| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.3 This is a standard M2 projectiles question requiring routine application of SUVAT equations to derive trajectory, solve a quadratic, and find velocity direction. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average but still requiring multiple steps and careful algebra. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| i | \(x = (7\cos 30)t\) | B1 |
| \(y = (7\sin 30)t - gt^2/2\) | B1 | |
| \(y = x\tan 30 - gx^2/(2 \times 7^2 \cos^2 30)\) | M1 | Attempt to eliminate \(t\) |
| \(y = x/\sqrt{3} - 2x^2/15\) or \(y = 0.577x - 0.133x^2\) aef | A1 | |
| [4] | ||
| ii | \(2x^2/15 - x/\sqrt{3} + 0.6 = 0\) or \(9.8t^2 - 7t + 1.2 = 0\) | M1 |
| \(x = 1.73 \text{ m}\) or \(\sqrt{3} \text{ m}\) | M1 | Solve 3 term Q.E. for \(x\) or \(t\) |
| \(2.6(0)\) m or \(3\sqrt{3}/2\) m | A1 | |
| [4] | ||
| iii | \(v^2 = (7\sin 30)^2 - 2 \times 9.8 \times 0.6\) | M1 |
| \(v = 0.7 \text{ ms}^{-1}\) | A1 | Use component of 7 |
| \(\tan \theta = 0.7/(7\cos 30)\) | M1 | |
| \(\theta = 6.59°\) to horizontal or \(83.4°\) to vertical | A1 | |
| [4] | ||
| OR | ||
| Attempt to differentiate equation of trajectory: \(\tan 30 - gx/(7^2 \cos^2 30)\) | M1 | Allow \(1/\sqrt{3} - 4x/15\) or \(y' = 0.577 - 0.267x\) |
| Substitute \(x = \sqrt{3}\) and equate to \(\tan \theta\) | M1 | |
| \(\theta = 6.59°\) to horizontal or \(83.4°\) to vertical | A1 | |
| [4] |
| **Part** | **Answer/Working** | **Marks** | **Guidance** |
|----------|-------------------|----------|-------------|
| i | $x = (7\cos 30)t$ | B1 | |
| | $y = (7\sin 30)t - gt^2/2$ | B1 | |
| | $y = x\tan 30 - gx^2/(2 \times 7^2 \cos^2 30)$ | M1 | Attempt to eliminate $t$ |
| | $y = x/\sqrt{3} - 2x^2/15$ or $y = 0.577x - 0.133x^2$ aef | A1 | |
| | | [4] | |
| ii | $2x^2/15 - x/\sqrt{3} + 0.6 = 0$ or $9.8t^2 - 7t + 1.2 = 0$ | M1 | Create a 3 term Q.E. in $x$ or $t$ with $y = 0.6$ |
| | $x = 1.73 \text{ m}$ or $\sqrt{3} \text{ m}$ | M1 | Solve 3 term Q.E. for $x$ or $t$ |
| | $2.6(0)$ m or $3\sqrt{3}/2$ m | A1 | |
| | | [4] | |
| iii | $v^2 = (7\sin 30)^2 - 2 \times 9.8 \times 0.6$ | M1 | Using $v^2 = u^2 - 2gs$ with $u$ a component of 7; can find $t$ first from their $x$ in (i), and then use $v = u + at$. |
| | $v = 0.7 \text{ ms}^{-1}$ | A1 | Use component of 7 |
| | $\tan \theta = 0.7/(7\cos 30)$ | M1 | |
| | $\theta = 6.59°$ to horizontal or $83.4°$ to vertical | A1 | |
| | | [4] | |
| **OR** | | | |
| | Attempt to differentiate equation of trajectory: $\tan 30 - gx/(7^2 \cos^2 30)$ | M1 | Allow $1/\sqrt{3} - 4x/15$ or $y' = 0.577 - 0.267x$ |
| | Substitute $x = \sqrt{3}$ and equate to $\tan \theta$ | M1 | |
| | $\theta = 6.59°$ to horizontal or $83.4°$ to vertical | A1 | |
| | | [4] | |5 A particle is projected with speed $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of elevation of $30 ^ { \circ }$ from a point $O$ and moves freely under gravity. The horizontal and vertically upwards displacements of the particle from $O$ at any subsequent time $t \mathrm {~s}$ are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively.\\
(i) Express $x$ and $y$ in terms of $t$ and hence find the equation of the trajectory of the particle.\\
(ii) Calculate the values of $x$ when $y = 0.6$.\\
(iii) Find the direction of motion of the particle when $y = 0.6$ and the particle is rising.
\hfill \mbox{\textit{OCR M2 2011 Q5 [12]}}