| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.8 This M3 projectile question requires deriving a quadratic in tan α from trajectory equations, then using discriminant conditions for real solutions—a multi-step problem involving algebraic manipulation and inequality reasoning beyond standard trajectory calculations, but follows established techniques for this module level. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05a Sine, cosine, tangent: definitions for all arguments3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Horizontal: \(2k = ut\cos\alpha\) so \(t = \frac{2k}{u\cos\alpha}\) | B1 | |
| Vertical: \(k = ut\sin\alpha - \frac{1}{2}(10)t^2\) | M1 | Using correct equation of motion vertically |
| Substituting \(t\): \(k = u\sin\alpha \cdot \frac{2k}{u\cos\alpha} - 5\left(\frac{2k}{u\cos\alpha}\right)^2\) | M1 A1 | Substituting expression for \(t\) |
| \(k = 2k\tan\alpha - \frac{20k^2}{u^2\cos^2\alpha}\) | A1 | |
| Using \(\sec^2\alpha = 1 + \tan^2\alpha\): \(k = 2k\tan\alpha - \frac{20k^2(1+\tan^2\alpha)}{u^2}\) | M1 | Using \(\frac{1}{\cos^2\alpha} = 1+\tan^2\alpha\) |
| \(u^2k = 2ku^2\tan\alpha - 20k^2(1+\tan^2\alpha)\) | ||
| \(20k\tan^2\alpha - 2u^2\tan\alpha + u^2 + 20k = 0\) | A1 | Fully correct equation shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| For real solutions in \(\tan\alpha\), discriminant \(\geq 0\) | M1 | |
| \((2u^2)^2 - 4(20k)(u^2+20k) \geq 0\) | M1 | Correct use of \(b^2 - 4ac \geq 0\) |
| \(4u^4 - 80ku^2 - 1600k^2 \geq 0\) | ||
| \(u^4 - 20ku^2 - 400k^2 \geq 0\) | A1 | Completion to given result |
# Question 3:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Horizontal: $2k = ut\cos\alpha$ so $t = \frac{2k}{u\cos\alpha}$ | B1 | |
| Vertical: $k = ut\sin\alpha - \frac{1}{2}(10)t^2$ | M1 | Using correct equation of motion vertically |
| Substituting $t$: $k = u\sin\alpha \cdot \frac{2k}{u\cos\alpha} - 5\left(\frac{2k}{u\cos\alpha}\right)^2$ | M1 A1 | Substituting expression for $t$ |
| $k = 2k\tan\alpha - \frac{20k^2}{u^2\cos^2\alpha}$ | A1 | |
| Using $\sec^2\alpha = 1 + \tan^2\alpha$: $k = 2k\tan\alpha - \frac{20k^2(1+\tan^2\alpha)}{u^2}$ | M1 | Using $\frac{1}{\cos^2\alpha} = 1+\tan^2\alpha$ |
| $u^2k = 2ku^2\tan\alpha - 20k^2(1+\tan^2\alpha)$ | | |
| $20k\tan^2\alpha - 2u^2\tan\alpha + u^2 + 20k = 0$ | A1 | Fully correct equation shown |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| For real solutions in $\tan\alpha$, discriminant $\geq 0$ | M1 | |
| $(2u^2)^2 - 4(20k)(u^2+20k) \geq 0$ | M1 | Correct use of $b^2 - 4ac \geq 0$ |
| $4u^4 - 80ku^2 - 1600k^2 \geq 0$ | | |
| $u^4 - 20ku^2 - 400k^2 \geq 0$ | A1 | Completion to given result |
I can see these are answer space pages (pages 7, 9, 10, 11) and a question page (page 8) from what appears to be an AQA Mechanics exam paper. However, these images show only the **question text and blank answer spaces** — they do not contain any mark scheme content.
To provide the mark scheme breakdown you're asking for, I would need images of the actual mark scheme document, which is a separate publication.
If you can share the mark scheme pages, I'd be happy to extract and format the content as requested.
Alternatively, I can help by **working through the solutions** to Question 4 based on the question text shown:
---3 (In this question, take $g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.)\\
A projectile is fired from a point $O$ with speed $u$ at an angle of elevation $\alpha$ above the horizontal so as to pass through a point $P$. The projectile travels in a vertical plane through $O$ and $P$. The point $P$ is at a horizontal distance $2 k$ from $O$ and at a vertical distance $k$ above $O$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ satisfies the equation
$$20 k \tan ^ { 2 } \alpha - 2 u ^ { 2 } \tan \alpha + u ^ { 2 } + 20 k = 0$$
\item Deduce that
$$u ^ { 4 } - 20 k u ^ { 2 } - 400 k ^ { 2 } \geqslant 0$$
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2012 Q3 [10]}}