| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile clearing obstacle |
| Difficulty | Standard +0.3 This is a standard M2 projectile question requiring application of trajectory equations with given initial conditions. Part (a) involves finding the minimum angle using the range equation (moderate algebra), while part (b) is a straightforward 'show that' calculation with given angle. The multi-step nature and need to handle the initial height adds slight complexity, but these are routine M2 techniques with no novel insight required. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | Min. \(\alpha\) when ball passes through \((12, -0.6)\): \(12 = 14\cos\alpha\) \(\therefore t = \frac{6}{7\cos\alpha}\) | M1 A1 |
| \(-0.6 = 14\sin\alpha - 4.9t^2\) | M1 | |
| Sub. in \(r\) giving \(-0.6 = 14(\frac{6}{7\cos\alpha})\sin\alpha - 4.9(\frac{6}{7\cos\alpha})^2\) | A1 | |
| \(-0.6 = 12\tan\alpha - 3.6\sec^2\alpha\) | M1 | |
| Use \(\sec^2\alpha \equiv 1 + \tan^2\alpha\) giving \(6\tan^2\alpha - 20\tan\alpha + 5 = 0\) | A1 | |
| Use of quad. form. giving \((\tan\alpha = 0.27\) and \(3.06)\); min. \(\alpha = 15°\) (nearest degree) | M1 A1 | |
| (b) | At\(\cos\alpha = 12\): \(12 = 14(\frac{3}{5})\) \(\therefore t = \frac{10}{7}\) | M1 A1 |
| Vert. disp.: \(ut\sin\alpha - \frac{1}{2}gt^2 = 14(\frac{10}{7})(\frac{4}{3}) - 4.9(\frac{10}{7})^2\) | M1 | |
| \(= 16 - 10 = 6\) | M1 A1 | |
| i.e. \(6 + 0.6\) above \(M\) \(\therefore 6.6 - 2.4 = 4.2\text{m}\) above crossbar | A1 | (14 marks total) |
(a) | Min. $\alpha$ when ball passes through $(12, -0.6)$: $12 = 14\cos\alpha$ $\therefore t = \frac{6}{7\cos\alpha}$ | M1 A1 |
| $-0.6 = 14\sin\alpha - 4.9t^2$ | M1 |
| Sub. in $r$ giving $-0.6 = 14(\frac{6}{7\cos\alpha})\sin\alpha - 4.9(\frac{6}{7\cos\alpha})^2$ | A1 |
| $-0.6 = 12\tan\alpha - 3.6\sec^2\alpha$ | M1 |
| Use $\sec^2\alpha \equiv 1 + \tan^2\alpha$ giving $6\tan^2\alpha - 20\tan\alpha + 5 = 0$ | A1 |
| Use of quad. form. giving $(\tan\alpha = 0.27$ and $3.06)$; min. $\alpha = 15°$ (nearest degree) | M1 A1 |
(b) | At$\cos\alpha = 12$: $12 = 14(\frac{3}{5})$ $\therefore t = \frac{10}{7}$ | M1 A1 |
| Vert. disp.: $ut\sin\alpha - \frac{1}{2}gt^2 = 14(\frac{10}{7})(\frac{4}{3}) - 4.9(\frac{10}{7})^2$ | M1 |
| $= 16 - 10 = 6$ | M1 A1 |
| i.e. $6 + 0.6$ above $M$ $\therefore 6.6 - 2.4 = 4.2\text{m}$ above crossbar | A1 | (14 marks total)
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f5449ec3-ead0-464f-9d03-f225cd21bca6-4_412_770_198_507}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A football player strikes a ball giving it an initial speed of $14 \mathrm {~ms} ^ { - 1 }$ at an angle $\alpha$ to the horizontal as shown in Figure 2. At the instant he strikes the ball it is 0.6 m vertically above the point $P$ on the ground. The trajectory of the ball is in a vertical plane containing $P$ and $M$, the middle of the goal-line. The distance between $P$ and $M$ is 12 m and the ground is horizontal.
Given that the ball passes over the point $M$ without bouncing,
\begin{enumerate}[label=(\alph*)]
\item find, to the nearest degree, the minimum value of $\alpha$.
Given that the crossbar of the goal is 2.4 m above $M$ and that $\tan \alpha = \frac { 4 } { 3 }$,
\item show that the ball passes 4.2 m vertically above the crossbar.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q6 [14]}}