| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| 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 u=29.4 m/s, and part (b) applies s=ut+½at² or v²=u²+2as. Both parts are routine recall with minimal problem-solving, making this easier than average A-level questions. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(u = 30\) | B1 | From \(v = u + at\) or equivalent. |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0 = 30^2 - 2 \times 10 \times s\) | M1 | Use of *suvat* formulae. |
| Greatest height \(= 45\text{ m}\) | A1 FT | FT \(\frac{u^2}{2g}\) with *their* \(u\) from (a). |
| Total: 2 |
## Question 1:
**Part (a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u = 30$ | **B1** | From $v = u + at$ or equivalent. |
| | **Total: 1** | |
**Part (b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0 = 30^2 - 2 \times 10 \times s$ | **M1** | Use of *suvat* formulae. |
| Greatest height $= 45\text{ m}$ | **A1 FT** | FT $\frac{u^2}{2g}$ with *their* $u$ from **(a)**. |
| | **Total: 2** | |
---1 A particle $P$ is projected vertically upwards with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point on the ground. $P$ reaches its greatest height after 3 s .
\begin{enumerate}[label=(\alph*)]
\item Find $u$.
\item Find the greatest height of $P$ above the ground.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q1 [3]}}