| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Moderate -0.3 This is a straightforward projectile motion question requiring standard kinematic equations. Part (a) involves routine substitution of given values into standard formulae (with sin θ = 12/13 provided, making cos θ = 5/13 easily found). Part (b) requires solving a quadratic equation after substituting y = 16. Part (c) is a standard modelling discussion. The question is slightly easier than average due to the convenient numbers (g = 10, Pythagorean triple) and the 'show that' structure guiding students through the solution. |
| Spec | 1.02z Models in context: use functions in modelling3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = ut\sin\theta - \frac{1}{2}gt^2\) stated and used | M1 | |
| \(y = 26 \times \frac{12}{13}t - 5t^2 = 24t - 5t^2\) | E1 | AG — given answer must be seen to score E1 |
| \(x = 26 \times \frac{5}{13}t = 10t\) | M1, A1 [4] | Use of \(\frac{5}{13}\); accept any form |
| Answer | Marks | Guidance |
|---|---|---|
| \(16 = 24t - 5t^2\) | M1 | Equating their \(y\) expression to 16 |
| Solving \(5t^2 - 24t + 16 = 0\), \(((5t-4)(t-4) = 0\) or ...) | M1 | Method giving 2 correct roots for their quadratic |
| \(t = 0.8\) or \(4\) | A1 | |
| Distances are \(10 \times 0.8 = 8\) m and \(10 \times 4 = 40\) m | B1FT [4] | FT only their \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| E.g. Air resistance should be included | B1 | |
| E.g. The balloon should not be treated as a particle | B1 [2] | Any two appropriate factors that would have an impact on the model |
## Question 7:
### Part (a):
$y = ut\sin\theta - \frac{1}{2}gt^2$ stated and used | **M1** |
$y = 26 \times \frac{12}{13}t - 5t^2 = 24t - 5t^2$ | **E1** | AG — given answer must be seen to score E1
$x = 26 \times \frac{5}{13}t = 10t$ | **M1, A1** [4] | Use of $\frac{5}{13}$; accept any form
### Part (b):
$16 = 24t - 5t^2$ | **M1** | Equating their $y$ expression to 16
Solving $5t^2 - 24t + 16 = 0$, $((5t-4)(t-4) = 0$ or ...) | **M1** | Method giving 2 correct roots for their quadratic
$t = 0.8$ or $4$ | **A1** |
Distances are $10 \times 0.8 = 8$ m and $10 \times 4 = 40$ m | **B1FT** [4] | FT only their $t$
### Part (c):
E.g. Air resistance should be included | **B1** |
E.g. The balloon should not be treated as a particle | **B1** [2] | Any two appropriate factors that would have an impact on the model
---7 In this question take $\boldsymbol { g } = \mathbf { 1 0 }$.\\
A small stone is projected from a point O with a speed of $26 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\theta$ above the horizontal. The initial velocity and part of the path of the stone are shown in Fig. 7.\\
You are given that $\sin \theta = \frac { 12 } { 13 }$.\\
After $t$ seconds the horizontal displacement of the stone from O is $x$ metres and the vertical displacement is $y$ metres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ff44367e-c992-4e79-b255-5a04e0b8e21e-07_419_479_904_248}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Using the standard model for projectile motion,
\begin{itemize}
\item show that $y = 24 t - 5 t ^ { 2 }$,
\item find an expression for $x$ in terms of $t$.
\end{itemize}
The stone passes through a point A . Point A is 16 m above the level of O .
\item Find the two possible horizontal distances of A from O .
A toy balloon is projected from O with the same initial velocity as the small stone.
\item Suggest two ways in which the standard model could be adapted.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 Q7 [10]}}