| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | November |
| Marks | 6 |
| 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 direct application of standard kinematic equations with constant acceleration. Part (i) uses v² = u² + 2as to find maximum height, and part (ii) uses s = ut + ½at² to find return time. Both are routine textbook exercises with no problem-solving insight required, making it easier than average. |
| Spec | 3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([0 = 6^2 - 2g \times s]\) | M1 | For using \(v^2 = u^2 + 2as\) |
| \(s = 1.8\) | A1 | |
| Total height \(= 2.3\text{ m}\) | B1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([6^2 = u^2 - 2g \times 0.5]\) | M1 | For using \(v^2 = u^2 + 2as\) to find the initial velocity |
| \(u^2 = 46\) | A1 | |
| \(0^2 = 46 - 2gs \rightarrow s =\) total height \(= 2.3\text{ m}\) | B1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([2.3 = 0 + 0.5gt^2]\) | M1 | For using \(s = ut + 0.5gt^2\) to find time to reach the ground |
| \(t = 0.678\) | A1 | |
| Total time \(= 2 \times 0.678 = 1.36\text{ s}\) | B1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([0 = \sqrt{46} - gt]\) | M1 | Using \(v = u - gt\) to find time taken to the highest point |
| \(t = \dfrac{\sqrt{46}}{10} = 0.678\) | A1 | |
| Total time \(= 2 \times 0.678 = 1.36\text{ s}\) | B1 | [3] |
## Question 3(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0 = 6^2 - 2g \times s]$ | M1 | For using $v^2 = u^2 + 2as$ |
| $s = 1.8$ | A1 | |
| Total height $= 2.3\text{ m}$ | B1 | [3] |
**Alternative for 3(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[6^2 = u^2 - 2g \times 0.5]$ | M1 | For using $v^2 = u^2 + 2as$ to find the initial velocity |
| $u^2 = 46$ | A1 | |
| $0^2 = 46 - 2gs \rightarrow s =$ total height $= 2.3\text{ m}$ | B1 | [3] |
## Question 3(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[2.3 = 0 + 0.5gt^2]$ | M1 | For using $s = ut + 0.5gt^2$ to find time to reach the ground |
| $t = 0.678$ | A1 | |
| Total time $= 2 \times 0.678 = 1.36\text{ s}$ | B1 | [3] |
**Alternative for 3(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0 = \sqrt{46} - gt]$ | M1 | Using $v = u - gt$ to find time taken to the highest point |
| $t = \dfrac{\sqrt{46}}{10} = 0.678$ | A1 | |
| Total time $= 2 \times 0.678 = 1.36\text{ s}$ | B1 | [3] |
---3 A particle $P$ is projected vertically upwards from a point $O$. When the particle is at a height of 0.5 m , its speed is $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find\\
(i) the greatest height reached by the particle above $O$,\\
(ii) the time after projection at which the particle returns to $O$.
\hfill \mbox{\textit{CAIE M1 2016 Q3 [6]}}