| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.8 This is a straightforward projectile motion question requiring only standard formula application: time of flight from vertical motion, range from horizontal motion, resultant velocity from components, and recognizing minimum speed occurs at maximum height. All parts follow directly from given information with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(s = ut + \frac{1}{2}at^2\) | ||
| \(0 = 2\frac{1}{2}ut - \frac{1}{2}gt^2\) | M1, A1 | full method required for time (equation of motion, or standard result) |
| \(0 = t\left(2\frac{1}{2}u - \frac{1}{2}gt\right)\) | m1 | |
| \(t = \frac{5u}{g}\) | A1 | Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(OA = 6u \times \frac{5u}{g}\) | M1 | |
| \(= \frac{30u^2}{g}\) | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\text{speed}^2 = (6u)^2 + \left(2\frac{1}{2}u\right)^2\) | M1 | |
| \(\text{speed} = 6\frac{1}{2}u\) | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Least speed, at top \(= 6u\) | B1 | Total: 1 |
## Question 5:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $s = ut + \frac{1}{2}at^2$ | | |
| $0 = 2\frac{1}{2}ut - \frac{1}{2}gt^2$ | M1, A1 | full method required for time (equation of motion, or standard result) |
| $0 = t\left(2\frac{1}{2}u - \frac{1}{2}gt\right)$ | m1 | |
| $t = \frac{5u}{g}$ | A1 | Total: 4 | if $g = 9.8$ used, lose last A1 |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $OA = 6u \times \frac{5u}{g}$ | M1 | |
| $= \frac{30u^2}{g}$ | A1 | Total: 2 | cao |
### Part (c)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\text{speed}^2 = (6u)^2 + \left(2\frac{1}{2}u\right)^2$ | M1 | |
| $\text{speed} = 6\frac{1}{2}u$ | A1 | Total: 2 | cao |
### Part (d)
| Working | Marks | Guidance |
|---------|-------|----------|
| Least speed, at top $= 6u$ | B1 | Total: 1 | |
---5 A golf ball is projected from a point $O$ with initial velocity $V$ at an angle $\alpha$ to the horizontal. The ball first hits the ground at a point $A$ which is at the same horizontal level as $O$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{c220e6c4-2676-4022-8301-7d720dc082b2-4_227_602_484_735}
It is given that $V \cos \alpha = 6 u$ and $V \sin \alpha = 2.5 u$.
\begin{enumerate}[label=(\alph*)]
\item Show that the time taken for the ball to travel from $O$ to $A$ is $\frac { 5 u } { g }$.
\item Find, in terms of $g$ and $u$, the distance $O A$.
\item Find $V$, in terms of $u$.
\item State, in terms of $u$, the least speed of the ball during its flight from $O$ to $A$.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2006 Q5 [9]}}