| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion down smooth slope |
| Difficulty | Easy -1.2 This is a straightforward application of constant acceleration kinematics on an inclined plane. Students need to resolve weight to find acceleration (g sin 30°), then apply standard SUVAT equations twice. It requires only routine mechanics techniques with no problem-solving insight or multi-step reasoning beyond basic substitution. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = g\sin30°\) | B1 | — |
| \(v_2^2 = 2(g\sin30°) \times 0.9\) or \(\frac{1}{2}mv_1^2 = mg(0.9\sin30°)\) or \(v_2 = (g\sin30°) \times 0.8\) | M1 | For using \(v^2 = 2as\) or \(\frac{1}{2}mv^2 = mgh\) or \(v = at\) |
| (i) Speed is \(3 \text{ ms}^{-1}\) or (ii) Speed is \(4\text{ms}^{-1}\) | A1, B1 | [4] |
$a = g\sin30°$ | B1 | —
$v_2^2 = 2(g\sin30°) \times 0.9$ or $\frac{1}{2}mv_1^2 = mg(0.9\sin30°)$ or $v_2 = (g\sin30°) \times 0.8$ | M1 | For using $v^2 = 2as$ or $\frac{1}{2}mv^2 = mgh$ or $v = at$
(i) Speed is $3 \text{ ms}^{-1}$ or (ii) Speed is $4\text{ms}^{-1}$ | A1, B1 | [4]
---1 A particle $P$ is released from rest at a point on a smooth plane inclined at $30 ^ { \circ }$ to the horizontal. Find the speed of $P$\\
(i) when it has travelled 0.9 m ,\\
(ii) 0.8 s after it is released.
\hfill \mbox{\textit{CAIE M1 2010 Q1 [4]}}