| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Free fall: time or distance |
| Difficulty | Moderate -0.3 Part (a) is a straightforward SUVAT application (v = u + at with given velocities). Part (b) requires recognizing that h must exceed the distance from M to N (using v² = u² + 2as), which is a simple inequality argument. Both parts are routine mechanics questions with clear methods and minimal problem-solving demand, making this slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = u + at\), with \(u=u\), \(v=3u\), \(a=g\) | M1 | Selects appropriate equation of constant acceleration to find \(t\) |
| \(3u = u + gt \Rightarrow \frac{3u-u}{g} = t \Rightarrow t = \frac{2u}{g}\) | R1 | Must clearly state \(u=u\), \(v=3u\), \(a=g\) and must see either \(3u=u+gt\) or \(\frac{3u-u}{g}=t\); AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(v^2 = u^2 + 2as\), with \(u=u\), \(v=3u\), \(a=g\) | M1 | Selects correct equation of constant acceleration; condone \(a=-g\) |
| \((3u)^2 = u^2 + 2gs \Rightarrow 9u^2 = u^2 + 2gs \Rightarrow 8u^2 = 2gs\) | A1 | Completes reasoned argument with at least one more intermediate step to obtain \(\frac{4u^2}{g}\) |
| \(MN = s = \frac{4u^2}{g}\); since \(N\) is above the ground then \(h > \frac{4u^2}{g}\) | R1 | Explains that \(MN\) has been found and \(N\) is not on the surface to justify \(h > \frac{4u^2}{g}\); AG |
## Question 13(a):
$v = u + at$, with $u=u$, $v=3u$, $a=g$ | M1 | Selects appropriate equation of constant acceleration to find $t$
$3u = u + gt \Rightarrow \frac{3u-u}{g} = t \Rightarrow t = \frac{2u}{g}$ | R1 | Must clearly state $u=u$, $v=3u$, $a=g$ and must see either $3u=u+gt$ or $\frac{3u-u}{g}=t$; AG
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## Question 13(b):
$v^2 = u^2 + 2as$, with $u=u$, $v=3u$, $a=g$ | M1 | Selects correct equation of constant acceleration; condone $a=-g$
$(3u)^2 = u^2 + 2gs \Rightarrow 9u^2 = u^2 + 2gs \Rightarrow 8u^2 = 2gs$ | A1 | Completes reasoned argument with at least one more intermediate step to obtain $\frac{4u^2}{g}$
$MN = s = \frac{4u^2}{g}$; since $N$ is above the ground then $h > \frac{4u^2}{g}$ | R1 | Explains that $MN$ has been found and $N$ is not on the surface to justify $h > \frac{4u^2}{g}$; AG
---13 A ball falls freely towards the Earth.\\
The ball passes through two different fixed points $M$ and $N$ before reaching the Earth's surface.
At $M$ the ball has velocity $u \mathrm {~ms} ^ { - 1 }$\\
At $N$ the ball has velocity $3 u \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
It can be assumed that:
\begin{itemize}
\item the motion is due to gravitational force only
\item the acceleration due to gravity remains constant throughout.
\end{itemize}
13
\begin{enumerate}[label=(\alph*)]
\item Show that the time taken for the ball to travel from $M$ to $N$ is $\frac { 2 u } { g }$ seconds.\\[0pt]
[2 marks]
13
\item Point $M$ is $h$ metres above the Earth.
Show that $h > \frac { 4 u ^ { 2 } } { g }$\\
Fully justify your answer.\\
The car is moving in a straight line.\\
The acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ of the car at time $t$ seconds is given by
$$a = 3 k t ^ { 2 } - 2 k t + 1$$
where $k$ is a constant.\\
When $t = 3$ the car has a velocity of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
Show that $k = \frac { 1 } { 3 }$
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2023 Q13 [5]}}