| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Velocity direction at specific time/point |
| Difficulty | Standard +0.3 This is a standard two-part projectile motion question requiring resolution of velocity components, application of constant acceleration equations, and solving a quadratic. While it involves multiple steps and careful calculation, the techniques are routine for M2 level with no novel problem-solving 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/Working | Mark | Guidance |
| \(V(vert) = 14\sin60 - 1.8g\) | B1 | \(-5.8756...\) |
| \(V^2 = (-)5.8756^2 + (14\cos60)^2\) | M1 | |
| \(V = 9.14\ \text{ms}^{-1}\) | A1 | \(9.1391...\) |
| \(\tan\theta = (-)5.8756/(14\cos60)\) | M1 | |
| \(\theta = 40.0°\) below horizontal | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-2 = (14\sin60)t - gt^2/2\) | M1 | \(-2 = ut - gt^2/2\) used vertically |
| \(5t^2 - 12.124t - 2 = 0\) | M1 | Solves correct 3 term quadratic |
| \(t = 2.58\) s | A1 [3] |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V(vert) = 14\sin60 - 1.8g$ | B1 | $-5.8756...$ |
| $V^2 = (-)5.8756^2 + (14\cos60)^2$ | M1 | |
| $V = 9.14\ \text{ms}^{-1}$ | A1 | $9.1391...$ |
| $\tan\theta = (-)5.8756/(14\cos60)$ | M1 | |
| $\theta = 40.0°$ below horizontal | A1 [5] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-2 = (14\sin60)t - gt^2/2$ | M1 | $-2 = ut - gt^2/2$ used vertically |
| $5t^2 - 12.124t - 2 = 0$ | M1 | Solves correct 3 term quadratic |
| $t = 2.58$ s | A1 [3] | |
---4 A small ball $B$ is projected from a point $O$ with speed $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $60 ^ { \circ }$ above the horizontal.\\
(i) Calculate the speed and direction of motion of $B$ for the instant 1.8 s after projection.
The point $O$ is 2 m above a horizontal plane.\\
(ii) Calculate the time after projection when $B$ reaches the plane.
\hfill \mbox{\textit{CAIE M2 2013 Q4 [8]}}