| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Vertical motion under gravity |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT question with three standard parts requiring direct application of kinematic equations. All parts involve routine substitution into v² = u² + 2as or s = ut + ½at², with no problem-solving insight needed beyond selecting the correct equation. Slightly easier than average due to the simplicity of downward motion (no direction changes) and clean numerical values. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v^2 = 14^2 + 2g \times 30\) | M1 | \(v^2 = u^2 +/- 2gs\); Using \(v^2 = u^2 + 2as\) |
| \(v = 28\) m s\(^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s = 14 \times 0.4 + g \times 0.4^2/2\) | M1 | |
| \(s = 6.384\) m | A1 | Accept 6.38 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(15 = 28t - gt^2/2\) | M1* | Uses \(s = vt +/- gt^2/2\); Accept cv(28) but not \(v=0\) |
| \(4.9t^2 - 28t + 15 = 0\) | D*M1 | Attempts to solve 3 term QE |
| \(t = (5.12)\ 0.598\) s | A1 | Ignore 5.12 if seen |
| *OR* | ||
| \(28^2 = u^2 + 2g \times 15\) | M1* | \(v^2 = 14^2 + 2g \times 15\); Accept cv(28) but not \(v=0\) |
| \(28 = \sqrt{490} + gt\) | D*M1 | |
| \(t = 0.598\) s | A1 | |
| *OR* | ||
| \(15 = 14t + gt^2/2\) | M1* | Attempts to solve 3 term QE; Accept cv(28) but not \(v=0\) |
| \(30 = (14+28)t/2\) | D*M1 | Finding total time |
| \(t = 0.598\) s | A1 |
## Question 1:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v^2 = 14^2 + 2g \times 30$ | M1 | $v^2 = u^2 +/- 2gs$; Using $v^2 = u^2 + 2as$ |
| $v = 28$ m s$^{-1}$ | A1 | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s = 14 \times 0.4 + g \times 0.4^2/2$ | M1 | |
| $s = 6.384$ m | A1 | Accept 6.38 |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $15 = 28t - gt^2/2$ | M1* | Uses $s = vt +/- gt^2/2$; Accept cv(28) but not $v=0$ |
| $4.9t^2 - 28t + 15 = 0$ | D*M1 | Attempts to solve 3 term QE |
| $t = (5.12)\ 0.598$ s | A1 | Ignore 5.12 if seen |
| *OR* | | |
| $28^2 = u^2 + 2g \times 15$ | M1* | $v^2 = 14^2 + 2g \times 15$; Accept cv(28) but not $v=0$ |
| $28 = \sqrt{490} + gt$ | D*M1 | |
| $t = 0.598$ s | A1 | |
| *OR* | | |
| $15 = 14t + gt^2/2$ | M1* | Attempts to solve 3 term QE; Accept cv(28) but not $v=0$ |
| $30 = (14+28)t/2$ | D*M1 | Finding total time |
| $t = 0.598$ s | A1 | |
---1 A particle $P$ is projected vertically downwards with speed $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point 30 m above the ground.\\
(i) Calculate the speed of $P$ when it reaches the ground.\\
(ii) Find the distance travelled by $P$ in the first 0.4 s of its motion.\\
(iii) Calculate the time taken for $P$ to travel the final 15 m of its descent.
\hfill \mbox{\textit{OCR M1 2015 Q1 [7]}}