| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.3 This is a standard M3 projectile question requiring derivation of trajectory equation (routine using parametric elimination), substitution into a quadratic in tan α, application of discriminant condition (b²≥4ac), and solving for minimum u. All steps are textbook-standard with clear signposting, making it slightly easier than average despite being multi-part. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown1.02g Inequalities: linear and quadratic in single variable1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=13.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y = ut\sin\alpha - \frac{1}{2}gt^2\) | M1 A1 | |
| \(x = ut\cos\alpha\) | M1 | |
| \(t = \frac{x}{u\cos\alpha}\) | A1 | |
| \(y = u\left(\frac{x}{u\cos\alpha}\right)\sin\alpha - \frac{1}{2}g\left(\frac{x}{u\cos\alpha}\right)^2\) | M1 | |
| \(y = x\tan\alpha - \frac{gx^2}{u^2\cos^2\alpha}\) | ||
| \(y = x\tan\alpha - \frac{gx^2}{2u^2}(1+\tan^2\alpha)\) | A1 | 6 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(1 = R\tan\alpha - \frac{10R^2}{2u^2}(1+\tan^2\alpha)\) | M1 | |
| \(5R^2\tan^2\alpha - u^2R\tan\alpha + 5R^2 + u^2 = 0\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| For real solutions of the quadratic: \(u^4R^2 - 20R^2(5R^2 + u^2) \geq 0\) | M1 | |
| \(R^2 \leq \frac{u^4 - 20u^2}{100}\) | ||
| \(R^2 \leq \frac{u^2(u^2-20)}{100}\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(5^2 \leq \frac{u^2(u^2-20)}{100}\) | ||
| \(u^4 - 20u^2 - 2500 \geq 0\) | M1 | Condone equation |
| \(u_{\min}^2 = 61.0\) (or \(10 + \sqrt{2600}\)) | A1 | |
| \(u_{\min} = 7.81\) | A1F | 3 marks |
# Question 5:
## Part 5(a):
| Working | Marks | Guidance |
|---------|-------|---------|
| $y = ut\sin\alpha - \frac{1}{2}gt^2$ | M1 A1 | |
| $x = ut\cos\alpha$ | M1 | |
| $t = \frac{x}{u\cos\alpha}$ | A1 | |
| $y = u\left(\frac{x}{u\cos\alpha}\right)\sin\alpha - \frac{1}{2}g\left(\frac{x}{u\cos\alpha}\right)^2$ | M1 | |
| $y = x\tan\alpha - \frac{gx^2}{u^2\cos^2\alpha}$ | | |
| $y = x\tan\alpha - \frac{gx^2}{2u^2}(1+\tan^2\alpha)$ | A1 | 6 marks | Answer given |
## Part 5(b)(i):
| Working | Marks | Guidance |
|---------|-------|---------|
| $1 = R\tan\alpha - \frac{10R^2}{2u^2}(1+\tan^2\alpha)$ | M1 | |
| $5R^2\tan^2\alpha - u^2R\tan\alpha + 5R^2 + u^2 = 0$ | A1 | 2 marks | Answer given |
## Part 5(b)(ii):
| Working | Marks | Guidance |
|---------|-------|---------|
| For real solutions of the quadratic: $u^4R^2 - 20R^2(5R^2 + u^2) \geq 0$ | M1 | |
| $R^2 \leq \frac{u^4 - 20u^2}{100}$ | | |
| $R^2 \leq \frac{u^2(u^2-20)}{100}$ | A1 | 2 marks | Answer given |
## Part 5(b)(iii):
| Working | Marks | Guidance |
|---------|-------|---------|
| $5^2 \leq \frac{u^2(u^2-20)}{100}$ | | |
| $u^4 - 20u^2 - 2500 \geq 0$ | M1 | Condone equation |
| $u_{\min}^2 = 61.0$ (or $10 + \sqrt{2600}$) | A1 | |
| $u_{\min} = 7.81$ | A1F | 3 marks | 3 sf required |
---5 A ball is projected with speed $u \mathrm {~ms} ^ { - 1 }$ at an angle of elevation $\alpha$ above the horizontal so as to hit a point $P$ on a wall. The ball travels in a vertical plane through the point of projection. During the motion, the horizontal and upward vertical displacements of the ball from the point of projection are $x$ metres and $y$ metres respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that, during the flight, the equation of the trajectory of the ball is given by
$$y = x \tan \alpha - \frac { g x ^ { 2 } } { 2 u ^ { 2 } } \left( 1 + \tan ^ { 2 } \alpha \right)$$
\item The ball is projected from a point 1 metre vertically below and $R$ metres horizontally from the point $P$.
\begin{enumerate}[label=(\roman*)]
\item By taking $g = 10 \mathrm {~ms} ^ { - 2 }$, show that $R$ satisfies the equation
$$5 R ^ { 2 } \tan ^ { 2 } \alpha - u ^ { 2 } R \tan \alpha + 5 R ^ { 2 } + u ^ { 2 } = 0$$
\item Hence, given that $u$ and $R$ are constants, show that, for $\tan \alpha$ to have real values, $R$ must satisfy the inequality
$$R ^ { 2 } \leqslant \frac { u ^ { 2 } \left( u ^ { 2 } - 20 \right) } { 100 }$$
\item Given that $R = 5$, determine the minimum possible speed of projection.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M3 2007 Q5 [13]}}