| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Speed at specific time or position |
| Difficulty | Moderate -0.3 This is a standard M1 projectiles question requiring routine application of SUVAT equations with resolved components. Part (a) uses v²=u²+2as at maximum height, part (b) solves s=ut+½at² (straightforward quadratic), and part (c) finds speed from horizontal and vertical components. All steps are textbook exercises with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(0^2 = (28\sin 50°)^2 + 2\times(-9.8)s\) | M1 | Equation to find max height, with \(v=0\), \(u=28\sin 50°\) or \(u=28\cos 50°\) and \(-9.8\) or \(-g\) |
| Correct equation | A1 | |
| \(s = \frac{(28\sin 50°)^2}{2\times 9.8} = 23.5\) m | dM1 | Solving for the height |
| Correct height, awrt 23.5 | A1 | Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(0 = 28\sin 50° - 9.8t\) | (M1) | Equation to find time to max height, \(v=0\) |
| \(t = \frac{28\sin 50°}{9.8} = 2.1887\) | (A1) | Correct time |
| \(s = 28\sin 50° \times 2.1887 - 4.9\times 2.1887^2 = 23.5\) | (dM1) | Finding height with their time, \(u=28\sin 50°\) or \(u=28\cos 50°\) and \(-4.9\) or \(-g/2\) |
| Correct height, awrt 23.5 | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2 = 28\sin 50°\, t - 4.9t^2\) | M1 | Quadratic equation in \(t\) with \(\pm 2\), \(u=28\sin 50°\) or \(u=28\cos 50°\) and \(-4.9\) or \(-g/2\) |
| Correct terms | A1 | |
| Correct signs for equation | A1 | |
| \(0 = 4.9t^2 - 28\sin 50°\, t + 2\) | ||
| \(t = 0.0953\) or \(t = 4.282\) | dM1 | Solving the quadratic equation |
| \(t = 4.282 = 4.28\) s (to 3 sf) AG | A1 | Correct larger time selected from two values. Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(0 = 28\sin 50° - 9.8t \Rightarrow t = \frac{28\sin 50°}{9.8} = 2.1887\) | (M1)(A1) | Calculation of two times whose sum or difference gives time of flight; correct time for zero vertical velocity or maximum height |
| \(23.5 = 28\sin 50°\, t - 4.9t^2 \Rightarrow t = 2.1887\) OR \(21.5 = 4.9t^2\) | (dM1) | Correct expression for time to fall |
| \(t = \sqrt{\frac{21.5}{4.9}} = 2.0947\) | (A1) | Correct time |
| \(2.1887 + 2.0947 = 4.2834 = 4.28\) (to 3 sf) AG | (A1) | Accept 4.29 if their answer rounds to 4.29 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(v_x = 28\cos 50° = 18.00 \text{ ms}^{-1}\) | B1 | Horizontal component, need not be evaluated |
| \(v_y = 28\sin 50° - 9.8\times 4.282 = -20.51 \text{ ms}^{-1}\) | M1 | Equation for vertical component with \(28\sin 50°\) (or \(28\cos 50°\) if \(\sin 50°\) used for horizontal), \(-9.8\) and awrt 4.28 |
| Correct vertical component, awrt \(\pm 20.5\) | A1 | |
| \(v = \sqrt{18.00^2 + 20.51^2} = 27.3 \text{ ms}^{-1}\) | dM1 | Finding speed with \(+\) sign inside square root |
| Correct speed, awrt 27.3 | A1F | Intermediate values can be implied by final answer. Total: 5 |
| Section Total: 14 | Paper Total: 75 |
## Question 8:
### Part (a): Maximum Height
| Working | Mark | Guidance |
|---------|------|----------|
| $0^2 = (28\sin 50°)^2 + 2\times(-9.8)s$ | M1 | Equation to find max height, with $v=0$, $u=28\sin 50°$ or $u=28\cos 50°$ and $-9.8$ or $-g$ |
| Correct equation | A1 | |
| $s = \frac{(28\sin 50°)^2}{2\times 9.8} = 23.5$ m | dM1 | Solving for the height |
| Correct height, awrt 23.5 | A1 | **Total: 4** |
**OR (alternative method):**
| Working | Mark | Guidance |
|---------|------|----------|
| $0 = 28\sin 50° - 9.8t$ | (M1) | Equation to find time to max height, $v=0$ |
| $t = \frac{28\sin 50°}{9.8} = 2.1887$ | (A1) | Correct time |
| $s = 28\sin 50° \times 2.1887 - 4.9\times 2.1887^2 = 23.5$ | (dM1) | Finding height with their time, $u=28\sin 50°$ or $u=28\cos 50°$ and $-4.9$ or $-g/2$ |
| Correct height, awrt 23.5 | (A1) | |
---
### Part (b): Time of Flight
| Working | Mark | Guidance |
|---------|------|----------|
| $2 = 28\sin 50°\, t - 4.9t^2$ | M1 | Quadratic equation in $t$ with $\pm 2$, $u=28\sin 50°$ or $u=28\cos 50°$ and $-4.9$ or $-g/2$ |
| Correct terms | A1 | |
| Correct signs for equation | A1 | |
| $0 = 4.9t^2 - 28\sin 50°\, t + 2$ | | |
| $t = 0.0953$ or $t = 4.282$ | dM1 | Solving the quadratic equation |
| $t = 4.282 = 4.28$ s (to 3 sf) **AG** | A1 | Correct larger time selected from two values. **Total: 5** |
**OR (alternative method):**
| Working | Mark | Guidance |
|---------|------|----------|
| $0 = 28\sin 50° - 9.8t \Rightarrow t = \frac{28\sin 50°}{9.8} = 2.1887$ | (M1)(A1) | Calculation of two times whose sum or difference gives time of flight; correct time for zero vertical velocity or maximum height |
| $23.5 = 28\sin 50°\, t - 4.9t^2 \Rightarrow t = 2.1887$ OR $21.5 = 4.9t^2$ | (dM1) | Correct expression for time to fall |
| $t = \sqrt{\frac{21.5}{4.9}} = 2.0947$ | (A1) | Correct time |
| $2.1887 + 2.0947 = 4.2834 = 4.28$ (to 3 sf) **AG** | (A1) | Accept 4.29 if their answer rounds to 4.29 |
---
### Part (c): Speed at Landing
| Working | Mark | Guidance |
|---------|------|----------|
| $v_x = 28\cos 50° = 18.00 \text{ ms}^{-1}$ | B1 | Horizontal component, need not be evaluated |
| $v_y = 28\sin 50° - 9.8\times 4.282 = -20.51 \text{ ms}^{-1}$ | M1 | Equation for vertical component with $28\sin 50°$ (or $28\cos 50°$ if $\sin 50°$ used for horizontal), $-9.8$ and awrt 4.28 |
| Correct vertical component, awrt $\pm 20.5$ | A1 | |
| $v = \sqrt{18.00^2 + 20.51^2} = 27.3 \text{ ms}^{-1}$ | dM1 | Finding speed with $+$ sign inside square root |
| Correct speed, awrt 27.3 | A1F | Intermediate values can be implied by final answer. **Total: 5** |
**Section Total: 14 | Paper Total: 75**8 A cricket ball is hit at ground level on a horizontal surface. It initially moves at $28 \mathrm {~ms} ^ { - 1 }$ at an angle of $50 ^ { \circ }$ above the horizontal.
\begin{enumerate}[label=(\alph*)]
\item Find the maximum height of the ball during its flight.
\item The ball is caught when it is at a height of 2 metres above ground level, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{8c6f9ac0-c24f-48d0-9fb2-883651e791d7-5_332_1070_1601_477}
Show that the time that it takes for the ball to travel from the point where it was hit to the point where it was caught is 4.28 seconds, correct to three significant figures.
\item Find the speed of the ball when it is caught.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2009 Q8 [14]}}