| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: speed at given height |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT application with vertical motion under gravity. Part (i) requires using v² = u² + 2as to find speed at half the maximum height, and part (ii) uses v = u + at. Both are standard textbook exercises requiring only direct substitution into familiar equations with no problem-solving insight needed. |
| Spec | 3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \([0 = 8^2 - 2gs]\) | M1 | For using \(0 = u^2 - 2gs\) |
| Maximum height is 3.2 m | A1 | |
| \([v^2 = 8^2 - 2g \times 1.6]\) | M1 | For using \(v^2 = u^2 - 2gs\) |
| Speed is 5.66 m s\(^{-1}\) | A1 | 4 marks total |
| (ii) \([5.65685... = 8 - 10t]\) | M1 | For using \(v = u - gt\) |
| Time is 0.234 s | A1 | 2 marks total |
**(i)** $[0 = 8^2 - 2gs]$ | M1 | For using $0 = u^2 - 2gs$
Maximum height is 3.2 m | A1 |
$[v^2 = 8^2 - 2g \times 1.6]$ | M1 | For using $v^2 = u^2 - 2gs$
Speed is 5.66 m s$^{-1}$ | A1 | 4 marks total
**(ii)** $[5.65685... = 8 - 10t]$ | M1 | For using $v = u - gt$
Time is 0.234 s | A1 | 2 marks total
---3 A particle $P$ is projected vertically upwards, from a point $O$, with a velocity of $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The point $A$ is the highest point reached by $P$. Find\\
(i) the speed of $P$ when it is at the mid-point of $O A$,\\
(ii) the time taken for $P$ to reach the mid-point of $O A$ while moving upwards.
\hfill \mbox{\textit{CAIE M1 2012 Q3 [6]}}