| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable force (velocity v) - use v dv/dx |
| Difficulty | Challenging +1.2 This is a Further Maths FM2 question requiring v dv/dx = F/m setup, separation of variables, and integration. While it involves multiple steps (finding k, deriving the exponential relationship, then time integration), the techniques are standard for FM2 students and the question provides clear guidance through parts (a) and (b). The mathematical manipulations are straightforward once the correct differential equation is established. |
| Spec | 1.08h Integration by substitution6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Form differential equation: \(0.5a = 0.5v\frac{dv}{dx} = -kv^2\) | M1 | 2.5 - Form differential equation in \(v\) and \(x\). Condone sign error |
| \(\Rightarrow \int \frac{1}{2v}dv = \int -k\,dx\) | M1 | 2.1 - Separate and integrate to form equation in \(v\) and \(x\). Condone missing constant of integration |
| \(\frac{1}{2}\ln v = -kx + C\) | A1 | 1.1b - Any equivalent form. Condone missing constant of integration |
| \(x=1, v=4\): \(\frac{1}{2}\ln 4 = -k + C\) and \(x=2, v=2\): \(\frac{1}{2}\ln 2 = -2k + C\) | M1 | 3.1a - Complete strategy to use the differential equation and boundary conditions to find \(v\) |
| \(\Rightarrow k = \frac{1}{2}(\ln 4 - \ln 2) = \frac{1}{2}\ln 2\), \(C = \frac{1}{2}\ln 8\): \(\ln v = -x\ln 2 + \ln 8\) | A1 | 1.1b - Correct expression in \(v\) and \(x\) in any form. Accept \(\ln v = -0.693...x + 2.079...\) |
| \(\ln v = x\ln\frac{1}{2} + \ln 8\), \(v = 8\times\left(\frac{1}{2}\right)^x\) | A1 | 2.2a - Expression in the required form. Do not need to see a separate statement of the values of \(a\) and \(b\) |
| \(\left(a = 8,\ b = \frac{1}{2}\right)\) | If mass is omitted from the differential equation can score M0M1A1M1A1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.5\frac{dv}{dt} = -kv^2\) (follow their \(k\)) | M1 | 2.5 - Differential equation in \(v\) and \(t\). Condone missing mass |
| \(\int\frac{1}{v^2}dv = \int -\ln 2\,dt \Rightarrow -\frac{1}{v} + C' = -t\ln 2\) | M1 | 2.1 - Separate and integrate |
| \(\Rightarrow \left[-\frac{1}{v}\right]_4^2 = \left[-t\ln 2\right]_0^T\) | M1 | 1.1b - Use limits on a definite integral or to find value of \(C'\) |
| \(-\frac{1}{2} + \frac{1}{4} = -T\ln 2\), \(T = \frac{1}{4\ln 2}\) * | A1* | 2.2a - Obtain given result from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = \frac{8}{2^x} = \frac{dx}{dt}\) (follow their \(v\)) | M1 | 2.5 |
| \(\int 2^x\,dx = \int 8\,dt \Rightarrow \frac{2^x}{\ln 2} = 8t + C'\) | M1 | 2.1 |
| \(\left[\frac{2^x}{\ln 2}\right]_1^2 = \left[8t\right]_0^T\) | M1 | 1.1b |
| \(\frac{1}{\ln 2}(4-2) = 8T\), \(T = \frac{1}{4\ln 2}\) * | A1* | 2.2a - If mass omitted can score M0M1M1A0 |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Form differential equation: $0.5a = 0.5v\frac{dv}{dx} = -kv^2$ | M1 | 2.5 - Form differential equation in $v$ and $x$. Condone sign error |
| $\Rightarrow \int \frac{1}{2v}dv = \int -k\,dx$ | M1 | 2.1 - Separate and integrate to form equation in $v$ and $x$. Condone missing constant of integration |
| $\frac{1}{2}\ln v = -kx + C$ | A1 | 1.1b - Any equivalent form. Condone missing constant of integration |
| $x=1, v=4$: $\frac{1}{2}\ln 4 = -k + C$ and $x=2, v=2$: $\frac{1}{2}\ln 2 = -2k + C$ | M1 | 3.1a - Complete strategy to use the differential equation and boundary conditions to find $v$ |
| $\Rightarrow k = \frac{1}{2}(\ln 4 - \ln 2) = \frac{1}{2}\ln 2$, $C = \frac{1}{2}\ln 8$: $\ln v = -x\ln 2 + \ln 8$ | A1 | 1.1b - Correct expression in $v$ and $x$ in any form. Accept $\ln v = -0.693...x + 2.079...$ |
| $\ln v = x\ln\frac{1}{2} + \ln 8$, $v = 8\times\left(\frac{1}{2}\right)^x$ | A1 | 2.2a - Expression in the required form. Do not need to see a separate statement of the values of $a$ and $b$ |
| $\left(a = 8,\ b = \frac{1}{2}\right)$ | | If mass is omitted from the differential equation can score M0M1A1M1A1A0 |
**(6 marks)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.5\frac{dv}{dt} = -kv^2$ (follow their $k$) | M1 | 2.5 - Differential equation in $v$ and $t$. Condone missing mass |
| $\int\frac{1}{v^2}dv = \int -\ln 2\,dt \Rightarrow -\frac{1}{v} + C' = -t\ln 2$ | M1 | 2.1 - Separate and integrate |
| $\Rightarrow \left[-\frac{1}{v}\right]_4^2 = \left[-t\ln 2\right]_0^T$ | M1 | 1.1b - Use limits on a definite integral or to find value of $C'$ |
| $-\frac{1}{2} + \frac{1}{4} = -T\ln 2$, $T = \frac{1}{4\ln 2}$ * | A1* | 2.2a - Obtain given result from correct working |
## Part (b) alt:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = \frac{8}{2^x} = \frac{dx}{dt}$ (follow their $v$) | M1 | 2.5 |
| $\int 2^x\,dx = \int 8\,dt \Rightarrow \frac{2^x}{\ln 2} = 8t + C'$ | M1 | 2.1 |
| $\left[\frac{2^x}{\ln 2}\right]_1^2 = \left[8t\right]_0^T$ | M1 | 1.1b |
| $\frac{1}{\ln 2}(4-2) = 8T$, $T = \frac{1}{4\ln 2}$ * | A1* | 2.2a - If mass omitted can score M0M1M1A0 |
**(4 marks) — Total: (10 marks)**
---\begin{enumerate}
\item A particle $P$ of mass 0.5 kg is moving along the positive $x$-axis in the direction of $x$ increasing. At time $t$ seconds $( t \geqslant 0 ) , P$ is $x$ metres from the origin $O$ and the speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The resultant force acting on $P$ is directed towards $O$ and has magnitude $k v ^ { 2 } \mathrm {~N}$, where $k$ is a positive constant.
\end{enumerate}
When $x = 1 , v = 4$ and when $x = 2 , v = 2$\\
(a) Show that $v = a b ^ { x }$, where $a$ and $b$ are constants to be found.
The time taken for the speed of $P$ to decrease from $4 \mathrm {~ms} ^ { - 1 }$ to $2 \mathrm {~ms} ^ { - 1 }$ is $T$ seconds.\\
(b) Show that $T = \frac { 1 } { 4 \ln 2 }$
\hfill \mbox{\textit{Edexcel FM2 2020 Q3 [10]}}