| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| 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 with straightforward application of SUVAT equations. Parts (a)-(d) involve routine calculations using horizontal motion (constant velocity) and vertical motion (constant acceleration). Part (e) requires solving a quadratic inequality, which is slightly more challenging but still standard. The elevated starting point adds minimal complexity. Overall, this is slightly easier than average due to its methodical, step-by-step nature with clear given values. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Horizontal distance: \(57.6 = p \times 3\) | M1 | |
| \(p = 19.2\) | A1 | |
| Total: (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(s = ut + \frac{1}{2}at^2\) for vertical displacement | M1 | |
| \(-0.9 = q \times 3 - \frac{1}{2}g \times 3^2\) | A1 | |
| \(-0.9 = 3q - \dfrac{9g}{2} = 3q - 44.1\) | ||
| \(q = \dfrac{43.2}{3} = 14.4\) | A1 cso | AG |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Initial speed \(= \sqrt{p^2 + 14.4^2}\) (with their \(p\)) | M1 | |
| \(= \sqrt{576} = \underline{24}\) (m s\(^{-1}\)) | A1 cao | |
| Total: (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\alpha = \dfrac{14.4}{p}\left(= \dfrac{3}{4}\right)\) (with their \(p\)) | B1 | |
| Total: (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When ball is 4 m above ground, \(3.1 = ut + \frac{1}{2}at^2\) used | M1 | |
| \(3.1 = 14.4t - \frac{1}{2}gt^2\) o.e. \((4.9t^2 - 14.4t + 3.1 = 0)\) | A1 | |
| \(\Rightarrow t = \dfrac{14.4 \pm \sqrt{(14.4)^2 - 4(4.9)(3.1)}}{2(4.9)}\) seen or implied | M1 | |
| \(t = \dfrac{14.4 \pm \sqrt{146.6}}{9.8} = 0.023389\ldots\) or \(2.70488\ldots\) | A1 | awrt 0.23 and 2.7 |
| Duration \(= 2.70488\ldots - 0.23389\ldots\) | M1 | |
| \(= 2.47\) or \(2.5\) (seconds) | A1 | |
| Total: (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1A1M1 as above | ||
| \(t = \dfrac{14.4 \pm \sqrt{146.6}}{9.8}\) | A1 | |
| Duration \(= 2 \times \dfrac{\sqrt{146.6}}{9.8}\) o.e. | M1 | |
| \(= 2.47\) or \(2.5\) (seconds) | A1 | |
| Total: (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. Variable \(g\), Air resistance, Speed of wind, Swing of ball, The ball is not a particle. | B1 | |
| Total: [15] |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal distance: $57.6 = p \times 3$ | M1 | |
| $p = 19.2$ | A1 | |
| **Total: (2)** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $s = ut + \frac{1}{2}at^2$ for vertical displacement | M1 | |
| $-0.9 = q \times 3 - \frac{1}{2}g \times 3^2$ | A1 | |
| $-0.9 = 3q - \dfrac{9g}{2} = 3q - 44.1$ | | |
| $q = \dfrac{43.2}{3} = 14.4$ | A1 cso | **AG** |
| **Total: (3)** | | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Initial speed $= \sqrt{p^2 + 14.4^2}$ (with their $p$) | M1 | |
| $= \sqrt{576} = \underline{24}$ (m s$^{-1}$) | A1 cao | |
| **Total: (2)** | | |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\alpha = \dfrac{14.4}{p}\left(= \dfrac{3}{4}\right)$ (with their $p$) | B1 | |
| **Total: (1)** | | |
### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When ball is 4 m above ground, $3.1 = ut + \frac{1}{2}at^2$ used | M1 | |
| $3.1 = 14.4t - \frac{1}{2}gt^2$ o.e. $(4.9t^2 - 14.4t + 3.1 = 0)$ | A1 | |
| $\Rightarrow t = \dfrac{14.4 \pm \sqrt{(14.4)^2 - 4(4.9)(3.1)}}{2(4.9)}$ seen or implied | M1 | |
| $t = \dfrac{14.4 \pm \sqrt{146.6}}{9.8} = 0.023389\ldots$ or $2.70488\ldots$ | A1 | awrt 0.23 and 2.7 |
| Duration $= 2.70488\ldots - 0.23389\ldots$ | M1 | |
| $= 2.47$ or $2.5$ (seconds) | A1 | |
| **Total: (6)** | | |
### Part (e) – Alternative Method:
| Answer/Working | Mark | Guidance |
|---|---|---|
| M1A1M1 as above | | |
| $t = \dfrac{14.4 \pm \sqrt{146.6}}{9.8}$ | A1 | |
| Duration $= 2 \times \dfrac{\sqrt{146.6}}{9.8}$ o.e. | M1 | |
| $= 2.47$ or $2.5$ (seconds) | A1 | |
| **Total: (6)** | | |
### Part (f):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. Variable $g$, Air resistance, Speed of wind, Swing of ball, The ball is not a particle. | B1 | |
| **Total: [15]** | | |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4c8ebad3-0ebb-4dfe-8036-54b651deb9cf-10_506_1361_205_299}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A cricket ball is hit from a point $A$ with velocity of $( p \mathbf { i } + q \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$, at an angle $\alpha$ above the horizontal. The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are respectively horizontal and vertically upwards. The point $A$ is 0.9 m vertically above the point $O$, which is on horizontal ground.
The ball takes 3 seconds to travel from $A$ to $B$, where $B$ is on the ground and $O B = 57.6 \mathrm {~m}$, as shown in Figure 3. By modelling the motion of the cricket ball as that of a particle moving freely under gravity,
\begin{enumerate}[label=(\alph*)]
\item find the value of $p$,
\item show that $q = 14.4$,
\item find the initial speed of the cricket ball,
\item find the exact value of $\tan \alpha$.
\item Find the length of time for which the cricket ball is at least 4 m above the ground.
\item State an additional physical factor which may be taken into account in a refinement of the above model to make it more realistic.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2009 Q6 [15]}}