| Exam Board | WJEC |
|---|---|
| Module | Further Unit 6 (Further Unit 6) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of velocity (separation of variables) |
| Difficulty | Standard +0.8 This is a standard Further Maths mechanics question involving variable resistance and differential equations. Part (a) requires applying F=ma with velocity-dependent resistance (straightforward). Part (b) involves separating variables and integrating (standard technique). Part (c) is simple substitution. The question follows a predictable structure with no novel insights required, but the topic itself (non-constant acceleration with differential equations) places it above average difficulty for A-level. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using N2L | M1 | |
| \(-0.2 - 0.03v = 9\frac{dv}{dt}\) | A1 | |
| \(900\frac{dv}{dt} = -(20+3v)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(900\int\frac{dv}{20+3v} = -\int dt\) | M1 | sep. var. |
| \(900\times\frac{1}{3}\ln(20+3v) = -t\ (+C)\) | A1, A1 | \(\ln(20+3v)\); all correct |
| When \(t=0\), \(v=20\): \(C = 300\ln 80\) | m1 | used |
| \(t = 300\ln(80) - 300\ln(20+3v) = 300\ln\!\left(\frac{80}{20+3v}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When body is at rest, \(v=0\): \(T = 300\ln(80) - 300\ln(20)\) | m1 | used |
| \(T = 300\ln(4) = \underline{416}\ \text{s}\) | A1 | cao |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using N2L | M1 | |
| $-0.2 - 0.03v = 9\frac{dv}{dt}$ | A1 | |
| $900\frac{dv}{dt} = -(20+3v)$ | A1 | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $900\int\frac{dv}{20+3v} = -\int dt$ | M1 | sep. var. |
| $900\times\frac{1}{3}\ln(20+3v) = -t\ (+C)$ | A1, A1 | $\ln(20+3v)$; all correct |
| When $t=0$, $v=20$: $C = 300\ln 80$ | m1 | used |
| $t = 300\ln(80) - 300\ln(20+3v) = 300\ln\!\left(\frac{80}{20+3v}\right)$ | A1 | |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| When body is at rest, $v=0$: $T = 300\ln(80) - 300\ln(20)$ | m1 | used |
| $T = 300\ln(4) = \underline{416}\ \text{s}$ | A1 | cao |3. A body, of mass 9 kg , is projected along a straight horizontal track with an initial speed of $20 \mathrm {~ms} ^ { - 1 }$. At time $t \mathrm {~s}$ the body experiences a resistance of magnitude $( 0.2 + 0.03 v ) \mathrm { N }$ where $v \mathrm {~ms} ^ { - 1 }$ is its speed.
\begin{enumerate}[label=(\alph*)]
\item Show that $v$ satisfies the differential equation
$$900 \frac { \mathrm {~d} v } { \mathrm {~d} t } = - ( 20 + 3 v )$$
\item Find an expression for $t$ in terms of $v$.
\item Calculate, to the nearest second, the time taken for the body to come to rest.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 6 Q3 [10]}}