| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: time at height |
| Difficulty | Moderate -0.3 Part (a) is a straightforward SUVAT substitution requiring students to use s = ut + ½at² with known values to find u. Part (b) requires understanding that the time difference between reaching height h twice relates to the period above that height, then solving a quadratic—standard mechanics technique but slightly less routine. Overall slightly easier than average due to clear setup and standard methods. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(24 = u \times 2 - \dfrac{1}{2}g \times 2^2\) | M1 | Use of \(s = ut + \frac{1}{2}at^2\) |
| \(u = 22\) | A1 | AG |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At maximum height \(0 = 22^2 - 2gs\) | M1 | Use of \(v^2 = u^2 + 2as\) to find maximum height |
| Maximum height \(s = 24.2\) m | A1 | |
| Height down \(= 0.5g \times 1.8^2\ (=16.2)\) | M1 | Find distance travelled down in 1.8 s |
| \(h = 8\) | A1 | |
| Alternative method: | ||
| \(0 = 22 - 10t\) | M1 | Use of \(v = u - gt\) with \(u = 22\) and \(v = 0\) to find time to reach maximum height |
| \(t = 2.2\) | A1 | |
| \(h = 22 \times (2.2 - 1.8) - \dfrac{1}{2}g \times (2.2 - 1.8)^2\) | M1 | Use of \(s = ut + \frac{1}{2}at^2\) to find value of \(h\) |
| \(h = 8\) | A1 | |
| Alternative method: | ||
| \(22t - \dfrac{1}{2}gt^2 = 22(t+3.6) - \dfrac{1}{2}g(t+3.6)^2\) | M1 | Use of \(s = ut + \frac{1}{2}at^2\) for times \(t\) and \(t+3.6\) to find time taken to reach height \(h\) |
| \(t = 0.4\) (or \(t + 3.6 = 4\)) | A1 | |
| \(h = 22 \times 0.4 - \dfrac{1}{2}g \times 0.4^2\) | M1 | Use of \(s = ut + \frac{1}{2}at^2\) to find value of \(h\) |
| \(h = 8\) | A1 | |
| 4 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $24 = u \times 2 - \dfrac{1}{2}g \times 2^2$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ |
| $u = 22$ | A1 | AG |
| | **2** | |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At maximum height $0 = 22^2 - 2gs$ | M1 | Use of $v^2 = u^2 + 2as$ to find maximum height |
| Maximum height $s = 24.2$ m | A1 | |
| Height down $= 0.5g \times 1.8^2\ (=16.2)$ | M1 | Find distance travelled down in 1.8 s |
| $h = 8$ | A1 | |
| **Alternative method:** | | |
| $0 = 22 - 10t$ | M1 | Use of $v = u - gt$ with $u = 22$ and $v = 0$ to find time to reach maximum height |
| $t = 2.2$ | A1 | |
| $h = 22 \times (2.2 - 1.8) - \dfrac{1}{2}g \times (2.2 - 1.8)^2$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ to find value of $h$ |
| $h = 8$ | A1 | |
| **Alternative method:** | | |
| $22t - \dfrac{1}{2}gt^2 = 22(t+3.6) - \dfrac{1}{2}g(t+3.6)^2$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ for times $t$ and $t+3.6$ to find time taken to reach height $h$ |
| $t = 0.4$ (or $t + 3.6 = 4$) | A1 | |
| $h = 22 \times 0.4 - \dfrac{1}{2}g \times 0.4^2$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ to find value of $h$ |
| $h = 8$ | A1 | |
| | **4** | |
---4 A particle is projected vertically upwards with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point on horizontal ground. After 2 seconds, the height of the particle above the ground is 24 m .
\begin{enumerate}[label=(\alph*)]
\item Show that $u = 22$.
\item The height of the particle above the ground is more than $h \mathrm {~m}$ for a period of 3.6 s .
Find $h$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q4 [6]}}