| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Vector form projectile motion |
| Difficulty | Moderate -0.3 This is a standard M2 projectile motion question using vector notation. Part (a) requires applying SUVAT equations with constant acceleration (routine derivation), parts (b-c) involve basic substitution and differentiation, and parts (d-e) require solving for when velocity components satisfy tan(45°)=1. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average for A-level. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using \(s = ut + \frac{1}{2}at^2\) | M1 | Method must be clear |
| \(\mathbf{r} = (3t)\mathbf{i} + (10 + 5t - 4.9t^2)\mathbf{j}\) | A1 A1 | Answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{j}\) component \(= 0\): \(10 + 5t - 4.9t^2\) | M1 | |
| quadratic formula: \(t = \frac{5 \pm \sqrt{25 + 196}}{9.8} = \frac{5 \pm \sqrt{221}}{9.8}\) | DM1 | |
| \(T = 2.03\)(s), \(2.0\) (s), positive solution only | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiating the position vector (or working from first principles) | M1 | |
| \(\mathbf{v} = 3\mathbf{i} + (5 - 9.8t)\mathbf{j}\) (ms\(^{-1}\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At \(B\) the \(\mathbf{j}\) component of velocity is the negative of the \(\mathbf{i}\) component: \(5 - 9.8t = -3\) | M1 | |
| \(8 = 9.8t\) | A1 | |
| \(t = 0.82\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{v} = 3\mathbf{i} - 3\mathbf{j}\), speed \(= \sqrt{3^2 + 3^2} = \sqrt{18} = 4.24\) (m s\(^{-1}\)) | M1A1 |
# Question 6:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $s = ut + \frac{1}{2}at^2$ | M1 | Method must be clear |
| $\mathbf{r} = (3t)\mathbf{i} + (10 + 5t - 4.9t^2)\mathbf{j}$ | A1 A1 | Answer given |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{j}$ component $= 0$: $10 + 5t - 4.9t^2$ | M1 | |
| quadratic formula: $t = \frac{5 \pm \sqrt{25 + 196}}{9.8} = \frac{5 \pm \sqrt{221}}{9.8}$ | DM1 | |
| $T = 2.03$(s), $2.0$ (s), positive solution only | A1 | |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiating the position vector (or working from first principles) | M1 | |
| $\mathbf{v} = 3\mathbf{i} + (5 - 9.8t)\mathbf{j}$ (ms$^{-1}$) | A1 | |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $B$ the $\mathbf{j}$ component of velocity is the negative of the $\mathbf{i}$ component: $5 - 9.8t = -3$ | M1 | |
| $8 = 9.8t$ | A1 | |
| $t = 0.82$ | | |
## Part (e):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{v} = 3\mathbf{i} - 3\mathbf{j}$, speed $= \sqrt{3^2 + 3^2} = \sqrt{18} = 4.24$ (m s$^{-1}$) | M1A1 | |
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\begin{enumerate}
\item \hspace{0pt} [In this question, the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are in a vertical plane, $\mathbf { i }$ being horizontal and $\mathbf { j }$ being vertically upwards.]
\end{enumerate}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c5760fa5-3c7f-4e29-87a2-b3b4145b9361-12_689_1042_360_459}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
At time $t = 0$, a particle $P$ is projected from the point $A$ which has position vector 10j metres with respect to a fixed origin $O$ at ground level. The ground is horizontal. The velocity of projection of $P$ is $( 3 \mathbf { i } + 5 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$, as shown in Figure 3. The particle moves freely under gravity and reaches the ground after $T$ seconds.\\
(a) For $0 \leqslant t \leqslant T$, show that, with respect to $O$, the position vector, $\mathbf { r }$ metres, of $P$ at time $t$ seconds is given by
$$\mathbf { r } = 3 t \mathbf { i } + \left( 10 + 5 t - 4.9 t ^ { 2 } \right) \mathbf { j }$$
(b) Find the value of $T$.\\
(c) Find the velocity of $P$ at time $t$ seconds $( 0 \leqslant t \leqslant T )$.
When $P$ is at the point $B$, the direction of motion of $P$ is $45 ^ { \circ }$ below the horizontal.\\
(d) Find the time taken for $P$ to move from $A$ to $B$.\\
(e) Find the speed of $P$ as it passes through $B$.
\hfill \mbox{\textit{Edexcel M2 2011 Q6 [12]}}