| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | March |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: max height |
| Difficulty | Moderate -0.8 This is a straightforward two-part SUVAT question requiring standard application of kinematic equations. Part (a) uses v²=u²+2as with v=0 at maximum height (routine calculation), and part (b) requires finding times when height equals 15m using s=ut+½at². Both parts are textbook exercises with no problem-solving insight needed, making this easier than average but not trivial due to the two-step nature of part (b). |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0 = u^2 - 2 \times 10 \times 20\) OR \(0 = u - 10t\) and \(20 = vt + \frac{1}{2} \times 10 \times t^2\) or \(20 = ut - \frac{1}{2} \times 10 \times t^2\) or \(20 = \frac{u+0}{2} \times t\) | M1 | Complete method to set up an equation in \(u\) only. Use of \(v^2 = u^2 + 2as\) or finding time to reach maximum height (\(t=2\)) and using this value to set up another equation in \(u\) only |
| \(u = 20\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(15 = 20t - \frac{1}{2} \times 10 \times t^2\) | M1 | Use of \(s = ut + \frac{1}{2}at^2\) and attempt to set up an equation from which a relevant \(t\) value can be found. Must be using *their* \(u\) and \(a = -10\) |
| \(t = 1\) or \(t = 3\) | A1 | |
| Total time \(= 2\) s | A1 | CWO |
| Alternative: \(5 = \frac{1}{2} \times 10 \times t^2\) | M1 | Use of \(s = ut + \frac{1}{2}at^2\) and attempt to set up equation from which a relevant \(t\) value can be found. Must be using \(u=0\) and \(a=10\) |
| \(t = 1\) | A1 | |
| Total time \(= 2\) s | A1 | CWO |
| 3 |
## Question 2:
### Part 2(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0 = u^2 - 2 \times 10 \times 20$ OR $0 = u - 10t$ and $20 = vt + \frac{1}{2} \times 10 \times t^2$ or $20 = ut - \frac{1}{2} \times 10 \times t^2$ or $20 = \frac{u+0}{2} \times t$ | M1 | Complete method to set up an equation in $u$ only. Use of $v^2 = u^2 + 2as$ or finding time to reach maximum height ($t=2$) and using this value to set up another equation in $u$ only |
| $u = 20$ | A1 | |
| | **2** | |
### Part 2(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $15 = 20t - \frac{1}{2} \times 10 \times t^2$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ and attempt to set up an equation from which a relevant $t$ value can be found. Must be using *their* $u$ and $a = -10$ |
| $t = 1$ or $t = 3$ | A1 | |
| Total time $= 2$ s | A1 | CWO |
| **Alternative:** $5 = \frac{1}{2} \times 10 \times t^2$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ and attempt to set up equation from which a relevant $t$ value can be found. Must be using $u=0$ and $a=10$ |
| $t = 1$ | A1 | |
| Total time $= 2$ s | A1 | CWO |
| | **3** | |
---2 A particle $P$ is projected vertically upwards from horizontal ground with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 } . P$ reaches a maximum height of 20 m above the ground.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $u$.
\item Find the total time for which $P$ is at least 15 m above the ground.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q2 [5]}}