| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle above horizontal |
| Difficulty | Moderate -0.8 This is a standard M1 projectile motion question with straightforward application of SUVAT equations. All parts follow routine procedures: using vertical motion equation for time, horizontal distance = velocity × time, finding speed from components, and angle from tan^(-1). The 'show that' in part (a) removes problem-solving difficulty. Easier than average A-level maths due to being purely procedural with no conceptual challenges. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Vertical: \(u_y = 12\sin 30° = 6\) m s\(^{-1}\) | B1 | Correct vertical component |
| Using \(s = u_y t - \frac{1}{2}gt^2\) with \(s = 1 - 1.5 = -0.5\) m | M1 | Correct use of \(s = ut + \frac{1}{2}at^2\) vertically, with correct displacement |
| \(-0.5 = 6t - \frac{1}{2}(9.8)t^2\) | A1 | Correct equation |
| \(4.9t^2 - 6t - 0.5 = 0\) | M1 | Forming correct quadratic and attempting to solve |
| \(t = \frac{6 \pm \sqrt{36 + 4(4.9)(0.5)}}{9.8}\) | ||
| \(t = 1.299... \approx 1.30\) s | A1 | Correct answer shown to 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u_x = 12\cos 30°\) | M1 | Use of horizontal distance \(= u_x \times t\) |
| \(x = 12\cos 30° \times 1.30 = 13.5\) m | A1 | Accept 13.5 or 13.6 m (use of \(t = 1.30\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Horizontal component: \(v_x = 12\cos 30°\) | B1 | Correct horizontal component (unchanged) |
| Vertical component: \(v_y = 6 - 9.8(1.30)\) | M1 | Use of \(v = u + at\) vertically |
| \(v_y = 6 - 12.74 = -6.74\) m s\(^{-1}\) | A1 | Correct value of \(v_y\) |
| Speed \(= \sqrt{(12\cos 30°)^2 + (-6.74)^2}\) | M1 | Use of Pythagoras |
| \(= \sqrt{108 + 45.4} \approx 12.4\) m s\(^{-1}\) | A1 | Correct speed (accept 12.3–12.4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan\theta = \frac{ | v_y | }{v_x} = \frac{6.74}{12\cos 30°}\) |
| \(\theta \approx 33.0°\) below horizontal | A1 | Correct angle (follow through from (c)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| No air resistance (only gravity acts) | B1 | Accept equivalent statements e.g. "gravity is the only force acting" |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Vertical: $u_y = 12\sin 30° = 6$ m s$^{-1}$ | B1 | Correct vertical component |
| Using $s = u_y t - \frac{1}{2}gt^2$ with $s = 1 - 1.5 = -0.5$ m | M1 | Correct use of $s = ut + \frac{1}{2}at^2$ vertically, with correct displacement |
| $-0.5 = 6t - \frac{1}{2}(9.8)t^2$ | A1 | Correct equation |
| $4.9t^2 - 6t - 0.5 = 0$ | M1 | Forming correct quadratic and attempting to solve |
| $t = \frac{6 \pm \sqrt{36 + 4(4.9)(0.5)}}{9.8}$ | | |
| $t = 1.299... \approx 1.30$ s | A1 | Correct answer shown to 3 s.f. |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u_x = 12\cos 30°$ | M1 | Use of horizontal distance $= u_x \times t$ |
| $x = 12\cos 30° \times 1.30 = 13.5$ m | A1 | Accept 13.5 or 13.6 m (use of $t = 1.30$) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Horizontal component: $v_x = 12\cos 30°$ | B1 | Correct horizontal component (unchanged) |
| Vertical component: $v_y = 6 - 9.8(1.30)$ | M1 | Use of $v = u + at$ vertically |
| $v_y = 6 - 12.74 = -6.74$ m s$^{-1}$ | A1 | Correct value of $v_y$ |
| Speed $= \sqrt{(12\cos 30°)^2 + (-6.74)^2}$ | M1 | Use of Pythagoras |
| $= \sqrt{108 + 45.4} \approx 12.4$ m s$^{-1}$ | A1 | Correct speed (accept 12.3–12.4) |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\theta = \frac{|v_y|}{v_x} = \frac{6.74}{12\cos 30°}$ | M1 | Correct use of components to find angle |
| $\theta \approx 33.0°$ below horizontal | A1 | Correct angle (follow through from (c)) |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| No air resistance (only gravity acts) | B1 | Accept equivalent statements e.g. "gravity is the only force acting" |7 An arrow is fired from a point at a height of 1.5 metres above horizontal ground. It has an initial velocity of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ above the horizontal. The arrow hits a target at a height of 1 metre above horizontal ground. The path of the arrow is shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{f30b02da-a41e-44cb-b45f-9e6a3a9d0528-18_341_1260_550_390}
Model the arrow as a particle.
\begin{enumerate}[label=(\alph*)]
\item Show that the time taken for the arrow to travel to the target is 1.30 seconds, correct to three significant figures.
\item Find the horizontal distance between the point where the arrow is fired and the target.
\item Find the speed of the arrow when it hits the target.
\item Find the angle between the velocity of the arrow and the horizontal when the arrow hits the target.
\item State one assumption that you have made about the forces acting on the arrow.\\
(1 mark)
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{f30b02da-a41e-44cb-b45f-9e6a3a9d0528-19_2486_1714_221_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{f30b02da-a41e-44cb-b45f-9e6a3a9d0528-20_2486_1714_221_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2011 Q7 [14]}}