| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle above horizontal |
| Difficulty | Standard +0.3 Part (a) is a standard 'show that' derivation of the trajectory equation using parametric equations—routine M2 bookwork. Part (b) applies this formula with given values requiring a quadratic solution. Part (c) adds a timing element but is straightforward once part (b) is solved. All parts use standard techniques with no novel insight required, making this slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Horiz: \(x = u \cos \alpha t\) | B1 | |
| Vert: \(y = u \sin \alpha t - \frac{1}{2}gt^2\) | M1 | |
| \(y = u \sin \alpha \times \frac{x}{u \cos \alpha} - \frac{1}{2}g \times \frac{x^2}{u^2 \cos^2 \alpha}\) | DM1 | |
| \(y = x \tan \alpha - \frac{gx^2}{2u^2 \cos^2 \alpha}\) ** | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = -7\): \(-7 = x \tan 45x - \frac{gx^2}{2 \times 7^2 \cos^2 45}\) | M1 A1 | |
| \(-7 = x - \frac{9.8x^2}{7^2}\) | ||
| \(-7 = x - \frac{x^2}{5}\) | M1 | |
| \(x^2 - 5x - 35 = 0\) | ||
| \(x = \frac{5 \pm \sqrt{25 + 4 \times 35}}{2}\) | M1 | |
| \(x = 8.92\) or 8.9 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Time to travel 8.922 m horizontally \(= \frac{8.922}{7 \cos 45} = 1.802\ldots\)s | M1 | |
| \(v = \frac{8.922}{1.402}\) | M1 A1 ft | |
| \(= 6.36\) or 6.4 (m s\(^{-1}\)) | A1 |
## Part (a):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| Horiz: $x = u \cos \alpha t$ | B1 | |
| Vert: $y = u \sin \alpha t - \frac{1}{2}gt^2$ | M1 | |
| $y = u \sin \alpha \times \frac{x}{u \cos \alpha} - \frac{1}{2}g \times \frac{x^2}{u^2 \cos^2 \alpha}$ | DM1 | |
| $y = x \tan \alpha - \frac{gx^2}{2u^2 \cos^2 \alpha}$ ** | A1 | |
## Part (b):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $y = -7$: $-7 = x \tan 45x - \frac{gx^2}{2 \times 7^2 \cos^2 45}$ | M1 A1 | |
| $-7 = x - \frac{9.8x^2}{7^2}$ | | |
| $-7 = x - \frac{x^2}{5}$ | M1 | |
| $x^2 - 5x - 35 = 0$ | | |
| $x = \frac{5 \pm \sqrt{25 + 4 \times 35}}{2}$ | M1 | |
| $x = 8.92$ or 8.9 | A1 | |
## Part (c):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| Time to travel 8.922 m horizontally $= \frac{8.922}{7 \cos 45} = 1.802\ldots$s | M1 | |
| $v = \frac{8.922}{1.402}$ | M1 A1 ft | |
| $= 6.36$ or 6.4 (m s$^{-1}$) | A1 | |A particle is projected from a point $O$ with speed $u$ at an angle of elevation $\alpha$ above the horizontal and moves freely under gravity. When the particle has moved a horizontal distance $x$, its height above $O$ is $y$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$y = x \tan \alpha - \frac{gx^2}{2u^2 \cos^2 \alpha}$$ [4]
\end{enumerate}
A girl throws a ball from a point $A$ at the top of a cliff. The point $A$ is 8 m above a horizontal beach. The ball is projected with speed 7 m s$^{-1}$ at an angle of elevation of 45°. By modelling the ball as a particle moving freely under gravity,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the horizontal distance of the ball from $A$ when the ball is 1 m above the beach. [5]
\end{enumerate}
A boy is standing on the beach at the point $B$ vertically below $A$. He starts to run in a straight line with speed $v$ m s$^{-1}$, leaving $B$ 0.4 seconds after the ball is thrown.
He catches the ball when it is 1 m above the beach.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the value of $v$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2011 Q8 [13]}}