 | |
| [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(t \to \infty\), \(e^{-4t} \to 0 \Rightarrow v \to 5/1 = 5\) | E1 | |
| When \(t = 0.5\), \(t = \frac{5(1-e^{-2})}{1+e^{-2}} = 3.8 \text{ ms}^{-1}\) | M1A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The first model | E1 | www |
| [1] |
# Question 2:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $t = 0$, $v = 5(1 - e^0) = 0$ | E1 | |
| As $t \to \infty$, $e^{-2t} \to 0 \Rightarrow v \to 5$ | E1 | |
| When $t = 0.5$, $v = 3.16 \text{ ms}^{-1}$ | B1 | |
| **[3]** | | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dv}{dt} = 5 \times (-2)e^{-2t} = 10e^{-2t}$ | B1 | |
| $10 - 2v = 10 - 10(1 - e^{-2t}) = 10e^{-2t}$ | M1 | |
| $\Rightarrow \frac{dv}{dt} = 10 - 2v$ | E1 | |
| **[3]** | | |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dv}{dt} = 10 - 0.4v^2$ | | |
| $\frac{10}{100 - 4v^2}\frac{dv}{dt} = 1$ | M1 | |
| $\Rightarrow \frac{10}{25 - v^2}\frac{dv}{dt} = 4$ | | |
| $\Rightarrow \frac{10}{(5-v)(5+v)}\frac{dv}{dt} = 4$ | E1 | |
| $\frac{10}{(5-v)(5+v)} = \frac{A}{5-v} + \frac{B}{5+v}$ | | |
| $10 = A(5+v) + B(5-v)$ | M1 | |
| $v = 5 \Rightarrow 10 = 10A \Rightarrow A = 1$ | A1 | for both $A=1, B=1$ |
| $v = -5 \Rightarrow 10 = 10B \Rightarrow B = 1$ | | |
| $\Rightarrow \frac{10}{(5-v)(5+v)} = \frac{1}{5-v} + \frac{1}{5+v}$ | | |
| $\Rightarrow \int\left(\frac{1}{5-v} + \frac{1}{5+v}\right)dv = 4\int dt$ | M1 | separating variables correctly and indicating integration |
| $\Rightarrow \ln(5+v) - \ln(5-v) = 4t + c$ | A1 | ft their $A,B$; condone absence of $c$ |
| when $t=0$, $v=0 \Rightarrow 0 = 4\times 0 + c \Rightarrow c = 0$ | A1 | ft finding $c$ from expression of correct form |
| $\Rightarrow \ln\left(\frac{5+v}{5-v}\right) = 4t$ | | |
| $\Rightarrow t = \frac{1}{4}\ln\left(\frac{5+v}{5-v}\right)$ | E1 | |
| **[8]** | | |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $t \to \infty$, $e^{-4t} \to 0 \Rightarrow v \to 5/1 = 5$ | E1 | |
| When $t = 0.5$, $t = \frac{5(1-e^{-2})}{1+e^{-2}} = 3.8 \text{ ms}^{-1}$ | M1A1 | |
| **[3]** | | |
## Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| The first model | E1 | www |
| **[1]** | | |
---2 A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after $t$ seconds it is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Its terminal (long-term) velocity is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
A model of the particle's motion is proposed. In this model, $v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)$.\\
(i) Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model.\\
(ii) Verify that $v$ satisfies the differential equation $\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 2 v$.
In a second model, $v$ satisfies the differential equation
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 0.4 v ^ { 2 }$$
As before, when $t = 0 , v = 0$.\\
(iii) Show that this differential equation may be written as
$$\frac { 10 } { ( 5 - v ) ( 5 + v ) } \frac { \mathrm { d } v } { \mathrm {~d} t } = 4$$
Using partial fractions, solve this differential equation to show that
$$t = \frac { 1 } { 4 } \ln \left( \frac { 5 + v } { 5 - v } \right)$$
This can be re-arranged to give $v = \frac { 5 \left( 1 - \mathrm { e } ^ { - 4 t } \right) } { 1 + \mathrm { e } ^ { - 4 t } }$. [You are not required to show this result.]\\
(iv) Verify that this model also gives a terminal velocity of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Calculate the velocity after 0.5 seconds as given by this model.
The velocity of the particle after 0.5 seconds is measured as $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(v) Which of the two models fits the data better?
\hfill \mbox{\textit{OCR MEI C4 Q2 [18]}}