| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two possible trajectories through point |
| Difficulty | Standard +0.3 This is a standard M2 projectile question requiring derivation of the trajectory equation (routine manipulation), solving a quadratic in tan θ (straightforward), and finding range difference. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^23.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x = 49\cos\theta \cdot t\) | B1 | |
| \(y = 49\sin\theta t - \frac{1}{2}.9.8.t^2\) | B1 | |
| \(y = x\tan\theta - 4.9x^2/49^2.\cos^2\theta\) | M1 | aef (eliminating t) |
| \(y = x\tan\theta - x^2(1 + \tan^2\theta)/490\) | A1 | 4 |
| (ii) \(30 = 70\tan\theta - 10(1 + \tan^2\theta)\) | M1 | |
| \(\tan\theta = (70 \pm \sqrt{3300}) \div 20\) | M1 | (6.37/0.628) |
| \(81.1°\) | A1 | \(\theta_1\) or \(\theta_2\) |
| \(32.1°\) | A1 | 4 |
| (iii) \(x^2(1 + \tan^2\theta)/490 = x\tan\theta\) | M1 | set y = 0 |
| \(x = 490\tan\theta/(1 + \tan^2\theta)\) | A1 | |
| \(x = 75.0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 221\) (220.6) | A1 | |
| \(d = 146 \text{ m}\) | A1 | 5 |
| (iii) Alternatively (\(1^{\text{st}}\) 2 marks) | ||
| \(t = 49\sin\theta/4.9\) and (9.88/5.31) | M1 | \(s = ut + \frac{1}{2}at^2\) and \(x = 49\cos\theta t\) or \(R = u^2\sin 2\theta/g\) (precise) 245sin20 |
| \(x = 49\cos\theta \cdot t\) | A1 |
(i) $x = 49\cos\theta \cdot t$ | B1 |
$y = 49\sin\theta t - \frac{1}{2}.9.8.t^2$ | B1 |
$y = x\tan\theta - 4.9x^2/49^2.\cos^2\theta$ | M1 | aef (eliminating t)
$y = x\tan\theta - x^2(1 + \tan^2\theta)/490$ | A1 | 4 | AG
(ii) $30 = 70\tan\theta - 10(1 + \tan^2\theta)$ | M1 |
$\tan\theta = (70 \pm \sqrt{3300}) \div 20$ | M1 | (6.37/0.628)
$81.1°$ | A1 | $\theta_1$ or $\theta_2$
$32.1°$ | A1 | 4 |
(iii) $x^2(1 + \tan^2\theta)/490 = x\tan\theta$ | M1 | set y = 0
$x = 490\tan\theta/(1 + \tan^2\theta)$ | A1 |
$x = 75.0$ | A1 |
# Question 8 (continued):
$x = 221$ (220.6) | A1 |
$d = 146 \text{ m}$ | A1 | 5 | $\checkmark$ | 13
(iii) **Alternatively ($1^{\text{st}}$ 2 marks)** | |
$t = 49\sin\theta/4.9$ and (9.88/5.31) | M1 | $s = ut + \frac{1}{2}at^2$ and $x = 49\cos\theta t$ or $R = u^2\sin 2\theta/g$ (precise) 245sin20
$x = 49\cos\theta \cdot t$ | A1 |8 A particle is projected with speed $49 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of elevation $\theta$ from a point $O$ on a horizontal plane, and moves freely under gravity. The horizontal and upward vertical displacements of the particle from $O$ at time $t$ seconds after projection are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively.\\
(i) Express $x$ and $y$ in terms of $\theta$ and $t$, and hence show that
$$y = x \tan \theta - \frac { x ^ { 2 } \left( 1 + \tan ^ { 2 } \theta \right) } { 490 } .$$
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{35477eb8-59e0-4de6-889c-1f5841f65eec-4_627_1249_1699_447}
\end{center}
The particle passes through the point where $x = 70$ and $y = 30$. The two possible values of $\theta$ are $\theta _ { 1 }$ and $\theta _ { 2 }$, and the corresponding points where the particle returns to the plane are $A _ { 1 }$ and $A _ { 2 }$ respectively (see diagram).\\
(ii) Find $\theta _ { 1 }$ and $\theta _ { 2 }$.\\
(iii) Calculate the distance between $A _ { 1 }$ and $A _ { 2 }$.
\hfill \mbox{\textit{OCR M2 2005 Q8 [13]}}