| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | November |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: max height |
| Difficulty | Easy -1.2 This is a straightforward two-part SUVAT question requiring direct application of standard kinematic equations. Part (a) uses v = u + at with v=0 at maximum height to find initial speed, and part (b) applies s = ut + ½at² or v² = u² + 2as. Both are routine textbook exercises with no problem-solving insight required, making this easier than average. |
| Spec | 3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = 30\) | B1 | Use \(v = u + at\) (or equivalent *suvat*) with \(v = 0\), \(a = -g\) and \(t = 3\) |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([0 = 30^2 + 2(-10)s]\) | M1 | Using \(v^2 = u^2 + 2as\) with \(a = -g\), \(v = 0\) and \(u =\) value from 1(a), or equivalent *suvat* method |
| Greatest height is 45 m | A1 | |
| [2] |
## Question 1:
**Part (a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = 30$ | **B1** | Use $v = u + at$ (or equivalent *suvat*) with $v = 0$, $a = -g$ and $t = 3$ |
| | **[1]** | |
**Part (b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[0 = 30^2 + 2(-10)s]$ | **M1** | Using $v^2 = u^2 + 2as$ with $a = -g$, $v = 0$ and $u =$ value from **1(a)**, or equivalent *suvat* method |
| Greatest height is 45 m | **A1** | |
| | **[2]** | |
---1 A particle $P$ is projected vertically upwards with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point on the ground. $P$ reaches its greatest height after 3 s .
\begin{enumerate}[label=(\alph*)]
\item Find $v$.
\item Find the greatest height of $P$ above the ground.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q1 [3]}}