| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: time at height |
| Difficulty | Standard +0.3 This is a straightforward SUVAT problem requiring standard application of kinematic equations for vertical motion under gravity. Part (i) uses v²=u²+2as to find initial speed, part (ii) solves a quadratic to find two times at the same height, and part (iii) reverses this process. All steps are routine with no novel insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(0 = u^2 - 2gs\) | |
| \(u^2 = 2 \times 10 \times 45\); speed is \(30\ ms^{-1}\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([40 = 30t - 5t^2 \Rightarrow t = 2, 4]\) | M1 | For using \(s = ut - \frac{1}{2}gt^2\) with \(s = 40\), \(u = 30\) and \(T = t_2 - t_1\) or \(s = ut + \frac{1}{2}gt^2\), \(s = 5\), \(u = 0\) and \(T = 2t\) |
| \([5 = \frac{1}{2}10t^2 \Rightarrow t = 1]\) | M1 | |
| Time above the ground is 2 s | A1ft | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5 = \frac{1}{2}10t^2 \Rightarrow t = 1\), the length of time required is 1 s | B1 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Max. height above top of cliff \(= \frac{1}{2}g(17 \div 4)\ (= 21.25)\) | B1 | |
| \([0 = V^2 - 2g(40 + 21.25)]\) | M1 | For using \(0 = u^2 - 2gs\) |
| Speed is \(35\ ms^{-1}\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(40 = Vt - 5t^2 \Rightarrow t_2 - t_1 = \frac{1}{2}(V/5 + \sqrt{V^2/25 - 32}) - \frac{1}{2}(V/5 - \sqrt{V^2/25 - 32})\) | |
| \(17 = V^2/25 - 32\) | A1 | |
| Speed is \(35\ ms^{-1}\) | A1 | [3] |
## Question 3:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $0 = u^2 - 2gs$ |
| $u^2 = 2 \times 10 \times 45$; speed is $30\ ms^{-1}$ | A1 | [2] |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[40 = 30t - 5t^2 \Rightarrow t = 2, 4]$ | M1 | For using $s = ut - \frac{1}{2}gt^2$ with $s = 40$, $u = 30$ and $T = t_2 - t_1$ or $s = ut + \frac{1}{2}gt^2$, $s = 5$, $u = 0$ and $T = 2t$ |
| $[5 = \frac{1}{2}10t^2 \Rightarrow t = 1]$ | M1 | |
| Time above the ground is 2 s | A1ft | [2] |
**Special Ruling** (assuming upward movement only, maximum mark 1):
### Part (ii) continued
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5 = \frac{1}{2}10t^2 \Rightarrow t = 1$, the length of time required is 1 s | B1 | B1 |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Max. height above top of cliff $= \frac{1}{2}g(17 \div 4)\ (= 21.25)$ | B1 | |
| $[0 = V^2 - 2g(40 + 21.25)]$ | M1 | For using $0 = u^2 - 2gs$ |
| Speed is $35\ ms^{-1}$ | A1 | [3] |
**Alternative for Part (iii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $40 = Vt - 5t^2 \Rightarrow t_2 - t_1 = \frac{1}{2}(V/5 + \sqrt{V^2/25 - 32}) - \frac{1}{2}(V/5 - \sqrt{V^2/25 - 32})$ |
| $17 = V^2/25 - 32$ | A1 | |
| Speed is $35\ ms^{-1}$ | A1 | [3] |
---3 The top of a cliff is 40 metres above the level of the sea. A man in a boat, close to the bottom of the cliff, is in difficulty and fires a distress signal vertically upwards from sea level. Find\\
(i) the speed of projection of the signal given that it reaches a height of 5 m above the top of the cliff,\\
(ii) the length of time for which the signal is above the level of the top of the cliff.
The man fires another distress signal vertically upwards from sea level. This signal is above the level of the top of the cliff for $\sqrt { } ( 17 ) \mathrm { s }$.\\
(iii) Find the speed of projection of the second signal.
\hfill \mbox{\textit{CAIE M1 2013 Q3 [7]}}