| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Moderate -0.3 This is a standard M2 projectile question requiring routine application of trajectory equations. Part (a) involves substituting x=10, y=2 into standard kinematic equations—a bookwork derivation. Part (b) requires finding u from the given equation with α=45°, then using v²=vₓ²+vᵧ² with standard velocity components. All techniques are direct applications of memorized formulas with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\rightarrow\ x = u\cos\alpha\, t = 10\) | M1A1 | |
| \(\uparrow\ y = u\sin\alpha\, t - \frac{1}{2}gt^2 = 2\) | M1A1 | |
| \(\Rightarrow t = \frac{10}{u\cos\alpha}\) | M1 | |
| \(2 = u\sin\alpha \times \frac{10}{u\cos\alpha} - \frac{g}{2} \times \frac{100}{u^2\cos^2\alpha} = 10\tan\alpha - \frac{50g}{u^2\cos^2\alpha}\) | A1 | (given answer) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2 = 10 \times 1 - \frac{100g \times 2}{2u^2 \times 1}\) | M1A1 | |
| \(u^2 = \frac{100g}{8},\ u = \sqrt{\frac{100g}{8}} = 11.1 \text{ (m s}^{-1})\) | A1 | |
| \(\frac{1}{2}mu^2 = m \times 9.8 \times 2 + \frac{1}{2}mv^2\) | M1A1 | |
| \(v = 9.1\text{ms}^{-1}\) | A1 |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\rightarrow\ x = u\cos\alpha\, t = 10$ | M1A1 | |
| $\uparrow\ y = u\sin\alpha\, t - \frac{1}{2}gt^2 = 2$ | M1A1 | |
| $\Rightarrow t = \frac{10}{u\cos\alpha}$ | M1 | |
| $2 = u\sin\alpha \times \frac{10}{u\cos\alpha} - \frac{g}{2} \times \frac{100}{u^2\cos^2\alpha} = 10\tan\alpha - \frac{50g}{u^2\cos^2\alpha}$ | A1 | (given answer) |
**Subtotal: (6)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2 = 10 \times 1 - \frac{100g \times 2}{2u^2 \times 1}$ | M1A1 | |
| $u^2 = \frac{100g}{8},\ u = \sqrt{\frac{100g}{8}} = 11.1 \text{ (m s}^{-1})$ | A1 | |
| $\frac{1}{2}mu^2 = m \times 9.8 \times 2 + \frac{1}{2}mv^2$ | M1A1 | |
| $v = 9.1\text{ms}^{-1}$ | A1 | |
**Subtotal: (6) Total: [12]**
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8e220b8a-46f1-4b9b-88a4-f032c7fbda50-09_323_1018_274_452}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A child playing cricket on horizontal ground hits the ball towards a fence 10 m away. The ball moves in a vertical plane which is perpendicular to the fence. The ball just passes over the top of the fence, which is 2 m above the ground, as shown in Figure 3.
The ball is modelled as a particle projected with initial speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from point $O$ on the ground at an angle $\alpha$ to the ground.
\begin{enumerate}[label=(\alph*)]
\item By writing down expressions for the horizontal and vertical distances, from $O$ of the ball $t$ seconds after it was hit, show that
$$2 = 10 \tan \alpha - \frac { 50 g } { u ^ { 2 } \cos ^ { 2 } \alpha }$$
Given that $\alpha = 45 ^ { \circ }$,
\item find the speed of the ball as it passes over the fence.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2009 Q6 [12]}}