| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.8 This is a straightforward projectiles question requiring standard SUVAT equations to derive a given result, then basic optimization using the range of sin²θ, and finally conceptual understanding that mass/size doesn't affect projectile motion. All parts are routine applications with no problem-solving insight required, making it easier than average. |
| Spec | 3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u = 7\sin\theta\) for vertical component of initial velocity | B1 (1.1b) | |
| Uses \(v^2 = u^2 + 2as\) with \(v=0\); \(0 = 49\sin^2\theta - 19.6h\) | M1 (3.3) | Or uses appropriate constant acceleration equations forming complete method to obtain \(h\) |
| \(h = 2.5\sin^2\theta\) | R1 (2.1) | Completes argument substituting \(s=h\), \(v=0\), \(u=7\sin\theta\), \(a=-9.8\); accept negative values used consistently; AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\theta = 60°\), substituted into formula | M1 (1.1a) | Substitutes any value of \(\theta\) in range \(0 < \theta \leq 60\) to obtain height greater than zero |
| \(h = 2.5\sin^2 60 = 1.9\) | A1 (2.2a) | AWRT 1.9 or \(\dfrac{15}{8}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Nisha is incorrect; the model ignores air resistance which gets greater as ball gets larger | E1 (3.5a) | Must state incorrect AND refer to air resistance or ball modelled as a particle |
## Question 13(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = 7\sin\theta$ for vertical component of initial velocity | B1 (1.1b) | |
| Uses $v^2 = u^2 + 2as$ with $v=0$; $0 = 49\sin^2\theta - 19.6h$ | M1 (3.3) | Or uses appropriate constant acceleration equations forming complete method to obtain $h$ |
| $h = 2.5\sin^2\theta$ | R1 (2.1) | Completes argument substituting $s=h$, $v=0$, $u=7\sin\theta$, $a=-9.8$; accept negative values used consistently; AG |
## Question 13(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\theta = 60°$, substituted into formula | M1 (1.1a) | Substitutes any value of $\theta$ in range $0 < \theta \leq 60$ to obtain height greater than zero |
| $h = 2.5\sin^2 60 = 1.9$ | A1 (2.2a) | AWRT 1.9 or $\dfrac{15}{8}$ |
## Question 13(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Nisha is incorrect; the model ignores air resistance which gets greater as ball gets larger | E1 (3.5a) | Must state incorrect AND refer to air resistance or ball modelled as a particle |13
\begin{enumerate}[label=(\alph*)]
\item Show that
$$h = 2.5 \sin ^ { 2 } \theta$$
13 In this question use $g = 9.8 \mathrm {~ms} ^ { - 2 }$
13
\item Hence, given that $0 ^ { \circ } \leq \theta \leq 60 ^ { \circ }$, find the maximum value of $h$.\\
13
\item Nisha claims that the larger the size of the ball, the greater the maximum vertical height will be.
State whether Nisha is correct, giving a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2022 Q13 [6]}}