| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical motion: velocity-time graph |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT question requiring reading values from velocity-time graphs and applying basic kinematic equations. Part (i) uses the fact that both particles have the same acceleration (g = 10 m/s²) to find V from the gradient. Parts (ii) and (iii) involve simple area calculations under the graph and reading off values. All steps are routine applications of standard mechanics techniques with no problem-solving insight required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Using \(g = \frac{0-V}{2-0}\) or \(0 = V - gt\) | M1 | |
| \(V = 20\) | A1 | [2] |
| Speed is 40 ms⁻¹ | B1 | [1] |
| (ii) Using \(h = \frac{1}{2} \times 4 \times 40\) or \(h = \frac{1}{2}gt^2\) or \(40^2 = 2gh\) | M1 | |
| Height is 80 m | A1 | [2] |
**(i)** Using $g = \frac{0-V}{2-0}$ or $0 = V - gt$ | M1 |
$V = 20$ | A1 | [2]
Speed is 40 ms⁻¹ | B1 | [1]
**(ii)** Using $h = \frac{1}{2} \times 4 \times 40$ or $h = \frac{1}{2}gt^2$ or $40^2 = 2gh$ | M1 |
Height is 80 m | A1 | [2]1\\
\includegraphics[max width=\textwidth, alt={}, center]{5125fab5-0be5-4904-afdf-93e91b16e773-2_608_831_258_657}
Two particles $P$ and $Q$ move vertically under gravity. The graphs show the upward velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ of the particles at time $t \mathrm {~s}$, for $0 \leqslant t \leqslant 4 . P$ starts with velocity $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $Q$ starts from rest.\\
(i) Find the value of $V$.
Given that $Q$ reaches the horizontal ground when $t = 4$, find\\
(ii) the speed with which $Q$ reaches the ground,\\
(iii) the height of $Q$ above the ground when $t = 0$.
\hfill \mbox{\textit{CAIE M1 2010 Q1 [5]}}