| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Vector form projectile motion |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question using vector notation. Part (a) requires setting up position equations and using the geometric constraint OB=2AB to find time. Part (b) involves calculating speed from velocity components. Part (c) uses the symmetry property that equal speeds occur at equal heights. While multi-step with 15 marks total, it follows routine M2 techniques without requiring novel insight—slightly easier than average A-level. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks |
|---|---|
| i \(\rightarrow\) distance = \(6t\) | B1 |
| j \(\uparrow\) distance = \(12t - \frac{1}{2}gt^2\) | M1 A1 |
| At B, \(2\left(12t - \frac{1}{2}gt^2\right) = 6t\) | M1 A1 |
| \((24 - 6)t = gt^2\) | DM1 |
| \(18 = gt\), \(t = \frac{18}{g}\) (= 1.84s) | A1 |
| Answer | Marks |
|---|---|
| i \(\rightarrow\) speed = 6 | B1 |
| j \(\uparrow\) velocity = \(12 - gt = -6\) | M1 A1 |
| \(\therefore\) speed at A \(= \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}(= 8.49)\) (ms\(^{-1}\)) | M1 A1 |
| Answer | Marks |
|---|---|
| \(\uparrow\) speed = \(12 - gt = +6\) | M1 A1 ft |
| \(t = \frac{6}{g}\) (= 0.61s) | A1 |
## (a)
| i $\rightarrow$ distance = $6t$ | B1 |
| j $\uparrow$ distance = $12t - \frac{1}{2}gt^2$ | M1 A1 |
| At B, $2\left(12t - \frac{1}{2}gt^2\right) = 6t$ | M1 A1 |
| $(24 - 6)t = gt^2$ | DM1 |
| $18 = gt$, $t = \frac{18}{g}$ (= 1.84s) | A1 |
**Total: 7 marks**
## (b)
| i $\rightarrow$ speed = 6 | B1 |
| j $\uparrow$ velocity = $12 - gt = -6$ | M1 A1 |
| $\therefore$ speed at A $= \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}(= 8.49)$ (ms$^{-1}$) | M1 A1 |
**Total: 5 marks**
## (c)
| $\uparrow$ speed = $12 - gt = +6$ | M1 A1 ft |
| $t = \frac{6}{g}$ (= 0.61s) | A1 |
**Total: 3 marks**
**Question 7 Total: 15 marks**[In this question, the unit vectors $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertical respectively.]
\includegraphics{figure_3}
The point $O$ is a fixed point on a horizontal plane. A ball is projected from $O$ with velocity $(6\mathbf{i} + 12\mathbf{j})$ m s$^{-1}$, and passes through the point $A$ at time $t$ seconds after projection. The point $B$ is on the horizontal plane vertically below $A$, as shown in Figure 3. It is given that $OB = 2AB$.
Find
\begin{enumerate}[label=(\alph*)]
\item the value of $t$,
[7]
\item the speed, $V$ m s$^{-1}$, of the ball at the instant when it passes through $A$.
[5]
\end{enumerate}
At another point $C$ on the path the speed of the ball is also $V$ m s$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the time taken for the ball to travel from $O$ to $C$.
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2012 Q7 [15]}}