| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Horizontal projection from height |
| Difficulty | Moderate -0.8 This is a straightforward two-part projectile motion question requiring standard SUVAT equations with horizontal and vertical components. Students apply v² = u² + 2as to find time of flight, then use constant horizontal velocity for range, and Pythagoras for final speed—all routine mechanics procedures with no problem-solving insight required. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(20 = gt^2 / 2\) (t = 2), \(x = 15 \times 2\), \(x = 30\) | M1, DM1, A1 | \(y = -gx^2 / (2 \times 15^2)\) use of trajectory equation; \(-20 = -10x^2 / (2 \times 15^2)\) [3] |
| (ii) \(v = (g \times 2) = 20\), \(v = \sqrt{(15^2 + 20^2)}\), \(v = 25\) | B1, M1, A1 | [3] |
**(i)** $20 = gt^2 / 2$ (t = 2), $x = 15 \times 2$, $x = 30$ | M1, DM1, A1 | $y = -gx^2 / (2 \times 15^2)$ use of trajectory equation; $-20 = -10x^2 / (2 \times 15^2)$ [3]
**(ii)** $v = (g \times 2) = 20$, $v = \sqrt{(15^2 + 20^2)}$, $v = 25$ | B1, M1, A1 | [3]
---2 A stone is thrown with speed $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ horizontally from the top of a vertical cliff 20 m above the sea. Calculate\\
(i) the distance from the foot of the cliff to the point where the stone enters the sea,\\
(ii) the speed of the stone when it enters the sea.
\hfill \mbox{\textit{CAIE M2 2011 Q2 [6]}}