| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Modelling assumptions and limitations |
| Difficulty | Challenging +1.2 This is a multi-part projectiles question requiring range equations and double angle identities. Part (a) involves equating ranges and algebraic manipulation to reach a given result (standard 'show that'). Parts (b) and (c) are straightforward applications. While it requires multiple steps and combines mechanics with trigonometry, the techniques are standard A-level fare with no novel insight required—moderately above average difficulty. |
| Spec | 1.05l Double angle formulae: and compound angle formulae3.02i Projectile motion: constant acceleration model |
| 19 |
|
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0 = t_P u\sin\theta - \dfrac{1}{2}gt_P^2\) | B1 (AO 1.1b) | States or uses appropriate component of either horizontal or vertical velocity |
| \(t_P = \dfrac{2u\sin\theta}{g}\) | M1 (AO 3.3) | Considers vertical motion and uses constant acceleration equation for one particle |
| \(t_Q = \dfrac{4u\sin 2\theta}{g}\) | A1 (AO 3.4) | Correct expressions for \(t_P\) and \(t_Q\); PI by correct equation for ranges |
| \(t_P u\cos\theta = t_Q \times 2u\cos 2\theta\) | B1 (AO 3.3) | States horizontal distance \(= ut\cos\theta\) or \(2ut\cos 2\theta\) |
| \(\dfrac{2u^2\sin\theta\cos\theta}{g} = \dfrac{8u^2\sin 2\theta\cos 2\theta}{g}\) | M1 (AO 3.4) | Equates expressions for ranges and substitutes \(t_P\) and \(t_Q\) |
| \(2\sin\theta\cos\theta = 8\sin 2\theta\cos 2\theta\), then using \(\sin 2\theta = 2\sin\theta\cos\theta\) gives \(\dfrac{1}{8} = \cos 2\theta\) | R1 (AO 2.1) | Completes reasoned argument by simplifying to equation in \(\theta\) only. AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cos\theta = \dfrac{3}{4}\) | B1 (AO 1.1b) | Obtains exact value for \(\cos\theta\), or \(\theta = 41.4°\) AWRT; PI by obtaining \(t = 1.2\) |
| \(t_P\cos\theta = t_Q \times 2\cos 2\theta\); \(\cos\theta = \tfrac{3}{4}\), \(\cos 2\theta = \tfrac{1}{8}\), \(t_P = 0.4\) so \(0.4 \times \tfrac{3}{4} = t_Q \times 2 \times \tfrac{1}{8}\) | M1 (AO 3.1b) | Uses \(t_P\cos\theta = t_Q \times 2\cos 2\theta\) and substitutes \(t_P = 0.4\) and \(\theta\) value |
| Completes substitution to find \(t_Q\) | M1 (AO 3.4) | Completes substitution in \(t_P u\cos\theta = t_Q \times 2u\cos 2\theta\) or uses \(t_P\cos\theta = t_Q \times 2\cos 2\theta\) |
| \(t_Q = 1.2\) seconds | A1 (AO 1.1b) | AWRT 1.2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X\) and \(Y\) are at the same height | E1 (AO 3.5b) | States any suitable assumption, e.g. acceleration is constant, \(X\) and \(Y\) at same height |
## Question 18(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0 = t_P u\sin\theta - \dfrac{1}{2}gt_P^2$ | B1 (AO 1.1b) | States or uses appropriate component of either horizontal or vertical velocity |
| $t_P = \dfrac{2u\sin\theta}{g}$ | M1 (AO 3.3) | Considers vertical motion and uses constant acceleration equation for one particle |
| $t_Q = \dfrac{4u\sin 2\theta}{g}$ | A1 (AO 3.4) | Correct expressions for $t_P$ and $t_Q$; PI by correct equation for ranges |
| $t_P u\cos\theta = t_Q \times 2u\cos 2\theta$ | B1 (AO 3.3) | States horizontal distance $= ut\cos\theta$ or $2ut\cos 2\theta$ |
| $\dfrac{2u^2\sin\theta\cos\theta}{g} = \dfrac{8u^2\sin 2\theta\cos 2\theta}{g}$ | M1 (AO 3.4) | Equates expressions for ranges and substitutes $t_P$ and $t_Q$ |
| $2\sin\theta\cos\theta = 8\sin 2\theta\cos 2\theta$, then using $\sin 2\theta = 2\sin\theta\cos\theta$ gives $\dfrac{1}{8} = \cos 2\theta$ | R1 (AO 2.1) | Completes reasoned argument by simplifying to equation in $\theta$ only. AG |
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## Question 18(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos\theta = \dfrac{3}{4}$ | B1 (AO 1.1b) | Obtains exact value for $\cos\theta$, or $\theta = 41.4°$ AWRT; PI by obtaining $t = 1.2$ |
| $t_P\cos\theta = t_Q \times 2\cos 2\theta$; $\cos\theta = \tfrac{3}{4}$, $\cos 2\theta = \tfrac{1}{8}$, $t_P = 0.4$ so $0.4 \times \tfrac{3}{4} = t_Q \times 2 \times \tfrac{1}{8}$ | M1 (AO 3.1b) | Uses $t_P\cos\theta = t_Q \times 2\cos 2\theta$ and substitutes $t_P = 0.4$ and $\theta$ value |
| Completes substitution to find $t_Q$ | M1 (AO 3.4) | Completes substitution in $t_P u\cos\theta = t_Q \times 2u\cos 2\theta$ or uses $t_P\cos\theta = t_Q \times 2\cos 2\theta$ |
| $t_Q = 1.2$ seconds | A1 (AO 1.1b) | AWRT 1.2 |
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## Question 18(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X$ and $Y$ are at the same height | E1 (AO 3.5b) | States any suitable assumption, e.g. acceleration is constant, $X$ and $Y$ at same height |
---18 Two particles, $P$ and $Q$, are projected at the same time from a fixed point $X$, on the ground, so that they travel in the same vertical plane.\\
$P$ is projected at an acute angle $\theta ^ { \circ }$ to the horizontal, with speed $u \mathrm {~ms} ^ { - 1 }$\\
$Q$ is projected at an acute angle $2 \theta ^ { \circ }$ to the horizontal, with speed $2 u \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
Both particles land back on the ground at exactly the same point, $Y$.\\
Resistance forces to motion may be ignored.\\
18
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\cos 2 \theta = \frac { 1 } { 8 }$$
18
\item $\quad P$ takes a total of 0.4 seconds to travel from $X$ to $Y$.\\
Find the time taken by $Q$ to travel from $X$ to $Y$.\\
18
\item State one modelling assumption you have chosen to make in this question.\\[0pt]
[1 mark]\\
\begin{center}
\begin{tabular}{|l|l|}
\hline
19 & \begin{tabular}{l}
Two skaters, Jo and Amba, are separately skating across a smooth, horizontal surface of ice. \\
Both are moving in the same direction, so that their paths are straight and are parallel to each other. \\
Jo is moving with constant velocity $( 2.8 \mathbf { i } + 9.6 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ \\
At time $t = 0$ seconds Amba is at position ( $2 \mathbf { i } - 7 \mathbf { j }$ ) metres and is moving with a constant speed of $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ \\
Explain why Amba's velocity must be in the form $k ( 2.8 \mathbf { i } + 9.6 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$, where $k$ is a constant. \\[0pt]
[1 mark] $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ \\
\end{tabular}\\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2021 Q18 [11]}}