| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of velocity (separation of variables) |
| Difficulty | Standard +0.8 Part (a) is trivial constant acceleration. Part (b) requires solving a first-order differential equation dv/dt = g - 0.1v, which is A-level Further Maths content involving separation of variables and natural logarithms. Part (c) requires interpretation of limiting behavior. This is moderately challenging for A-level due to the differential equations component, though the DE itself is standard once recognized. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v = u + at\), with \(u = 0\), \(t = 2\), \(a = g\) | M1 (AO 1.1a) | Selects and uses appropriate constant acceleration equation, or uses integration with \(a = g\) to obtain \(v = gt + c\) |
| \(v = 0 + g \times 2 = 2g\) m s\(^{-1}\) | R1 (AO 2.1) | Must state \(u = 0\) and substitute correct values; must use consistent signs for \(v\) and \(a\). OR substitute \(v=0\), \(t=0\) into \(v = gt + c\) to get \(c = 0\), concluding \(v = 2g\). AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a = \dfrac{dv}{dt}\) or \(v = \int a \, dt\) | B1 (AO 3.4) | States or uses either form |
| \(\dfrac{dv}{dt} = g - 0.1v\) | M1 (AO 3.1a) | Forms differential equation using \(\dfrac{dv}{dt}\) |
| \(\displaystyle\int \dfrac{1}{g - 0.1v} \, dv = \int dt\) | M1 (AO 1.1a) | Separates variables and integrates one side correctly |
| \(-10\ln(g - 0.1v) = t + c\) | A1 (AO 1.1b) | Integral of form \(k\ln(g - 0.1v) = t + c\); condone missing constant |
| When \(t = 0\), \(v = 0\): \(-10\ln g = c\) | M1 (AO 3.4) | Substitutes initial conditions to find \(c\) |
| \(-10\ln\dfrac{g - 0.1v}{g} = t\) | A1 (AO 1.1b) | Correct constant of integration |
| \(v = 10g(1 - e^{-0.1t})\) | A1 (AO 1.1b) | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Under Andy's model the velocity keeps increasing | E1 (AO 3.5a) | Explains that as \(t\) becomes large, Andy's model has an increasing velocity |
| Under Amy's model the velocity approaches an upper limit | E1 (AO 3.5a) | Explains that as \(t\) becomes large, Amy's model reaches an upper limit |
## Question 17(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v = u + at$, with $u = 0$, $t = 2$, $a = g$ | M1 (AO 1.1a) | Selects and uses appropriate constant acceleration equation, or uses integration with $a = g$ to obtain $v = gt + c$ |
| $v = 0 + g \times 2 = 2g$ m s$^{-1}$ | R1 (AO 2.1) | Must state $u = 0$ and substitute correct values; must use consistent signs for $v$ and $a$. OR substitute $v=0$, $t=0$ into $v = gt + c$ to get $c = 0$, concluding $v = 2g$. AG |
---
## Question 17(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a = \dfrac{dv}{dt}$ or $v = \int a \, dt$ | B1 (AO 3.4) | States or uses either form |
| $\dfrac{dv}{dt} = g - 0.1v$ | M1 (AO 3.1a) | Forms differential equation using $\dfrac{dv}{dt}$ |
| $\displaystyle\int \dfrac{1}{g - 0.1v} \, dv = \int dt$ | M1 (AO 1.1a) | Separates variables and integrates one side correctly |
| $-10\ln(g - 0.1v) = t + c$ | A1 (AO 1.1b) | Integral of form $k\ln(g - 0.1v) = t + c$; condone missing constant |
| When $t = 0$, $v = 0$: $-10\ln g = c$ | M1 (AO 3.4) | Substitutes initial conditions to find $c$ |
| $-10\ln\dfrac{g - 0.1v}{g} = t$ | A1 (AO 1.1b) | Correct constant of integration |
| $v = 10g(1 - e^{-0.1t})$ | A1 (AO 1.1b) | OE |
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## Question 17(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Under Andy's model the velocity keeps increasing | E1 (AO 3.5a) | Explains that as $t$ becomes large, Andy's model has an increasing velocity |
| Under Amy's model the velocity approaches an upper limit | E1 (AO 3.5a) | Explains that as $t$ becomes large, Amy's model reaches an upper limit |
---17 A ball is released from a great height so that it falls vertically downwards towards the surface of the Earth.
17
\begin{enumerate}[label=(\alph*)]
\item Using a simple model, Andy predicts that the velocity of the ball, exactly 2 seconds after being released from rest, is $2 g \mathrm {~m} \mathrm {~s} ^ { - 1 }$
Show how Andy has obtained his prediction.\\
17
\item Using a refined model, Amy predicts that the ball's acceleration, $a \mathrm {~ms} ^ { - 2 }$, at time $t$ seconds after being released from rest is
$$a = g - 0.1 v$$
where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the velocity of the ball at time $t$ seconds.
Find an expression for $v$ in terms of $t$.\\
17
\item Comment on the value of $v$ for the two models as $t$ becomes large.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2021 Q17 [11]}}