| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Moderate -0.3 This is a standard two-equation projectile problem where students substitute known values (t=0.4s, x=12m, y=5m) into kinematic equations to find initial speed and angle, then use velocity components to find direction. It requires routine application of projectile formulas with straightforward algebra, making it slightly easier than average but not trivial due to the two-part calculation and trigonometric manipulation involved. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(5 = 0.4(V\sin\alpha) - g \times 0.4^2/2\) | M1 | \(\alpha\) is the angle of projection |
| \(V\sin\alpha = 14.5\) | A1 | |
| \(0.4(V\cos\alpha) = 12\) hence \(V\cos\alpha = 30\) | B1 | Or \(\tan\alpha = 14.5/30\) |
| \(V = \sqrt{(30^2 + 14.5^2)}\) | M1 | \(\alpha = 25.8°\) |
| \(V = 33.3\) | A1 | \(V = 33.3\) |
| \(\alpha = 25.8°\) | B1 [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v = 14.5 - 0.4g\) | B1 | \(v = \sqrt{(14.5^2 - 2g \times 5)}\) |
| \(\tan\theta = (14.5 - 0.4g)/30\) | M1 | \(\tan\theta = \sqrt{(14.5^2 - 2g \times 5)}/30\) |
| \(\theta = \tan^{-1}0.35 = 19.3°\) with the horizontal | A1 [3] | |
| OR | ||
| \(dy/dx = x\tan\alpha - gx^2\sec^2\alpha/(2V^2)\) | M1 | For differentiating the trajectory equation |
| \(\tan\theta = \tan 25.8° - 10 \times 12\sec^2 25.8°/33.3^2\) | M1 | For attempting to substitute \(x\), \(\alpha\) and \(v\) |
| \(\theta = 19.3°\) with the horizontal | A1 |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $5 = 0.4(V\sin\alpha) - g \times 0.4^2/2$ | M1 | $\alpha$ is the angle of projection |
| $V\sin\alpha = 14.5$ | A1 | |
| $0.4(V\cos\alpha) = 12$ hence $V\cos\alpha = 30$ | B1 | Or $\tan\alpha = 14.5/30$ |
| $V = \sqrt{(30^2 + 14.5^2)}$ | M1 | $\alpha = 25.8°$ |
| $V = 33.3$ | A1 | $V = 33.3$ |
| $\alpha = 25.8°$ | B1 [6] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v = 14.5 - 0.4g$ | B1 | $v = \sqrt{(14.5^2 - 2g \times 5)}$ |
| $\tan\theta = (14.5 - 0.4g)/30$ | M1 | $\tan\theta = \sqrt{(14.5^2 - 2g \times 5)}/30$ |
| $\theta = \tan^{-1}0.35 = 19.3°$ with the horizontal | A1 [3] | |
| **OR** | | |
| $dy/dx = x\tan\alpha - gx^2\sec^2\alpha/(2V^2)$ | M1 | For differentiating the trajectory equation |
| $\tan\theta = \tan 25.8° - 10 \times 12\sec^2 25.8°/33.3^2$ | M1 | For attempting to substitute $x$, $\alpha$ and $v$ |
| $\theta = 19.3°$ with the horizontal | A1 | |
---6 A particle $P$ is projected from a point $O$ on horizontal ground. 0.4 s after the instant of projection, $P$ is 5 m above the ground and a horizontal distance of 12 m from $O$.\\
(i) Calculate the initial speed and the angle of projection of $P$.\\
(ii) Find the direction of motion of the particle 0.4 s after the instant of projection.
\hfill \mbox{\textit{CAIE M2 2011 Q6 [9]}}