| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.8 This is a straightforward projectile trajectory question requiring only basic quadratic manipulation: completing the square or differentiation for maximum height, and solving a quadratic equation for range. Both parts are standard textbook exercises with no problem-solving insight needed, making it easier than average but not trivial since it requires correct application of techniques. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks |
|---|---|
| (ii) | dy |
| Answer | Marks |
|---|---|
| Distance = 8 m | M1 |
| Answer | Marks |
|---|---|
| A1 | [2] |
| [2] | Solving trajectory derivative = 0 |
Question 2:
--- 2 (i)
(ii) ---
2 (i)
(ii) | dy
= 1.2 – 2(0.15x) = 0
dx
y = 2.4
OR
Greatest height at half range (ii)
y = 2.4
Twice 'x' for greatest height
Distance = 8 m
OR
2
1.2x – 0.15 x = 0
Distance = 8 m | M1
A1
M1
A1
M1
A1
M1
A1 | [2]
[2] | Solving trajectory derivative = 0
From x = 4
2
Uses y = 1.2x – 0.15 x with x =
distance in (ii) ÷ 2
Using 'x' from (i)
No ft
Solves quadratic with y = 0A stone is projected from a point $O$ on horizontal ground. The equation of the trajectory of the stone is
$$y = 1.2x - 0.15x^2,$$
where $x$ m and $y$ m are respectively the horizontal and vertically upwards displacements of the stone from $O$. Find
\begin{enumerate}[label=(\roman*)]
\item the greatest height of the stone, [2]
\item the distance from $O$ of the point where the stone strikes the ground. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2015 Q2 [4]}}