| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Vector form projectile motion |
| Difficulty | Moderate -0.8 This is a straightforward application of standard projectile motion formulas in vector form. Part (a) requires writing down the kinematic equation r = r₀ + ut + ½at², and part (b) involves setting y = 0 to find landing time, then calculating distance. Both parts are routine calculations with no problem-solving insight required, making it easier than average but not trivial since it requires correct vector notation and multi-step calculation. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 14t\) | B1 | Must be \(x = ...\) or seen as first component of vector; Do not award for expression adding vector to scalar; SC1 for \(\begin{pmatrix}7t - 4.9t^2 + 5 \\ 14t\end{pmatrix}\) or \(\begin{pmatrix}14t - 4.9t^2 + 5 \\ 7t\end{pmatrix}\) |
| \(y = 7t - \frac{1}{2}gt^2 + 5\) | M1 | Allow without \(+5\), or if \(-5\) seen; Do not award for expression adding vector to scalar |
| Position vector is \(\begin{pmatrix}14t \\ 7t - 4.9t^2 + 5\end{pmatrix}\) | A1 | Must be a single vector; Accept \(\frac{1}{2}g\) in final answer; Accept \(14t\mathbf{i} + (7t - 4.9t^2 + 5)\mathbf{j}\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Lands when \(y = 0\): \(7t - 4.9t^2 + 5 = 0\) | M1 | Award for correct quadratic or attempt to find value of \(t\) when their quadratic \(y=0\) |
| \(t = 1.95\) | A1 | cao |
| gives \(x = 14t = 27.3\) m | B1 | FT their \(t\) and their linear expression for \(x\); ISW where candidates find distance from point of projection |
| [3] |
# Question 7:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 14t$ | B1 | Must be $x = ...$ or seen as first component of vector; Do not award for expression adding vector to scalar; SC1 for $\begin{pmatrix}7t - 4.9t^2 + 5 \\ 14t\end{pmatrix}$ or $\begin{pmatrix}14t - 4.9t^2 + 5 \\ 7t\end{pmatrix}$ |
| $y = 7t - \frac{1}{2}gt^2 + 5$ | M1 | Allow without $+5$, or if $-5$ seen; Do not award for expression adding vector to scalar |
| Position vector is $\begin{pmatrix}14t \\ 7t - 4.9t^2 + 5\end{pmatrix}$ | A1 | Must be a single vector; Accept $\frac{1}{2}g$ in final answer; Accept $14t\mathbf{i} + (7t - 4.9t^2 + 5)\mathbf{j}$ |
| **[3]** | | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Lands when $y = 0$: $7t - 4.9t^2 + 5 = 0$ | M1 | Award for correct quadratic or attempt to find value of $t$ when their quadratic $y=0$ |
| $t = 1.95$ | A1 | cao |
| gives $x = 14t = 27.3$ m | B1 | FT their $t$ and their linear expression for $x$; ISW where candidates find distance from point of projection |
| **[3]** | | |
---7 In this question the $x$ - and $y$-directions are horizontal and vertically upwards respectively and the origin is on horizontal ground.\\
A ball is thrown from a point 5 m above the origin with an initial velocity $\binom { 14 } { 7 } \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the position vector of the ball at time $t \mathrm {~s}$ after it is thrown.
\item Find the distance between the origin and the point at which the ball lands on the ground.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2022 Q7 [6]}}