| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.8 This is a straightforward application of standard projectile motion formulas (max height = u²sin²θ/2g, range = u²sin2θ/g) with all values given directly. Requires only substitution into well-known equations with no problem-solving or conceptual challenges, making it easier than average but not trivial since students must recall and apply the correct formulas. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Vertical motion \(u_y = 100\sin 25°\), \(v_y = 0\) | M1 | Use of suvat equation(s) with \(v = 0\) leading to a value for \(s\); allow sign errors |
| \(v^2 = u^2 + 2as \Rightarrow 0 = (100\sin 25°)^2 - 2 \times 9.8 \times s\) | B1 | Correct component of velocity soi |
| \(s = \frac{42.26^2}{19.6} = 91.1\text{ m}\) (3sf) | A1 [3] | cao; e.g. from \(t = 4.31\) s |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Vertical motion \(u_y = 100\sin 25°\), \(y = 0\) | M1 | Use of suvat equation(s) with \(y = 0\) and their \(u_y \neq 100\) leading to a value for \(t\); allow sign errors |
| \(s = ut + \frac{1}{2}at^2 \Rightarrow 0 = (100\sin 25°)t - 4.9t^2\) | ||
| \(t(100\sin 25° - 4.9t) = 0 \Rightarrow t = 0\) or \(8.62...\) | A1 | Correct value or expression for \(t\) from correct working; can be BC |
| \(x = (100\cos 25°) \times 8.62... = 781.678...\) | M1 | Use of horizontal motion equation with \(u_x = 100\cos 25°\); allow sin/cos interchange throughout part (b) |
| Range is \(782\text{ m}\) (3sf) | A1 [4] | FT their value for \(t\) |
## Question 9:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Vertical motion $u_y = 100\sin 25°$, $v_y = 0$ | M1 | Use of suvat equation(s) with $v = 0$ leading to a value for $s$; allow sign errors |
| $v^2 = u^2 + 2as \Rightarrow 0 = (100\sin 25°)^2 - 2 \times 9.8 \times s$ | B1 | Correct component of velocity soi |
| $s = \frac{42.26^2}{19.6} = 91.1\text{ m}$ (3sf) | A1 [3] | cao; e.g. from $t = 4.31$ s |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Vertical motion $u_y = 100\sin 25°$, $y = 0$ | M1 | Use of suvat equation(s) with $y = 0$ and their $u_y \neq 100$ leading to a value for $t$; allow sign errors |
| $s = ut + \frac{1}{2}at^2 \Rightarrow 0 = (100\sin 25°)t - 4.9t^2$ | | |
| $t(100\sin 25° - 4.9t) = 0 \Rightarrow t = 0$ or $8.62...$ | A1 | Correct value or expression for $t$ from correct working; can be BC |
| $x = (100\cos 25°) \times 8.62... = 781.678...$ | M1 | Use of horizontal motion equation with $u_x = 100\cos 25°$; allow sin/cos interchange throughout part (b) |
| Range is $782\text{ m}$ (3sf) | A1 [4] | FT their value for $t$ |
---9 A cannonball is fired from a point on horizontal ground at $100 \mathrm {~ms} ^ { - 1 }$ at an angle of $25 ^ { \circ }$ above the horizontal. Ignoring air resistance, calculate
\begin{enumerate}[label=(\alph*)]
\item the greatest height the cannonball reaches,
\item the range of the cannonball.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2019 Q9 [7]}}