| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.3 This is a standard M2 projectile question requiring routine application of kinematic equations and trajectory derivation. Part (i) involves basic substitution to eliminate t, part (ii) uses the given angle of elevation with tan(30°), and part (iii) requires differentiation of the trajectory equation. All techniques are standard textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x = (10\cos45°)t\) and \(y = (10\sin45°)t - gt^2/2\) | B1 | |
| \(y = \frac{10\sin45°}{10\cos45°}x - 10\!\left(\frac{x}{10\cos45°}\right)^{\!2}/2\) | M1 | |
| \(y = x - x^2/10\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y/x = \tan30°\) | M1 | |
| \(1 - x/10 = \tan30°\) | A1 | |
| \(x = 4.23\) | A1 [3] | 4.2264… |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(dy/dx = 1 - 2x/10\) | M1 | \(4.2264 = (10\cos45°)t\) |
| \(\tan\theta = dy/dx\) | B1 | \(t = 0.5977\) |
| \(\tan\theta = 1 - 2\times4.23/10\ (= 0.15472\ldots)\) | M1 | \(\tan\theta = \dfrac{10\sin45° - 10\times0.5977}{10\cos45°}\) |
| \(\theta = 8.79°\) | A1 [4] |
## Question 7:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = (10\cos45°)t$ and $y = (10\sin45°)t - gt^2/2$ | B1 | |
| $y = \frac{10\sin45°}{10\cos45°}x - 10\!\left(\frac{x}{10\cos45°}\right)^{\!2}/2$ | M1 | |
| $y = x - x^2/10$ | A1 [3] | |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y/x = \tan30°$ | M1 | |
| $1 - x/10 = \tan30°$ | A1 | |
| $x = 4.23$ | A1 [3] | 4.2264… |
### Part (iii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $dy/dx = 1 - 2x/10$ | M1 | $4.2264 = (10\cos45°)t$ |
| $\tan\theta = dy/dx$ | B1 | $t = 0.5977$ |
| $\tan\theta = 1 - 2\times4.23/10\ (= 0.15472\ldots)$ | M1 | $\tan\theta = \dfrac{10\sin45° - 10\times0.5977}{10\cos45°}$ |
| $\theta = 8.79°$ | A1 [4] | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{af7d1fc8-5552-48b8-a359-895b2b5d3d6c-4_433_841_255_653}
A particle $P$ is projected from a point $O$ with initial speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $45 ^ { \circ }$ above the horizontal. $P$ subsequently passes through the point $A$ which is at an angle of elevation of $30 ^ { \circ }$ from $O$ (see diagram). 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) Write down expressions for $x$ and $y$ in terms of $t$, and hence obtain the equation of the trajectory of $P$.\\
(ii) Calculate the value of $x$ when $P$ is at $A$.\\
(iii) Find the angle the trajectory makes with the horizontal when $P$ is at $A$.
\hfill \mbox{\textit{CAIE M2 2010 Q7 [10]}}