| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Finding angle given constraints |
| Difficulty | Moderate -0.3 This is a straightforward two-part projectiles question using standard formulae. Part (i) requires applying the maximum height formula (v²sin²θ = 2gh) to find θ, and part (ii) uses the range formula. Both are direct applications of memorized results with minimal problem-solving, making it slightly easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((20\sin\theta)^2 - 2g(2.44) = 0\) | M1 | Use \(v^2 = u^2 + 2as\) vertically with \(v = 0\) |
| \(\theta = 20.2\) | A1 | \(\theta = 20.22908\ldots\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(20\sin cv(\theta)t - \frac{1}{2}gt^2 = 0\) AND range \(= 20\,cv(t)\cos cv(\theta)\) | M1 | Use \(s = ut + \frac{1}{2}at^2\) vertically with \(s=0\) OR use \(v = u + at\) and doubles \(t\) AND horizontally with time found from vertical. \((t = 1.4113\ldots\) s or \(1.4093\ldots\)s (from 20.2)) |
| Range \(= 26.5\) m | A1 | Range \(= 26.48541\ldots\) m or \(26.45387\ldots\)m (from 20.2) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{20^2\sin(2\times cv(\theta))}{g}\) | M1 | Use of range formula |
| Range \(= 26.5\) m | A1 | Range \(= 26.48541\ldots\) m or \(26.45387\ldots\)m (from 20.2) |
| [2] |
## Question 1:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(20\sin\theta)^2 - 2g(2.44) = 0$ | M1 | Use $v^2 = u^2 + 2as$ vertically with $v = 0$ |
| $\theta = 20.2$ | A1 | $\theta = 20.22908\ldots$ |
| **[2]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $20\sin cv(\theta)t - \frac{1}{2}gt^2 = 0$ AND range $= 20\,cv(t)\cos cv(\theta)$ | M1 | Use $s = ut + \frac{1}{2}at^2$ vertically with $s=0$ OR use $v = u + at$ and doubles $t$ AND horizontally with time found from vertical. $(t = 1.4113\ldots$ s or $1.4093\ldots$s (from 20.2)) |
| Range $= 26.5$ m | A1 | Range $= 26.48541\ldots$ m or $26.45387\ldots$m (from 20.2) |
| **[2]** | | |
**OR**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{20^2\sin(2\times cv(\theta))}{g}$ | M1 | Use of range formula |
| Range $= 26.5$ m | A1 | Range $= 26.48541\ldots$ m or $26.45387\ldots$m (from 20.2) |
| **[2]** | | |
---1 A football is kicked from horizontal ground with speed $20 \mathrm {~ms} ^ { - 1 }$ at an angle of $\theta ^ { \circ }$ above the horizontal. The greatest height the football reaches above ground level is 2.44 m . By modelling the football as a particle and ignoring air resistance, find\\
(i) the value of $\theta$,\\
(ii) the range of the football.
\hfill \mbox{\textit{OCR M2 2014 Q1 [4]}}