| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Energy methods in projectiles |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question with three routine parts: (a) uses energy conservation (a direct formula application), (b) requires resolving velocity components at C using the given angle, and (c) involves standard projectile equations to find range. While multi-step with 14 marks total, each part follows textbook methods without requiring novel insight or complex problem-solving. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(u = \sqrt{306.25} = 17.5\) ★ | M1 A1=A1, A1 | cso; 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\psi = \arccos \frac{14}{24.5} \approx 55°\) | B1, M1 A1 | accept 55.2°; (0.96 rads, or 0.963 rads); 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\rightarrow\) \(BD = 14 \times 4\frac{2}{3}\) (14 × t) fit their t | B1, M1 A1, A1, M1 A1ft, A1 | = 60 (m) only; 7 marks; 14 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(u^2 = 306.25 \Rightarrow u = 17.5\) ★ | M1 A1,A1, A1 | cso; 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\psi = \arctan \frac{\sqrt{404.25}}{14} \approx 55°\) | B1, M1 A1 | accept 55.2°; 3 marks |
| Answer | Marks |
|---|---|
| \(BD = 60\) (m) | M1, B1,A1, A1, M1 A1ft, A1 |
(a) **Energy:** $\frac{1}{2}m(24.5^2 - u^2) = mg \times 15$
$u^2 = 24.5^2 - 30g = 306.25$
$u = \sqrt{306.25} = 17.5$ ★ | M1 A1=A1, A1 | cso; 4 marks
(b) $\rightarrow$ $u_x = u \cos \psi = 17.5 \times 0.8 = 14$
$\psi = \arccos \frac{14}{24.5} \approx 55°$ | B1, M1 A1 | accept 55.2°; (0.96 rads, or 0.963 rads); 3 marks
(c) $\uparrow$ $u_y = u \sin \theta = 17.5 \times 0.6 = 10.5$
$s = ut + \frac{1}{2}at^2 \Rightarrow -45 = 10.5t - 4.9t^2$
leading to $t = 4.3$, awrt $t = 4.3$ or $t = 4\frac{2}{3}$
$\rightarrow$ $BD = 14 \times 4\frac{2}{3}$ (14 × t) fit their t | B1, M1 A1, A1, M1 A1ft, A1 | = 60 (m) only; 7 marks; 14 marks total
**Alternative for (a):**
$\rightarrow u_x = u \cos \theta = 0.8u$, $\uparrow u_y = u \sin \theta = 0.6u$
$v_y^2 = 0.36u^2 + 2 \times 9.8 \times 15 = 0.36u^2 + 294$
$24.5^2 = u_x^2 + v_y^2 = 0.64u^2 + 0.36u^2 + 294$
$u^2 = 306.25 \Rightarrow u = 17.5$ ★ | M1 A1,A1, A1 | cso; 4 marks
**Alternative for (b):** $\rightarrow u_x = u \cos \theta = 17.5 \times 0.8 = 14$
$\uparrow v_y^2 = u^2 \sin^2 \theta + 2 \times 9.8 \times 15 = 404.25$
$\psi = \arctan \frac{\sqrt{404.25}}{14} \approx 55°$ | B1, M1 A1 | accept 55.2°; 3 marks
**Alternative for (c)** Use of $y = x \tan \theta - \frac{g \sec^2 \theta}{2u^2}x^2$
$-45 = \frac{3}{4}x \cdot \frac{g}{2 \times 17.5^2} \times \frac{25}{16}x^2$
$x^2 - 30x - 1800 = 0$ o.e.
Factors or quadratic formula
$BD = 60$ (m) | M1, B1,A1, A1, M1 A1ft, A1 |\includegraphics{figure_3}
A particle $P$ is projected from a point $A$ with speed $u$ m s$^{-1}$ at an angle of elevation $\theta$, where $\cos \theta = \frac{4}{5}$. The point $B$, on horizontal ground, is vertically below $A$ and $AB = 45$ m. After projection, $P$ moves freely under gravity passing through a point $C$, 30 m above the ground, before striking the ground at the point $D$, as shown in Figure 3.
Given that $P$ passes through $C$ with speed 24.5 m s$^{-1}$,
\begin{enumerate}[label=(\alph*)]
\item using conservation of energy, or otherwise, show that $u = 17.5$, [4]
\item find the size of the angle which the velocity of $P$ makes with the horizontal as $P$ passes through $C$, [3]
\item find the distance $BD$. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2007 Q7 [14]}}