Question 2 - A-Level Maths

Edexcel FM2 2023 June — Question 2 8 marks

Exam Board	Edexcel
Module	FM2 (Further Mechanics 2)
Year	2023
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Variable Force
Type	Force depends on velocity v
Difficulty	Standard +0.8 This is a Further Maths FM2 variable force question requiring separation of variables, integration with logarithms, and a second integration using substitution. While the techniques are standard for FM2 (F=ma with velocity-dependent force), it requires careful algebraic manipulation and the 'show that' in part (a) demands precise working. The two-part structure with exact answers (involving ln) is typical of FM2 but more demanding than standard A-level mechanics.
Spec	1.08h Integration by substitution 6.06a Variable force: dv/dt or v*dv/dx methods

A particle of mass 2 kg is moving in a straight line on a smooth horizontal surface under the action of a horizontal force of magnitude \(F\) newtons.

At time \(t\) seconds \(( t > 0 )\),

the particle is moving with speed \(v \mathrm {~ms} ^ { - 1 }\)
\(F = 2 + v\)

The time taken for the speed of the particle to increase from \(5 \mathrm {~ms} ^ { - 1 }\) to \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is \(T\) seconds.

Show that \(T = 2 \ln \frac { 12 } { 7 }\) The distance moved by the particle as its speed increases from \(5 \mathrm {~ms} ^ { - 1 }\) to \(10 \mathrm {~ms} ^ { - 1 }\) is \(D\) metres.
Find the exact value of \(D\).

Show mark scheme Show mark scheme source

Question 2(a):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
\(2\frac{dv}{dt} = 2 + v\)	M1	2.1 - Obtain a differential equation in \(v\) and \(t\); allow \(\pm\) for acceleration
\(\Rightarrow \int \frac{2}{2+v}\,dv = \int 1\,dt\)	DM1	1.1b - Separate variables and integrate
\(2\ln	2+v	= t(+c)\)
\(T = 2\ln 12 - 2\ln 7 = 2\ln\frac{12}{7}\) *	A1*	2.2a - Use limits to eliminate constant of integration to obtain given answer from full correct working

Question 2(a) alt (Integrating Factor method):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
\(\frac{dv}{dt} - \frac{1}{2}v = 1\)	M1
Use integrating factor \(e^{-\frac{1}{2}t}\) and integrate	DM1
Obtain \(ve^{-\frac{1}{2}t} = -2e^{-\frac{1}{2}t}(+C)\)	A1
\(\Rightarrow v = -2 + 7e^{\frac{1}{2}t}\), \(T = 2\ln\frac{12}{7}\) *	A1*

Question 2(a) alt (Auxiliary Equation method):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
\(\frac{dv}{dt} - \frac{1}{2}v = 1\)	M1
AE: \(\lambda - \frac{1}{2} = 0\), so CF is \(v = Ae^{\frac{1}{2}t}\) and PI is \(v = -2\)	DM1
\(v = Ae^{\frac{1}{2}t} - 2\)	A1
\(t=0, v=5 \Rightarrow A=7 \Rightarrow 10 = 7e^{\frac{1}{2}T} - 2 \Rightarrow T = 2\ln\frac{12}{7}\)	A1

Question 2(b):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
\(v\frac{dv}{dx} = 1 + \frac{1}{2}v\)	M1	3.4 - Obtain a differential equation in \(v\) and \(x\); allow \(\pm\) for acceleration
\(\Rightarrow \int \frac{2v}{2+v}\,dv = \int 1\,dx = \int 2 - \frac{4}{2+v}\,dv\)	DM1	1.1b - Separate variables and integrate; if using IBP must complete integration
\(x(+A) = 2v - 4\ln	2+v	\)
\(D = 2(10-5) - 4\ln 12 + 4\ln 7 = 10 - 4\ln\left(\frac{12}{7}\right)\)	A1	1.1b - Use limits to obtain exact distance from exact working; any equivalent simplified form

Question 2(b) alt (using \(x\) in terms of \(t\)):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
\(\frac{dx}{dt} = -2 + 7e^{\frac{1}{2}t}\)	M1
Integrate	DM1
Obtain \(x = -2t + 14e^{\frac{1}{2}t}(+A)\)	A1
\(D = -4\ln\left(\frac{12}{7}\right) + 14 \times \frac{12}{7} - 14 = 10 - 4\ln\left(\frac{12}{7}\right)\)	A1

Question 2(b) alt (Second order ODE):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
\(2\ddot{x} - \dot{x} = 2\)	M1
AE: \(2m^2 - m = 0\), CF: \(x = Ae^{\frac{1}{2}t} + B\), PI: \(x = -2t\)	M1
\(x = Ae^{\frac{1}{2}t} + B - 2t\)	A1
\(t=0, x=0, \dot{x}=5 \Rightarrow x = 14e^{\frac{1}{2}t} - 14 - 2t \Rightarrow D = 10 - 4\ln\left(\frac{12}{7}\right)\)	A1

# Question 2(a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $2\frac{dv}{dt} = 2 + v$ | M1 | 2.1 - Obtain a differential equation in $v$ and $t$; allow $\pm$ for acceleration |
| $\Rightarrow \int \frac{2}{2+v}\,dv = \int 1\,dt$ | DM1 | 1.1b - Separate variables and integrate |
| $2\ln|2+v| = t(+c)$ | A1 | 1.1b - Or equivalent unsimplified form; allow without modulus signs |
| $T = 2\ln 12 - 2\ln 7 = 2\ln\frac{12}{7}$ * | A1* | 2.2a - Use limits to eliminate constant of integration to obtain given answer from full correct working |

---

# Question 2(a) alt (Integrating Factor method):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dv}{dt} - \frac{1}{2}v = 1$ | M1 | |
| Use integrating factor $e^{-\frac{1}{2}t}$ and integrate | DM1 | |
| Obtain $ve^{-\frac{1}{2}t} = -2e^{-\frac{1}{2}t}(+C)$ | A1 | |
| $\Rightarrow v = -2 + 7e^{\frac{1}{2}t}$, $T = 2\ln\frac{12}{7}$ * | A1* | |

---

# Question 2(a) alt (Auxiliary Equation method):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dv}{dt} - \frac{1}{2}v = 1$ | M1 | |
| AE: $\lambda - \frac{1}{2} = 0$, so CF is $v = Ae^{\frac{1}{2}t}$ and PI is $v = -2$ | DM1 | |
| $v = Ae^{\frac{1}{2}t} - 2$ | A1 | |
| $t=0, v=5 \Rightarrow A=7 \Rightarrow 10 = 7e^{\frac{1}{2}T} - 2 \Rightarrow T = 2\ln\frac{12}{7}$ | A1 | |

---

# Question 2(b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $v\frac{dv}{dx} = 1 + \frac{1}{2}v$ | M1 | 3.4 - Obtain a differential equation in $v$ and $x$; allow $\pm$ for acceleration |
| $\Rightarrow \int \frac{2v}{2+v}\,dv = \int 1\,dx = \int 2 - \frac{4}{2+v}\,dv$ | DM1 | 1.1b - Separate variables and integrate; if using IBP must complete integration |
| $x(+A) = 2v - 4\ln|2+v|$ | A1 | 1.1b - Or equivalent unsimplified form; allow without modulus signs |
| $D = 2(10-5) - 4\ln 12 + 4\ln 7 = 10 - 4\ln\left(\frac{12}{7}\right)$ | A1 | 1.1b - Use limits to obtain exact distance from exact working; any equivalent simplified form |

---

# Question 2(b) alt (using $x$ in terms of $t$):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = -2 + 7e^{\frac{1}{2}t}$ | M1 | |
| Integrate | DM1 | |
| Obtain $x = -2t + 14e^{\frac{1}{2}t}(+A)$ | A1 | |
| $D = -4\ln\left(\frac{12}{7}\right) + 14 \times \frac{12}{7} - 14 = 10 - 4\ln\left(\frac{12}{7}\right)$ | A1 | |

---

# Question 2(b) alt (Second order ODE):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $2\ddot{x} - \dot{x} = 2$ | M1 | |
| AE: $2m^2 - m = 0$, CF: $x = Ae^{\frac{1}{2}t} + B$, PI: $x = -2t$ | M1 | |
| $x = Ae^{\frac{1}{2}t} + B - 2t$ | A1 | |
| $t=0, x=0, \dot{x}=5 \Rightarrow x = 14e^{\frac{1}{2}t} - 14 - 2t \Rightarrow D = 10 - 4\ln\left(\frac{12}{7}\right)$ | A1 | |

---

Show LaTeX source

\begin{enumerate}
  \item A particle of mass 2 kg is moving in a straight line on a smooth horizontal surface under the action of a horizontal force of magnitude $F$ newtons.
\end{enumerate}

At time $t$ seconds $( t > 0 )$,

\begin{itemize}
  \item the particle is moving with speed $v \mathrm {~ms} ^ { - 1 }$
  \item $F = 2 + v$
\end{itemize}

The time taken for the speed of the particle to increase from $5 \mathrm {~ms} ^ { - 1 }$ to $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is $T$ seconds.\\
(a) Show that $T = 2 \ln \frac { 12 } { 7 }$

The distance moved by the particle as its speed increases from $5 \mathrm {~ms} ^ { - 1 }$ to $10 \mathrm {~ms} ^ { - 1 }$ is $D$ metres.\\
(b) Find the exact value of $D$.

\hfill \mbox{\textit{Edexcel FM2 2023 Q2 [8]}}

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