| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.3 This is a standard projectiles question requiring routine application of kinematic equations and algebraic manipulation. Part (i) involves deriving the trajectory equation by eliminating t from parametric equations—a textbook exercise. Part (ii) is straightforward substitution into the derived equation. The calculations are mechanical with no novel insight required, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = (25\cos30)t\) | B1 | Horizontal motion |
| \(y = (25\sin30)t - gt^2/2\) | B1 | Vertical motion |
| \(y = (25\sin30)x/(25\cos30) - 5[x/(25\cos30)]^2\) | M1 | Attempts to eliminate \(t\) |
| \(y = \dfrac{x}{\sqrt{3}} - \dfrac{4x^2}{375}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(5 = x/\sqrt{3} - 4x^2/375\) (leads to \(4x^2 - 216.5x + 1875 = 0\)) | M1 | Substitutes \(y = 5\) into the trajectory equation |
| \(x = 43.3, 10.8\) | A1 | Solves the quadratic equation |
| Distance \(= 43.3 - 10.8 = 32.5 \text{ m}\) | A1 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = (25\cos30)t$ | B1 | Horizontal motion |
| $y = (25\sin30)t - gt^2/2$ | B1 | Vertical motion |
| $y = (25\sin30)x/(25\cos30) - 5[x/(25\cos30)]^2$ | M1 | Attempts to eliminate $t$ |
| $y = \dfrac{x}{\sqrt{3}} - \dfrac{4x^2}{375}$ | A1 | AG |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5 = x/\sqrt{3} - 4x^2/375$ (leads to $4x^2 - 216.5x + 1875 = 0$) | M1 | Substitutes $y = 5$ into the trajectory equation |
| $x = 43.3, 10.8$ | A1 | Solves the quadratic equation |
| Distance $= 43.3 - 10.8 = 32.5 \text{ m}$ | A1 | |
---4 A particle $P$ is projected with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ above the horizontal from a point $O$ on horizontal ground. At time $t \mathrm {~s}$ after projection the horizontal and vertically upwards 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 trajectory of $P$ is
$$y = \frac { x } { \sqrt { 3 } } - \frac { 4 x ^ { 2 } } { 375 }$$
(ii) Find the horizontal distance between the two points at which $P$ is 5 m above the ground.\\
\hfill \mbox{\textit{CAIE M2 2017 Q4 [7]}}