| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Standard +0.3 This is a standard M3 projectiles question requiring derivation of the trajectory equation from parametric equations (routine elimination of t), then solving a quadratic inequality and converting back to time. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(x = 2t\) | M1 | |
| \(y = -\frac{1}{2}gt^2 + 10t\) | M1 | |
| \(t = \frac{x}{2}\) | ||
| \(y = -\frac{1}{2}g\left(\frac{x}{2}\right)^2 + 10\left(\frac{x}{2}\right)\) | m1 | |
| \(y = -\frac{g}{8}x^2 + 5x\) | A1 | AG |
| Subtotal | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(1 = -\frac{g}{8}x^2 + 5x\) | M1 | |
| \(gx^2 - 40x + 8 = 0\) | ||
| \(x = \frac{40 \pm \sqrt{(-40)^2 - 4\times 8g}}{2g}\) | M1 | |
| \(x = 3.871,\ 0.211\) | A1 | A1 for both answers |
| Distance \(= 3.66\) m | A1 | |
| Subtotal | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(t = \frac{3.66}{2}\) | M1 | |
| \(t = 1.83\) sec | A1 | |
| Subtotal | 2 | |
| Total | 10 |
## Question 2:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $x = 2t$ | M1 | |
| $y = -\frac{1}{2}gt^2 + 10t$ | M1 | |
| $t = \frac{x}{2}$ | | |
| $y = -\frac{1}{2}g\left(\frac{x}{2}\right)^2 + 10\left(\frac{x}{2}\right)$ | m1 | |
| $y = -\frac{g}{8}x^2 + 5x$ | A1 | AG |
| **Subtotal** | **4** | |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $1 = -\frac{g}{8}x^2 + 5x$ | M1 | |
| $gx^2 - 40x + 8 = 0$ | | |
| $x = \frac{40 \pm \sqrt{(-40)^2 - 4\times 8g}}{2g}$ | M1 | |
| $x = 3.871,\ 0.211$ | A1 | A1 for both answers |
| Distance $= 3.66$ m | A1 | |
| **Subtotal** | **4** | |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $t = \frac{3.66}{2}$ | M1 | |
| $t = 1.83$ sec | A1 | |
| **Subtotal** | **2** | |
| **Total** | **10** | |
---2 A particle is projected from a point $O$ on a horizontal plane and has initial velocity components of $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $10 \mathrm {~ms} ^ { - 1 }$ parallel to and perpendicular to the plane respectively. At time $t$ seconds after projection, the horizontal and upward vertical distances of the particle from the point $O$ are $x$ metres and $y$ metres respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that $x$ and $y$ satisfy the equation
$$y = - \frac { g } { 8 } x ^ { 2 } + 5 x$$
\item By using the equation in part (a), find the horizontal distance travelled by the particle whilst it is more than 1 metre above the plane.
\item Hence find the time for which the particle is more than 1 metre above the plane.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2009 Q2 [10]}}