| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Maximum range or optimal angle |
| Difficulty | Standard +0.8 Part (a) is a standard derivation requiring parametric equations and double angle formula (routine for A-level mechanics). Part (b) requires setting up and solving an inequality comparing maximum height to range, involving trigonometric manipulation and the condition tan(2θ) < 2, which demands more problem-solving insight than typical projectile questions. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertical motion with \(u\sin\theta\), \(a = -g\), \(s = 0\): \(0 = u\sin\theta\, t - \frac{1}{2}gt^2\) | M1 | 2.1 |
| \(t = 0,\ \frac{2u\sin\theta}{g}\) | ||
| Horizontal distance \(R\): \(t = \frac{2u\sin\theta}{g}\), \(R = u\cos\theta \times \frac{2u\sin\theta}{g}\) | M1 | 2.1 |
| \(R = \frac{2u^2\sin\theta\cos\theta}{g}\) | A1 | 2.1 |
| Alternative: Substitute \(t = \frac{x}{u\cos\theta}\) to form trajectory \(y = u\sin\theta \times \frac{x}{u\cos\theta} - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2\) | M1 | |
| Equate \(y\) to zero and attempt to rearrange | M1 | |
| \(R = \frac{2u^2\sin\theta\cos\theta}{g}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Max height \(H\) when \(v_y = 0\) | ||
| \(0 = (u\sin\theta)^2 - 2gH\) | M1 | 3.1b — *suvat* equation(s) with \(v_y = 0\) leading to an equation for \(H\) not involving \(t\). Allow sin/cos interchange if consistent with their (a) |
| \(H = \dfrac{u^2\sin^2\theta}{2g}\) | A1 | 1.1b — correct expression for \(H\) |
| Max height exceeds range when: | ||
| \(\dfrac{u^2\sin^2\theta}{2g} > \dfrac{2u^2\sin\theta\cos\theta}{g}\) | M1 | 1.1a — Compares their \(H\) with given \(R\). Allow = used to find boundary value |
| \(\tan\theta > 4\) | M1 | 1.1a — simplifies the inequality to an inequality for \(\tan\theta\) (or equation) |
| \(76.0° < \theta \, [< 90°]\) | A1 | 1.1b — must be an inequality for \(\theta\). Do not penalise for omission of \(90°\) |
| Total | [5] |
## Question 15(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical motion with $u\sin\theta$, $a = -g$, $s = 0$: $0 = u\sin\theta\, t - \frac{1}{2}gt^2$ | M1 | 2.1 | Suvat equations used with $u\sin\theta$ leading to an expression for the time of flight. Do not allow for time to the top unless subsequently doubled |
| $t = 0,\ \frac{2u\sin\theta}{g}$ | | | |
| Horizontal distance $R$: $t = \frac{2u\sin\theta}{g}$, $R = u\cos\theta \times \frac{2u\sin\theta}{g}$ | M1 | 2.1 | Horizontal motion with $u\cos\theta$, $a=0$ and their expression for $t$. Allow sin/cos interchange if consistent with their vertical equation |
| $R = \frac{2u^2\sin\theta\cos\theta}{g}$ | A1 | 2.1 | Convincing argument AG. Note M0M1A0 for sin/cos interchange |
| **Alternative:** Substitute $t = \frac{x}{u\cos\theta}$ to form trajectory $y = u\sin\theta \times \frac{x}{u\cos\theta} - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2$ | M1 | | Allow equivalent formula quoted |
| Equate $y$ to zero and attempt to rearrange | M1 | | Allow sin/cos interchange if consistent with their horizontal equation |
| $R = \frac{2u^2\sin\theta\cos\theta}{g}$ | A1 | | Convincing argument AG |
**Total: [3]**
## Question 15(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Max height $H$ when $v_y = 0$ | | |
| $0 = (u\sin\theta)^2 - 2gH$ | M1 | 3.1b — *suvat* equation(s) with $v_y = 0$ leading to an equation for $H$ not involving $t$. Allow sin/cos interchange if consistent with their (a) |
| $H = \dfrac{u^2\sin^2\theta}{2g}$ | A1 | 1.1b — correct expression for $H$ |
| Max height exceeds range when: | | |
| $\dfrac{u^2\sin^2\theta}{2g} > \dfrac{2u^2\sin\theta\cos\theta}{g}$ | M1 | 1.1a — Compares their $H$ with given $R$. Allow = used to find boundary value |
| $\tan\theta > 4$ | M1 | 1.1a — simplifies the inequality to an inequality for $\tan\theta$ (or equation) |
| $76.0° < \theta \, [< 90°]$ | A1 | 1.1b — must be an inequality for $\theta$. Do not penalise for omission of $90°$ |
| **Total** | **[5]** | |15 A projectile is launched from a point on level ground with an initial velocity $u$ at an angle $\theta$ above the horizontal.
\begin{enumerate}[label=(\alph*)]
\item Show that the range of the projectile is given by $\frac { 2 u ^ { 2 } \sin \theta \cos \theta } { g }$.
\item Determine the set of values of $\theta$ for which the maximum height of the projectile is greater than the range, where $\theta$ is an acute angle. Give your answer in degrees.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2023 Q15 [8]}}