| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Horizontal projection from height |
| Difficulty | Moderate -0.8 This is a straightforward projectiles question requiring standard application of kinematic equations for horizontal projection. Part (i) involves deriving the trajectory equation by eliminating time from x=ut and y=-½gt², while part (ii) substitutes a given horizontal distance to find height. Both parts are routine textbook exercises with no problem-solving insight required, making this easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \pm \dfrac{10x^2}{2 \times 8^2 \times 1^2}\) | M1 | For correct substitution into trajectory formula (in booklet) |
| or \(x = 8t\), \(y = \pm \frac{1}{2}gt^2\) | M1 | For an attempt to eliminate \(t\) |
| \(y = -\dfrac{5}{64}x^2\) | A1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(h = -y(24)\) or substituting \(t=3\) in \(y = \frac{1}{2}gt^2\) | |
| Height \(= 45\)m | A1 (2) |
## Question 1:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \pm \dfrac{10x^2}{2 \times 8^2 \times 1^2}$ | M1 | For correct substitution into trajectory formula (in booklet) |
| or $x = 8t$, $y = \pm \frac{1}{2}gt^2$ | M1 | For an attempt to eliminate $t$ |
| $y = -\dfrac{5}{64}x^2$ | A1 (2) | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $h = -y(24)$ or substituting $t=3$ in $y = \frac{1}{2}gt^2$ |
| Height $= 45$m | A1 (2) | |
---1 A stone is projected horizontally with speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $O$ at the top of a vertical cliff. The horizontal and vertically upward displacements of the stone from $O$ are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively.\\
(i) Find the equation of the stone's trajectory.
The stone enters the sea at a horizontal distance of 24 m from the base of the cliff.\\
(ii) Find the height above sea level of the top of the cliff.
\hfill \mbox{\textit{CAIE M2 2006 Q1 [4]}}