| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Perpendicular velocity directions |
| Difficulty | Standard +0.8 This is a multi-part projectiles question requiring standard techniques for parts (a) and (b), but part (c) requires students to recognize that perpendicular velocity directions means the dot product of velocity vectors equals zero, then solve for time. This conceptual leap beyond routine projectile calculations elevates it above average difficulty. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = (25\sin\alpha)^2 - 2gs\) | M1 | A complete method using *suvat* to find \(s\) |
| \(s = 400 \div 19.6 \approx 20.4\) | A1 | Correct expression in \(s\) only |
| Height above ground \(= 10 + 400 \div 19.6 = 30\) or \(30.4\) m | A1 (3) | 30 or 30.4 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(10 = -25 \times \frac{4}{5}t + \frac{1}{2} \times gt^2\) | M1 | A complete method using *suvat* to find total time from \(A\) to \(B\). Condone sign slips. |
| A1 | Correctly substituted equation in \(t\) | |
| \(4.9t^2 - 20t - 10 = 0\), \(\quad t = \frac{20 \pm \sqrt{400 + 4 \times 4.9 \times 10}}{2 \times 4.9}\) | DM1 | Dependent on preceding M1. Solve for \(t\) |
| \(t = 4.531\ldots\) s | A1 | |
| Horizontal distance \(= 25\cos\alpha \cdot t\ (= 15t\ \text{m})\) | M1 | |
| \(= 68\) m | A1 (6) | 68 or 68.0 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(C\): horizontal speed \(= 15\) m s\(^{-1}\) | ||
| Vertical speed \(= \dfrac{15}{\tan\alpha} = 11.25\) | M1 | Use similar triangles, or equivalent, to find vertical speed at \(C\) |
| A1 | ||
| \(11.25 = -20 + gt\) | DM1 | Use *suvat* to find time from \(A\) to \(C\). Dependent on preceding M1 |
| \(t = \dfrac{20 + 11.25}{9.8} = 3.2\) or \(3.19\) | A1 (4) | 3.2 or 3.19 only |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = (25\sin\alpha)^2 - 2gs$ | M1 | A complete method using *suvat* to find $s$ |
| $s = 400 \div 19.6 \approx 20.4$ | A1 | Correct expression in $s$ only |
| Height above ground $= 10 + 400 \div 19.6 = 30$ or $30.4$ m | A1 (3) | 30 or 30.4 only |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $10 = -25 \times \frac{4}{5}t + \frac{1}{2} \times gt^2$ | M1 | A complete method using *suvat* to find total time from $A$ to $B$. Condone sign slips. |
| | A1 | Correctly substituted equation in $t$ |
| $4.9t^2 - 20t - 10 = 0$, $\quad t = \frac{20 \pm \sqrt{400 + 4 \times 4.9 \times 10}}{2 \times 4.9}$ | DM1 | Dependent on preceding M1. Solve for $t$ |
| $t = 4.531\ldots$ s | A1 | |
| Horizontal distance $= 25\cos\alpha \cdot t\ (= 15t\ \text{m})$ | M1 | |
| $= 68$ m | A1 (6) | 68 or 68.0 only |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $C$: horizontal speed $= 15$ m s$^{-1}$ | | |
| Vertical speed $= \dfrac{15}{\tan\alpha} = 11.25$ | M1 | Use similar triangles, or equivalent, to find vertical speed at $C$ |
| | A1 | |
| $11.25 = -20 + gt$ | DM1 | Use *suvat* to find time from $A$ to $C$. Dependent on preceding M1 |
| $t = \dfrac{20 + 11.25}{9.8} = 3.2$ or $3.19$ | A1 (4) | 3.2 or 3.19 only |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{82cadc37-4cb0-455e-9531-e09ec0c19533-11_711_917_219_561}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A particle $P$ is projected from a point $A$ with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of elevation $\alpha$, where $\sin \alpha = \frac { 4 } { 5 }$. The point $A$ is 10 m vertically above the point $O$ which is on horizontal ground, as shown in Figure 4. The particle $P$ moves freely under gravity and reaches the ground at the point $B$.
Calculate
\begin{enumerate}[label=(\alph*)]
\item the greatest height above the ground of $P$, as it moves from $A$ to $B$,
\item the distance $O B$.
The point $C$ lies on the path of $P$. The direction of motion of $P$ at $C$ is perpendicular to the direction of motion of $P$ at $A$.
\item Find the time taken by $P$ to move from $A$ to $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q6 [13]}}