| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2023 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Vector form projectile motion |
| Difficulty | Standard +0.3 This is a standard M2 projectile motion question using vector notation. Part (a) requires finding time to maximum height using v=u+at. Part (b) involves substituting into position equations. Part (c) uses velocity components and arctangent. Part (d) requires understanding perpendicular vectors (dot product = 0). All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10b 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 |
| \(4 - gT_1 = 0\) or \(T_1 = \dfrac{\sqrt{32}\sin 45°}{g}\) | M1 | Complete method using *suvat* |
| \(T_1 = 0.408\ (0.41)\) | A1 | 3 sf or 2 sf only. Not \(\frac{20}{49}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Height of \(Q\) above \(P\): | M1 | Complete method using *suvat* and 7 and 4 for initial vertical components |
| \(h = \left(7T_1 - \frac{1}{2}gT_1^2\right) - \left(4T_1 - \frac{1}{2}gT_1^2\right)\ \ (= 3T_1)\) | A1 | Correct unsimplified expression in \(T_1\) or their \(T_1\). They do not need to have substituted for \(T_1\). \((2.0408\ldots - 0.8163\ldots)\) |
| \(h = 1.2\ \text{(m)}\) | A1ft | 2 sf only \((3 \times \text{their}\ T_1)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct time for \(P\) to reach \(B\): \(\left(\frac{40}{49},\ 0.816,\ \text{or}\ \frac{8}{g}\ \text{or better}\right)\) | B1 | Seen or implied |
| Vertical component of speed \(= 7 - g \times 2T_1\ (= -1)\) | M1 | Complete method using *suvat* with \(2T_1\) or their \(t\) for the time at \(B\). M0 if not using 7 |
| \(\tan\alpha = \pm\dfrac{\text{their}\ 1}{5}\) | M1 | Correct use of their 1 and 5 to find an equation in a relevant angle (e.g. \(90 - \alpha\)) |
| \(\alpha = 11°\) | A1 | 11 or better (e.g. 11.3) |
| If they use \(T_1\) in place of \(2T_1\) can score B0M0M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Form an equation in \(T_2\) only | M1 | Complete method using *suvat* and perpendicular gradients. e.g. \(\begin{pmatrix}5\\ 7\end{pmatrix}\cdot\begin{pmatrix}5\\ 7 - gT_2\end{pmatrix} = 0\). Condone sign errors. (Vertical component of speed \(= \pm\frac{25}{7}\)). (Perpendicular direction is downwards at 35.5° to the horizontal) |
| \(-\dfrac{25}{7} = 7 - gT_2\) | A1 | Correct unsimplified equation |
| \(T_2 = 1.08\) or \(T_2 = 1.1\) | A1 | 3 sf or 2 sf only |
| Total: [3] | Overall: (12) |
# Question 4:
## Part 4a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4 - gT_1 = 0$ or $T_1 = \dfrac{\sqrt{32}\sin 45°}{g}$ | M1 | Complete method using *suvat* |
| $T_1 = 0.408\ (0.41)$ | A1 | 3 sf or 2 sf only. Not $\frac{20}{49}$ |
**Total: [2]**
## Part 4b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Height of $Q$ above $P$: | M1 | Complete method using *suvat* and 7 and 4 for initial vertical components |
| $h = \left(7T_1 - \frac{1}{2}gT_1^2\right) - \left(4T_1 - \frac{1}{2}gT_1^2\right)\ \ (= 3T_1)$ | A1 | Correct unsimplified expression in $T_1$ or their $T_1$. They do not need to have substituted for $T_1$. $(2.0408\ldots - 0.8163\ldots)$ |
| $h = 1.2\ \text{(m)}$ | A1ft | 2 sf only $(3 \times \text{their}\ T_1)$ |
**Total: [3]**
## Part 4c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct time for $P$ to reach $B$: $\left(\frac{40}{49},\ 0.816,\ \text{or}\ \frac{8}{g}\ \text{or better}\right)$ | B1 | Seen or implied |
| Vertical component of speed $= 7 - g \times 2T_1\ (= -1)$ | M1 | Complete method using *suvat* with $2T_1$ or their $t$ for the time at $B$. M0 if not using 7 |
| $\tan\alpha = \pm\dfrac{\text{their}\ 1}{5}$ | M1 | Correct use of their 1 and 5 to find an equation in a relevant angle (e.g. $90 - \alpha$) |
| $\alpha = 11°$ | A1 | 11 or better (e.g. 11.3) |
| **If they use $T_1$ in place of $2T_1$ can score B0M0M1A0** | | |
**Total: [4]**
## Part 4d:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Form an equation in $T_2$ only | M1 | Complete method using *suvat* and perpendicular gradients. e.g. $\begin{pmatrix}5\\ 7\end{pmatrix}\cdot\begin{pmatrix}5\\ 7 - gT_2\end{pmatrix} = 0$. Condone sign errors. (Vertical component of speed $= \pm\frac{25}{7}$). (Perpendicular direction is downwards at 35.5° to the horizontal) |
| $-\dfrac{25}{7} = 7 - gT_2$ | A1 | Correct unsimplified equation |
| $T_2 = 1.08$ or $T_2 = 1.1$ | A1 | 3 sf or 2 sf only |
**Total: [3] | Overall: (12)**
---\begin{enumerate}
\item \hspace{0pt} [In this question $\mathbf { i }$ and $\mathbf { j }$ are unit vectors, with $\mathbf { i }$ horizontal and $\mathbf { j }$ vertical.]
\end{enumerate}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f180f5f0-43c5-4365-b0d8-7284220b481e-12_278_891_294_587}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The fixed points $A$ and $B$ lie on horizontal ground.\\
At time $t = 0$, a particle $P$ is projected from $A$ with velocity ( $4 \mathbf { i } + 4 \mathbf { j }$ ) $\mathrm { ms } ^ { - 1 }$\\
Particle $P$ moves freely under gravity and hits the ground at $B$, as shown in Figure 3 .\\
At time $T _ { 1 }$ seconds, $P$ is at its highest point above the ground.\\
(a) Find the value of $T _ { 1 }$
At time $t = 0$, a particle $Q$ is also projected from $A$ but with velocity $( 5 \mathbf { i } + 7 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$ Particle $Q$ moves freely under gravity.\\
(b) Find the vertical distance between $Q$ and $P$ at time $T _ { 1 }$ seconds, giving your answer to 2 significant figures.
At the instant when particle $P$ reaches $B$, particle $Q$ is moving at $\alpha ^ { \circ }$ below the horizontal.\\
(c) Find the value of $\alpha$.
At time $T _ { 2 }$ seconds, the direction of motion of $Q$ is perpendicular to the initial direction of motion of $Q$.\\
(d) Find the value of $T _ { 2 }$
\hfill \mbox{\textit{Edexcel M2 2023 Q4 [12]}}