| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.3 This is a straightforward projectile motion problem requiring standard equations with given time and horizontal distance. Part (a) uses horizontal motion (constant velocity), part (b) uses vertical motion with SUVAT, and part (c) is recall. The small angle and realistic context add minor complexity, but the solution path is direct and algorithmic—slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Horizontal: \(V\cos2° \times 1.8 = 420\) | M1 | Horizontal equation with \(t = 1.8\) |
| \(V = \dfrac{420}{1.8\cos2°}\) | A1 | Correct expression |
| \(V \approx 233 \text{ m s}^{-1}\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Vertical displacement: \(y = V\sin2° \times 1.8 - \frac{1}{2}(9.8)(1.8)^2\) | M1 | Using \(s = ut + \frac{1}{2}at^2\) vertically |
| \(y = 233\sin2° \times 1.8 - 4.9(3.24)\) | A1 | Correct substitution |
| \(y \approx 14.64 - 15.876 \approx -1.24\) m | A1 | |
| Distance \(AC = | y | \approx 1.24\) m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| No air resistance (or gravity only acts on bullet) | B1 | Accept equivalent statements |
## Question 6:
**(a)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Horizontal: $V\cos2° \times 1.8 = 420$ | M1 | Horizontal equation with $t = 1.8$ |
| $V = \dfrac{420}{1.8\cos2°}$ | A1 | Correct expression |
| $V \approx 233 \text{ m s}^{-1}$ | A1 | Correct value |
**(b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Vertical displacement: $y = V\sin2° \times 1.8 - \frac{1}{2}(9.8)(1.8)^2$ | M1 | Using $s = ut + \frac{1}{2}at^2$ vertically |
| $y = 233\sin2° \times 1.8 - 4.9(3.24)$ | A1 | Correct substitution |
| $y \approx 14.64 - 15.876 \approx -1.24$ m | A1 | |
| Distance $AC = |y| \approx 1.24$ m | A1 | |
**(c)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| No air resistance (or gravity only acts on bullet) | B1 | Accept equivalent statements |
---6 A bullet is fired from a rifle at a target, which is at a distance of 420 metres from the rifle. The bullet leaves the rifle travelling at $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and at an angle of $2 ^ { \circ }$ above the horizontal. The centre of the target, $C$, is at the same horizontal level as the rifle. The bullet hits the target at the point $A$, which is on a vertical line through $C$. The bullet takes 1.8 seconds to reach the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Find $V$, showing clearly how you obtain your answer.
\item Find the distance between $A$ and $C$.
\item State one assumption that you have made about the forces acting on the bullet.\\[0pt]
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2014 Q6 [8]}}