| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Vertical motion under gravity |
| Difficulty | Moderate -0.8 This is a straightforward three-part SUVAT question requiring only direct application of standard kinematic equations with no problem-solving insight. Part (i) uses v=u+at with v=0 at maximum height, part (ii) uses s=ut+½at² or v²=u²+2as, and part (iii) applies the same equations for the descent. All steps are routine textbook exercises that any competent M1 student would recognize immediately. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(u = 4.905\) | Accept 4.9(0) or 4.91 | |
| (ii) \(s = 1.226...\) | Accept 1.23 but not 1.2 | |
| (iii) \(v = 5.8851...\) | Accept 5.89 but not 5.9 |
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) $u = 4.905$ | | Accept 4.9(0) or 4.91 |
| (ii) $s = 1.226...$ | | Accept 1.23 but not 1.2 |
| (iii) $v = 5.8851...$ | | Accept 5.89 but not 5.9 |
*(Notes apply when using $g = 9.81$)*
---2 A particle $P$ is projected vertically upwards and reaches its greatest height 0.5 s after the instant of projection. Calculate\\
(i) the speed of projection of $P$,\\
(ii) the greatest height of $P$ above the point of projection.
It is given that the point of projection is 0.539 m above the ground.\\
(iii) Find the speed of $P$ immediately before it strikes the ground.
\hfill \mbox{\textit{OCR M1 2013 Q2 [8]}}