| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.3 Part (i) is a standard textbook exercise requiring routine application of projectile motion equations (x = ut cos θ, y = ut sin θ - ½gt²) and elimination of t to get the trajectory equation. Part (ii) requires finding where dy/dx = tan(-15°), which adds a modest problem-solving element but remains straightforward. Overall slightly easier than average due to the routine nature of the techniques involved. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 25\cos 30t\) | B1 | Use horizontal motion |
| \(y = 25\sin 30t - \frac{gt^2}{2}\) | B1 | Use vertical motion |
| \(y = 25\sin 30\left(\frac{x}{25\cos 30}\right) - \frac{g\left(\frac{x}{25\cos 30}\right)^2}{2}\) | M1 | Eliminate \(t\) |
| \(y = \frac{x}{\sqrt{3}} - \frac{4x^2}{375}\) or \(y = 0.577x - 0.0107x^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{1}{\sqrt{3}} - \frac{8x}{375}\) or \(\frac{dy}{dx} = 0.577 - 0.0214x\) | M1A1 | Differentiate the equation from part (i) to find the gradient |
| \(-\tan 15 = \frac{1}{\sqrt{3}} - \frac{8x}{375}\) or \(-\tan 15 = 0.577 - 0.0214x\) | M1 | Attempt to solve |
| \(x = 39.6\) or \(x = 39.5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tan 15 = \frac{v_y}{v_x} = \frac{v_y}{12.5\sqrt{3}}\) | M1 | |
| \(v_y = 12.5\sqrt{3}\tan 15 (= 5.8)\) downwards | A1 | |
| \(-5.8 = 12.5 - 10t\) leading to \(t = 1.83\) | M1 | Vertical motion using \(v = u + at\) |
| \(X = 1.83 \times \frac{25\sqrt{3}}{2} = 39.6\) | A1 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 25\cos 30t$ | B1 | Use horizontal motion |
| $y = 25\sin 30t - \frac{gt^2}{2}$ | B1 | Use vertical motion |
| $y = 25\sin 30\left(\frac{x}{25\cos 30}\right) - \frac{g\left(\frac{x}{25\cos 30}\right)^2}{2}$ | M1 | Eliminate $t$ |
| $y = \frac{x}{\sqrt{3}} - \frac{4x^2}{375}$ or $y = 0.577x - 0.0107x^2$ | A1 | |
---
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{1}{\sqrt{3}} - \frac{8x}{375}$ or $\frac{dy}{dx} = 0.577 - 0.0214x$ | M1A1 | Differentiate the equation from part (i) to find the gradient |
| $-\tan 15 = \frac{1}{\sqrt{3}} - \frac{8x}{375}$ or $-\tan 15 = 0.577 - 0.0214x$ | M1 | Attempt to solve |
| $x = 39.6$ or $x = 39.5$ | A1 | |
**Alternative method for question 4(ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan 15 = \frac{v_y}{v_x} = \frac{v_y}{12.5\sqrt{3}}$ | M1 | |
| $v_y = 12.5\sqrt{3}\tan 15 (= 5.8)$ downwards | A1 | |
| $-5.8 = 12.5 - 10t$ leading to $t = 1.83$ | M1 | Vertical motion using $v = u + at$ |
| $X = 1.83 \times \frac{25\sqrt{3}}{2} = 39.6$ | A1 | |
---4 A small ball 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 the ball from $O$ are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively.\\
(i) Express $x$ and $y$ in terms of $t$ and hence find the equation of the trajectory of the ball.\\
(ii) Find $x$ for the position of the ball when its path makes an angle of $15 ^ { \circ }$ below the horizontal. [4]\\
\hfill \mbox{\textit{CAIE M2 2019 Q4 [8]}}