| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two possible trajectories through point |
| Difficulty | Standard +0.3 This is a standard M2 projectile question requiring systematic application of trajectory equations. Part (a) involves algebraic manipulation of the standard projectile formula to reach a given quadratic—routine for M2 students. Parts (b)-(d) follow directly once the quadratic is solved. The 'show that' format and multi-step nature add slight complexity, but this represents a typical exam question testing core mechanics techniques without requiring novel insight or particularly challenging problem-solving. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05o Trigonometric equations: solve in given intervals3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \((\rightarrow)\sqrt{27ag}\cos\theta \cdot t = 9a\) | M1 | Horizontal motion. Condone trig confusion. |
| A1 | ||
| \((\uparrow)\sqrt{27ag}\sin\theta \cdot t - \frac{1}{2}gt^2 = 6a\) | M1 | Vertical motion. Condone sign errors and trig confusion. |
| A1 | ||
| \((\uparrow)\sqrt{27ag}\sin\theta \cdot \frac{9a}{\sqrt{27ag}\cos\theta} - \frac{1}{2}g\left(\frac{9a}{\sqrt{27ag}\cos\theta}\right)^2 = 6a\) | DM1 | Substitute for t (unsimplified). Dependent on both previous M marks |
| \(9a \tan\theta - \frac{1}{2}g.81 \cdot \frac{(1 + \tan^2\theta)}{27ag} = 6a\) | DM1 | Express all trig terms in terms of tan. Dependent on preceding M. |
| \(\tan^2\theta - 6\tan\theta + 5 = 0\) | A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan^2\theta - 6\tan\theta + 5 = 0\) | ||
| \((\tan\theta - 1)(\tan\theta - 5) = 0\) | M1 | Method to find one root of the quadratic |
| \(\tan\theta_2 = 1\) or \(\tan\theta = 5\) | A1 A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = \frac{9a}{\sqrt{27ag}\cos\theta} = \frac{9a}{\sqrt{27ag}} \times \frac{\sqrt{26}}{1}\) | M1 | Use \(\tan\theta =\) their 5 to find t. |
| A1 ft | Correct unsimplified. Correct \(\cos\theta\) for their \(\tan\theta\) | |
| \(= \sqrt{\frac{81a^2 \cdot 26}{27a}} = \sqrt{\frac{78a}{g}}\) *Answer given* | A1 | (3) |
| [16] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}m(27ag - v^2) = mg \cdot 6a\) | M1 A1 | Conservation of energy. Requires all 3 terms. Condone sign error. Correct equation |
| \(v = \sqrt{15ag}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v^2 = (\sqrt{27ag}\cos\theta)^2 + \left(\sqrt{27ag}\sin\theta - g \cdot \sqrt{\frac{78a}{g}}\right)^2\) | M1 | Horizontal and vertical components and Pythagoras. Condone trig confusion. |
| \(= \left(\frac{27ag}{26}\right) + \sqrt{\frac{27ag}{26} - \sqrt{78ag}}^2\left(\frac{27}{26} + \frac{363}{26}\right)\) | A1 | Correctly substituted |
| \(v = \sqrt{15ag}\) | A1 | (3) |
| [16] |
**Part (a):**
| $(\rightarrow)\sqrt{27ag}\cos\theta \cdot t = 9a$ | M1 | Horizontal motion. Condone trig confusion. |
| | A1 | |
| $(\uparrow)\sqrt{27ag}\sin\theta \cdot t - \frac{1}{2}gt^2 = 6a$ | M1 | Vertical motion. Condone sign errors and trig confusion. |
| | A1 | |
| $(\uparrow)\sqrt{27ag}\sin\theta \cdot \frac{9a}{\sqrt{27ag}\cos\theta} - \frac{1}{2}g\left(\frac{9a}{\sqrt{27ag}\cos\theta}\right)^2 = 6a$ | DM1 | Substitute for t (unsimplified). Dependent on both previous M marks |
| $9a \tan\theta - \frac{1}{2}g.81 \cdot \frac{(1 + \tan^2\theta)}{27ag} = 6a$ | DM1 | Express all trig terms in terms of tan. Dependent on preceding M. |
| $\tan^2\theta - 6\tan\theta + 5 = 0$ | A1 | (7) |
**Part (b):**
| $\tan^2\theta - 6\tan\theta + 5 = 0$ | | |
| $(\tan\theta - 1)(\tan\theta - 5) = 0$ | M1 | Method to find one root of the quadratic |
| $\tan\theta_2 = 1$ or $\tan\theta = 5$ | A1 A1 (3) | |
**Part (c):**
| $t = \frac{9a}{\sqrt{27ag}\cos\theta} = \frac{9a}{\sqrt{27ag}} \times \frac{\sqrt{26}}{1}$ | M1 | Use $\tan\theta =$ their 5 to find t. |
| | A1 ft | Correct unsimplified. Correct $\cos\theta$ for their $\tan\theta$ |
| $= \sqrt{\frac{81a^2 \cdot 26}{27a}} = \sqrt{\frac{78a}{g}}$ *Answer given* | A1 | (3) |
| | [16] | |
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# Question 7 (continued):
**Part (d):**
| $\frac{1}{2}m(27ag - v^2) = mg \cdot 6a$ | M1 A1 | Conservation of energy. Requires all 3 terms. Condone sign error. Correct equation |
| $v = \sqrt{15ag}$ | A1 | (3) |
**Or (d):**
| $v^2 = (\sqrt{27ag}\cos\theta)^2 + \left(\sqrt{27ag}\sin\theta - g \cdot \sqrt{\frac{78a}{g}}\right)^2$ | M1 | Horizontal and vertical components and Pythagoras. Condone trig confusion. |
| $= \left(\frac{27ag}{26}\right) + \sqrt{\frac{27ag}{26} - \sqrt{78ag}}^2\left(\frac{27}{26} + \frac{363}{26}\right)$ | A1 | Correctly substituted |
| $v = \sqrt{15ag}$ | A1 | (3) |
| | [16] | |\includegraphics{figure_4}
A small ball is projected from a fixed point $O$ so as to hit a target $T$ which is at a horizontal distance $9a$ from $O$ and at a height $6a$ above the level of $O$. The ball is projected with speed $\sqrt{(27ag)}$ at an angle $\theta$ to the horizontal, as shown in Figure 4. The ball is modelled as a particle moving freely under gravity.
\begin{enumerate}[label=(\alph*)]
\item Show that tan$^2 \theta - 6$ tan $\theta + 5 = 0$ [7]
\end{enumerate}
The two possible angles of projection are $\theta_1$ and $\theta_2$, where $\theta_1 > \theta_2$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find tan $\theta_1$ and tan $\theta_2$. [3]
\end{enumerate}
The particle is projected at the larger angle $\theta_1$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Show that the time of flight from $O$ to $T$ is $\sqrt{\left(\frac{78a}{g}\right)}$. [3]
\item Find the speed of the particle immediately before it hits $T$. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q7 [16]}}