| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle above horizontal |
| Difficulty | Standard +0.3 This is a standard M2 projectile motion question requiring application of SUVAT equations in 2D. Part (a) uses horizontal motion (straightforward), part (b) uses vertical motion with a quadratic (routine), and part (c) finds the angle using velocity components (standard technique). All steps are textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(uT = 10\) | M1; M1; A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \((\uparrow): -4 = u \sin \alpha T - \frac{1}{2}gT^2\) | M1 A1 | |
| \(-4 = u \times \frac{3}{5} \times \frac{10}{u} - \frac{1}{2} \times 9.8 \times \left(\frac{10}{u}\right)^2\) | M1 | |
| \(u = 7\) | M1 A1 | (7) |
| \(v_H = u \cos \alpha = \frac{28}{5}\) | B1 ft | |
| \(v_V^2 = (-u \sin \alpha) + 2g \times 4\) | M1 | |
| \(\Rightarrow v_V = 9.8\) (= \(\frac{49}{5}\)) | A1 ft | |
| \(\tan \phi = \frac{49/5}{28/5} = \frac{7}{4}\) | M1 A1 cao | (5) |
## Part (a)
$(\rightarrow): u \cos \alpha \times T = 8$
$u \times \frac{4}{5} \times T = 8$
$uT = 10$ | M1; M1; A1 | (2)
## Part (b)
$(\uparrow): -4 = u \sin \alpha T - \frac{1}{2}gT^2$ | M1 A1
$-4 = u \times \frac{3}{5} \times \frac{10}{u} - \frac{1}{2} \times 9.8 \times \left(\frac{10}{u}\right)^2$ | M1
$u = 7$ | M1 A1 | (7)
$v_H = u \cos \alpha = \frac{28}{5}$ | B1 ft
$v_V^2 = (-u \sin \alpha) + 2g \times 4$ | M1
$\Rightarrow v_V = 9.8$ (= $\frac{49}{5}$) | A1 ft
$\tan \phi = \frac{49/5}{28/5} = \frac{7}{4}$ | M1 A1 cao | (5)
**Total: (12 marks)**
---\includegraphics{figure_3}
A ball is thrown from a point 4 m above horizontal ground. The ball is projected at an angle $\alpha$ above the horizontal, where $\tan \alpha = \frac{4}{3}$. The ball hits the ground at a point which is a horizontal distance 8 m from its point of projection, as shown in Fig. 3. The initial speed of the ball is $u$ m s$^{-1}$ and the time of flight is $T$ seconds.
\begin{enumerate}[label=(\alph*)]
\item Prove that $uT = 10$.
[2]
\item Find the value of $u$.
[5]
\end{enumerate}
As the ball hits the ground, its direction of motion makes an angle $\phi$ with the horizontal.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find $\tan \phi$.
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2003 Q5 [12]}}