| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Horizontal projection from height |
| Difficulty | Standard +0.3 This is a straightforward M2 projectiles question requiring standard application of SUVAT equations in two dimensions. Part (a) uses vertical motion to find time, (b) uses horizontal motion to find distance, and (c) requires resolving velocities at a specific point. All steps are routine with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \((\downarrow)\): \(u_y = 25\sin 30° (= 12.5)\) | B1 | |
| \(12 = 12.5t + 4.9t^2\) | −1 each error | |
| Leading to \(t = 0.743, 0.74\) | M1 A2 (1, 0) A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \((\rightarrow)\): \(u_x = 25\cos 30° \left(= \frac{25\sqrt{3}}{2} \approx 21.65\right)\) | B1 | |
| \(OB = 25\cos 30° \times t (\approx 16.094 58)\) | ft their (a) | |
| \(TB \approx 1.1\) (m) | M1 A1 ft awrt 1.09 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \((\rightarrow)\): \(15 = u_x \times t \Rightarrow t = \frac{15}{u_x}\left(= \frac{15}{2\sqrt{3}} \approx 0.693 \text{ or } 0.69\right)\) | M1 A1 | |
| either \((\downarrow)\): \(v_y = 12.5 + 9.8t (\approx 19.2896)\) | M1 | |
| \(V^2 = u_x^2 + v_y^2 (\approx 840.840)\) | ||
| \(V \approx 29\) (ms\(^{-1}\)), 29.0 | M1 A1 | [5] [14] |
| or \((\downarrow)\): \(s_y = 12.5t + 4.9t^2 (\approx 11.0)\) | M1 | |
| \(\frac{1}{2}m \times 25^2 + mg \times s_y = \frac{1}{2}mv^2\) | ||
| \(V = 29\) (ms\(^{-1}\)), 29.0 | M1 A1 |
**Part (a):**
$(\downarrow)$: $u_y = 25\sin 30° (= 12.5)$ | B1 |
$12 = 12.5t + 4.9t^2$ | | −1 each error
Leading to $t = 0.743, 0.74$ | M1 A2 (1, 0) A1 | [5]
**Part (b):**
$(\rightarrow)$: $u_x = 25\cos 30° \left(= \frac{25\sqrt{3}}{2} \approx 21.65\right)$ | B1 |
$OB = 25\cos 30° \times t (\approx 16.094 58)$ | | ft their (a)
$TB \approx 1.1$ (m) | M1 A1 ft awrt 1.09 | [4]
**Part (c):**
$(\rightarrow)$: $15 = u_x \times t \Rightarrow t = \frac{15}{u_x}\left(= \frac{15}{2\sqrt{3}} \approx 0.693 \text{ or } 0.69\right)$ | M1 A1 |
either $(\downarrow)$: $v_y = 12.5 + 9.8t (\approx 19.2896)$ | M1 |
$V^2 = u_x^2 + v_y^2 (\approx 840.840)$ | |
$V \approx 29$ (ms$^{-1}$), 29.0 | M1 A1 | [5] [14]
or $(\downarrow)$: $s_y = 12.5t + 4.9t^2 (\approx 11.0)$ | M1 |
$\frac{1}{2}m \times 25^2 + mg \times s_y = \frac{1}{2}mv^2$ | |
$V = 29$ (ms$^{-1}$), 29.0 | M1 A1 |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a2738ce4-4dc5-4cd1-ac3d-0c3fcf21ea71-11_755_1073_246_287}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A ball is thrown from a point $A$ at a target, which is on horizontal ground. The point $A$ is 12 m above the point $O$ on the ground. The ball is thrown from $A$ with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ below the horizontal. The ball is modelled as a particle and the target as a point $T$. The distance $O T$ is 15 m . The ball misses the target and hits the ground at the point $B$, where $O T B$ is a straight line, as shown in Figure 4. Find
\begin{enumerate}[label=(\alph*)]
\item the time taken by the ball to travel from $A$ to $B$,
\item the distance $T B$.
The point $X$ is on the path of the ball vertically above $T$.
\item Find the speed of the ball at $X$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2008 Q7 [14]}}