| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Standard +0.3 This is a standard projectile motion question requiring routine application of kinematic equations to derive trajectory and solve for range. Part (i) involves straightforward substitution to eliminate t, and part (ii) is a simple quadratic equation. The 45° angle simplifies calculations, and all steps follow textbook methods with no novel problem-solving required. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x = (25\cos45)t\) | B1 | |
| \(y = (25\sin45)t - gt^2/2\) | B1 | |
| \(y = x(25\sin45)/(25\cos45) - g[x/(25\cos45)]^2/2\) | M1 | Eliminates \(t\) between 2 simultaneous equations |
| \(y = x - 0.016x^2\) | A1 | [4] |
| (ii) \(2.4 = x - 0.016x^2\) | M1 | Creates and attempts to solve a quadratic equation (\(x = 2.5, 60\)) |
| Distance \(= 57.5 \text{ m}\) | A1 | [2] |
**(i)** $x = (25\cos45)t$ | B1 |
$y = (25\sin45)t - gt^2/2$ | B1 |
$y = x(25\sin45)/(25\cos45) - g[x/(25\cos45)]^2/2$ | M1 | Eliminates $t$ between 2 simultaneous equations
$y = x - 0.016x^2$ | A1 | [4]
**(ii)** $2.4 = x - 0.016x^2$ | M1 | Creates and attempts to solve a quadratic equation ($x = 2.5, 60$)
Distance $= 57.5 \text{ m}$ | A1 | [2]3 A particle $P$ is projected with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $45 ^ { \circ }$ above the horizontal from a point $O$ on horizontal ground. At time $t \mathrm {~s}$ after projection the horizontal and vertically upward displacements of $P$ from $O$ are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively.\\
(i) Express $x$ and $y$ in terms of $t$ and hence show that the equation of the path of $P$ is $y = x - 0.016 x ^ { 2 }$.\\
(ii) Calculate the horizontal distance between the two positions at which $P$ is 2.4 m above the ground.
\hfill \mbox{\textit{CAIE M2 2011 Q3 [6]}}