| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 10 |
| 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 standard SUVAT equations with vertical components. Part (a) uses v²=u²+2as at maximum height (v=0), and part (b) solves s=ut+½at² for time. Both are direct applications of memorized formulas with no problem-solving insight needed, making it easier than average but not trivial due to the multi-step nature and angle resolution 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 |
| \(0^2 = (50\sin 40°)^2 + 2 \times (-9.8)h\) | M1A1 | Equation for \(h\) with \(v=0\) and a component of velocity. Correct equation |
| \(h = \frac{(50\sin 40°)^2}{2 \times 9.8} = 52.7\) | dM1, A1 | Solving for \(h\); Correct \(h\) |
| Alt: \(0 = 50\sin 40° - 9.8t\) | (M1) | Equation for \(t\) with \(v=0\) and a component of velocity |
| \(t = \frac{50\sin 40°}{9.8} = 3.280\) | (A1) | Correct \(t\) |
| \(h = 50\sin 40° \times 3.280 - \frac{1}{2} \times 9.8 \times 3.280^2\) | (dM1) | Expression for \(h\) with a component of velocity |
| \(= 52.7\) (ALLOW 52.6) | (A1) | Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(6 = 50\sin 40°\,t - 4.9t^2\) | M1A1 | Forming a quadratic in \(t\). Correct terms with any signs |
| \(0 = 4.9t^2 - 50\sin 40°\,t + 6\) | A1 | Correct equation |
| \(t = \frac{50\sin 40° \pm \sqrt{(50\sin 40°)^2 - 4 \times 4.9 \times 6}}{2 \times 4.9}\) | dM1 | Solving quadratic |
| \(= 0.192\) or \(6.37\) | ||
| \(t = 6.37\) | A2 | Total: 6 |
| Alt: \(46.7 = 4.9t_1^2\) | (M1) | Finding two times |
| (dM1) | Equation for time to go down | |
| \(t_1 = 3.087\) | (A1) | Correct time |
| \(t_2 = 3.280\) | (A1) | Time to go up |
| \(t = 3.087 + 3.280 = 6.37\) | (A2) | Correct total |
## Question 7:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $0^2 = (50\sin 40°)^2 + 2 \times (-9.8)h$ | M1A1 | Equation for $h$ with $v=0$ and a component of velocity. Correct equation |
| $h = \frac{(50\sin 40°)^2}{2 \times 9.8} = 52.7$ | dM1, A1 | Solving for $h$; Correct $h$ |
| **Alt:** $0 = 50\sin 40° - 9.8t$ | (M1) | Equation for $t$ with $v=0$ and a component of velocity |
| $t = \frac{50\sin 40°}{9.8} = 3.280$ | (A1) | Correct $t$ |
| $h = 50\sin 40° \times 3.280 - \frac{1}{2} \times 9.8 \times 3.280^2$ | (dM1) | Expression for $h$ with a component of velocity |
| $= 52.7$ (ALLOW 52.6) | (A1) | **Total: 4** | Correct $h$ |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $6 = 50\sin 40°\,t - 4.9t^2$ | M1A1 | Forming a quadratic in $t$. Correct terms with any signs |
| $0 = 4.9t^2 - 50\sin 40°\,t + 6$ | A1 | Correct equation |
| $t = \frac{50\sin 40° \pm \sqrt{(50\sin 40°)^2 - 4 \times 4.9 \times 6}}{2 \times 4.9}$ | dM1 | Solving quadratic |
| $= 0.192$ or $6.37$ | | |
| $t = 6.37$ | A2 | **Total: 6** | Correct solution selected |
| **Alt:** $46.7 = 4.9t_1^2$ | (M1) | Finding two times |
| | (dM1) | Equation for time to go down |
| $t_1 = 3.087$ | (A1) | Correct time |
| $t_2 = 3.280$ | (A1) | Time to go up |
| $t = 3.087 + 3.280 = 6.37$ | (A2) | Correct total |
---7 A golf ball is struck from a point on horizontal ground so that it has an initial velocity of $50 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $40 ^ { \circ }$ above the horizontal.
Assume that the golf ball is a particle and its weight is the only force that acts on it once it is moving.
\begin{enumerate}[label=(\alph*)]
\item Find the maximum height of the golf ball.
\item After it has reached its maximum height, the golf ball descends but hits a tree at a point which is at a height of 6 metres above ground level.\\
\includegraphics[max width=\textwidth, alt={}, center]{965a176a-848c-478d-a748-80fc9dfe4399-5_289_1358_813_335}
\begin{displayquote}
Find the time that it takes for the ball to travel from the point where it was struck to the tree.
\end{displayquote}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2007 Q7 [10]}}