| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Horizontal projection from height |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question with two routine parts: (a) uses horizontal projection with given speed to find range, requiring basic SUVAT in two dimensions; (b) uses range formula with given angle and distance to find speed. Both parts are textbook exercises requiring straightforward application of standard methods with no novel insight, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\rightarrow 12.6t = x\) | B1 | |
| \(\downarrow 0.1 = 4.9t^2\) | B1 | |
| \(\Rightarrow 0.1 = 4.9 \times \frac{x^2}{12.6^2}\) | M1 | |
| \(\Rightarrow x = 1.8 \text{ m}\) | A1 | |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\rightarrow u\cos\alpha \cdot t = 2.5\) | M1 A1 | |
| \(\uparrow u\sin\alpha \cdot t = \frac{1}{2}gt^2\) | M1 A1 | |
| \(u \cdot \frac{24}{25} t = 2.5\) | ||
| \(u \cdot \frac{7}{25} = 4.9 \cdot \frac{2.5 \cdot 25}{24u}\) | ||
| \(u^2 = \frac{4.9 \times 2.5 \times 25^2}{7 \times 24}\) | ||
| \(\Rightarrow u \approx 6.75 \text{ or } 6.8 \text{ ms}^{-1}\) | M1 A1 | |
| (6) |
## Question 4:
### Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $\rightarrow 12.6t = x$ | B1 | |
| $\downarrow 0.1 = 4.9t^2$ | B1 | |
| $\Rightarrow 0.1 = 4.9 \times \frac{x^2}{12.6^2}$ | M1 | |
| $\Rightarrow x = 1.8 \text{ m}$ | A1 | |
| | **(4)** | |
### Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $\rightarrow u\cos\alpha \cdot t = 2.5$ | M1 A1 | |
| $\uparrow u\sin\alpha \cdot t = \frac{1}{2}gt^2$ | M1 A1 | |
| $u \cdot \frac{24}{25} t = 2.5$ | | |
| $u \cdot \frac{7}{25} = 4.9 \cdot \frac{2.5 \cdot 25}{24u}$ | | |
| $u^2 = \frac{4.9 \times 2.5 \times 25^2}{7 \times 24}$ | | |
| $\Rightarrow u \approx 6.75 \text{ or } 6.8 \text{ ms}^{-1}$ | M1 A1 | |
| | **(6)** | |
---4. A darts player throws darts at a dart board which hangs vertically. The motion of a dart is modelled as that of a particle moving freely under gravity. The darts move in a vertical plane which is perpendicular to the plane of the dart board. A dart is thrown horizontally with speed $12.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It hits the board at a point which is 10 cm below the level from which it was thrown.
\begin{enumerate}[label=(\alph*)]
\item Find the horizontal distance from the point where the dart was thrown to the dart board.
The darts player moves his position. He now throws a dart from a point which is at a horizontal distance of 2.5 m from the board. He throws the dart at an angle of elevation $\alpha$ to the horizontal, where $\tan \alpha = \frac { 7 } { 24 }$. This dart hits the board at a point which is at the same level as the point from which it was thrown.
\item Find the speed with which the dart is thrown.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2005 Q4 [10]}}