| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Standard +0.3 This is a standard M1 projectile motion question with projection from an elevated point. Part (a) is a 'show that' requiring horizontal motion equation (straightforward substitution). Parts (b) and (c) use standard SUVAT equations with given time. All techniques are routine for M1 students—no novel problem-solving or geometric insight required, just systematic application of projectile formulae. Slightly easier than average due to the guided structure and 'show that' reducing algebraic burden. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(57 = 24\cos 40° \times t\) | M1, A1 | Component attempted and acceleration \(= 0\); all correct |
| \(t = 3.10\) sec | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(h = 24\sin 40° \times 3.1 - \frac{1}{2} \times 9.8 \times 3.1^2\) | M1, A1 | Component attempted and acceleration \(= 9.8\); all correct |
| \(h = 0.734\) m | A1F | FT one slip e.g. \(+9.8\) used; accept 2 s.f. answer, AWRT \(0.71 - 0.74\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| horizontal: \(u = 24\cos 40° = 18.39 \text{ ms}^{-1}\) | B1 | Seen anywhere in (c), accept 18.4 |
| vertical: \(v = 24\sin 40° - 9.8 \times 3.1\) | M1 | Component attempted and acceleration \(= 9.8\) |
| \(v = -14.95 \text{ ms}^{-1}\) | A1 | (Accept \(-15.0\)) |
| \(V = \sqrt{(18.39)^2 + (-14.95)^2}\) | M1 | Use of candidate's \(u\) and new \(v\) (when \(t = 3.1\)) |
| \(V = 23.7 \text{ ms}^{-1}\) | A1F | FT use of candidate's \(u\) and \(v\) and new \(v\) when \(t = 3.1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\tan\theta = \frac{14.95}{18.39}\) | M1 | Use of candidate's \(u\) and \(v\); accept inverted ratio |
| \(\theta = 39.1°\) or \(39.2°\); also \(140.8°\) or \(140.9°\); accept \(\pm\) | A1F | FT use of candidates \(u\) and \(v\) and \(V\) |
## Question 7:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $57 = 24\cos 40° \times t$ | M1, A1 | Component attempted and acceleration $= 0$; all correct |
| $t = 3.10$ sec | A1 | CAO | **Total: 3** |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $h = 24\sin 40° \times 3.1 - \frac{1}{2} \times 9.8 \times 3.1^2$ | M1, A1 | Component attempted and acceleration $= 9.8$; all correct |
| $h = 0.734$ m | A1F | FT one slip e.g. $+9.8$ used; accept 2 s.f. answer, AWRT $0.71 - 0.74$ | **Total: 3** |
### Part (c)(i)
| Working | Marks | Guidance |
|---------|-------|----------|
| horizontal: $u = 24\cos 40° = 18.39 \text{ ms}^{-1}$ | B1 | Seen anywhere in (c), accept 18.4 |
| vertical: $v = 24\sin 40° - 9.8 \times 3.1$ | M1 | Component attempted and acceleration $= 9.8$ |
| $v = -14.95 \text{ ms}^{-1}$ | A1 | (Accept $-15.0$) |
| $V = \sqrt{(18.39)^2 + (-14.95)^2}$ | M1 | Use of candidate's $u$ and new $v$ (when $t = 3.1$) |
| $V = 23.7 \text{ ms}^{-1}$ | A1F | FT use of candidate's $u$ and $v$ and new $v$ when $t = 3.1$ | **Total: 5** |
### Part (c)(ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\tan\theta = \frac{14.95}{18.39}$ | M1 | Use of candidate's $u$ and $v$; accept inverted ratio |
| $\theta = 39.1°$ or $39.2°$; also $140.8°$ or $140.9°$; accept $\pm$ | A1F | FT use of candidates $u$ and $v$ and $V$ | **Total: 2** |
---7 A golf ball is struck from a point $O$ with velocity $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $40 ^ { \circ }$ to the horizontal. The ball first hits the ground at a point $P$, which is at a height $h$ metres above the level of $O$.\\
\includegraphics[max width=\textwidth, alt={}, center]{cfe0bdbc-35e3-485f-a922-b652a72f4c95-5_318_990_484_543}
The horizontal distance between $O$ and $P$ is 57 metres.
\begin{enumerate}[label=(\alph*)]
\item Show that the time that the ball takes to travel from $O$ to $P$ is 3.10 seconds, correct to three significant figures.
\item Find the value of $h$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the speed with which the ball hits the ground at $P$.
\item Find the angle between the direction of motion and the horizontal as the ball hits the ground at $P$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2006 Q7 [13]}}