| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.8 This is a standard M1 projectile question requiring routine application of SUVAT equations and projectile formulas. Part (a) is a 'show that' using v=u+at at maximum height, part (b) applies standard maximum height and range formulas, and part (c) asks for standard modeling assumptions. All steps are textbook exercises with no problem-solving insight required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Vertical component of velocity: \(u_y = 14.7\sin\alpha\) | B1 | |
| At max height, vertical velocity \(= 0\): \(0 = 14.7\sin\alpha - 9.8t\) | M1 | Using \(v = u + at\) vertically |
| \(t = \frac{14.7\sin\alpha}{9.8} = \frac{3\sin\alpha}{2}\) | A1 | Shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(s = u_y t - \frac{1}{2}gt^2\) at max height | M1 | |
| \(7 = 14.7\sin\alpha \cdot \frac{3\sin\alpha}{2} - \frac{1}{2}(9.8)\left(\frac{3\sin\alpha}{2}\right)^2\) | M1 | Substituting time |
| \(7 = \frac{44.1\sin^2\alpha}{2} - \frac{4.9 \times 9\sin^2\alpha}{4}\) | M1 | Expanding |
| \(7 = 22.05\sin^2\alpha - 11.025\sin^2\alpha = 11.025\sin^2\alpha\) | A1 | |
| \(\sin^2\alpha = \frac{7}{11.025}\), \(\sin\alpha = \frac{2\sqrt{7}}{14.7} \cdot ...\), \(\alpha = 41.8°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Time of flight \(= 2 \times \frac{3\sin\alpha}{2} = 3\sin\alpha\) | M1 | By symmetry |
| Horizontal velocity \(= 14.7\cos\alpha\) | M1 | |
| Range \(= 14.7\cos\alpha \times 3\sin\alpha = 44.1\sin\alpha\cos\alpha\) | A1 | Evaluating with found \(\alpha\) gives \(OA \approx 29.4\) m |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| No air resistance | B1 | |
| Ball treated as a particle | B1 |
# Question 8:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Vertical component of velocity: $u_y = 14.7\sin\alpha$ | B1 | |
| At max height, vertical velocity $= 0$: $0 = 14.7\sin\alpha - 9.8t$ | M1 | Using $v = u + at$ vertically |
| $t = \frac{14.7\sin\alpha}{9.8} = \frac{3\sin\alpha}{2}$ | A1 | Shown |
## Part (b)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $s = u_y t - \frac{1}{2}gt^2$ at max height | M1 | |
| $7 = 14.7\sin\alpha \cdot \frac{3\sin\alpha}{2} - \frac{1}{2}(9.8)\left(\frac{3\sin\alpha}{2}\right)^2$ | M1 | Substituting time |
| $7 = \frac{44.1\sin^2\alpha}{2} - \frac{4.9 \times 9\sin^2\alpha}{4}$ | M1 | Expanding |
| $7 = 22.05\sin^2\alpha - 11.025\sin^2\alpha = 11.025\sin^2\alpha$ | A1 | |
| $\sin^2\alpha = \frac{7}{11.025}$, $\sin\alpha = \frac{2\sqrt{7}}{14.7} \cdot ...$, $\alpha = 41.8°$ | A1 | |
## Part (b)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| Time of flight $= 2 \times \frac{3\sin\alpha}{2} = 3\sin\alpha$ | M1 | By symmetry |
| Horizontal velocity $= 14.7\cos\alpha$ | M1 | |
| Range $= 14.7\cos\alpha \times 3\sin\alpha = 44.1\sin\alpha\cos\alpha$ | A1 | Evaluating with found $\alpha$ gives $OA \approx 29.4$ m |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| No air resistance | B1 | |
| Ball treated as a particle | B1 | |
The images you've shared show only blank answer/continuation pages (pages 17-20) from what appears to be an AQA mathematics exam paper (P27841/Jun10/MM1B). These pages contain:
- Page 17: Blank lined continuation sheet with "END OF QUESTIONS"
- Pages 18-20: Blank pages stating "There are no questions printed on this page" and "DO NOT WRITE ON THIS PAGE / ANSWER IN THE SPACES PROVIDED"
**These pages do not contain any mark scheme content.** They are blank answer pages from the question paper itself, not a mark scheme document.
To extract mark scheme content, you would need to share the actual **mark scheme document** for this paper, which is a separate publication. Could you share those pages instead?8 A ball is struck so that it leaves a horizontal surface travelling at $14.7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\alpha$ above the horizontal. The path of the ball is shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-16_293_1364_461_347}
\begin{enumerate}[label=(\alph*)]
\item Show that the ball takes $\frac { 3 \sin \alpha } { 2 }$ seconds to reach its maximum height.
\item The ball reaches a maximum height of 7 metres.
\begin{enumerate}[label=(\roman*)]
\item Find $\alpha$.
\item Find the range, $O A$.
\end{enumerate}\item State two assumptions that you needed to make in order to answer the earlier parts of this question.
\includegraphics[max width=\textwidth, alt={}, center]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-17_2347_1691_223_153}\\
\includegraphics[max width=\textwidth, alt={}, center]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-18_2488_1719_219_150}\\
\includegraphics[max width=\textwidth, alt={}, center]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-19_2488_1719_219_150}\\
\includegraphics[max width=\textwidth, alt={}, center]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-20_2505_1734_212_138}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2010 Q8 [13]}}