| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Speed at specific time or position |
| Difficulty | Moderate -0.3 This is a straightforward projectiles question requiring standard kinematic equations to find speed components at a given time, then using symmetry to find the second time, plus a distance calculation. All steps are routine applications of well-practiced formulas with no conceptual challenges beyond basic projectile motion understanding. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V\sin\theta = 50\sin30 - 4g\) | B1 | \(-15\) |
| \(V^2 = (50\cos30)^2 + (-15)^2\) | M1 | |
| \(V = 45.8\text{ ms}^{-1}\) | A1 | |
| \(15 = 50\sin30 - gt\) | M1 | |
| \(t = 1\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(OP^2 = (4\times50\cos30)^2 + \left(4\times50\sin30 - \frac{4^2g}{2}\right)^2\) | M1 | |
| \(OP = 174\text{ m}\) | A1 | [2] |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V\sin\theta = 50\sin30 - 4g$ | B1 | $-15$ |
| $V^2 = (50\cos30)^2 + (-15)^2$ | M1 | |
| $V = 45.8\text{ ms}^{-1}$ | A1 | |
| $15 = 50\sin30 - gt$ | M1 | |
| $t = 1$ | A1 | [5] |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $OP^2 = (4\times50\cos30)^2 + \left(4\times50\sin30 - \frac{4^2g}{2}\right)^2$ | M1 | |
| $OP = 174\text{ m}$ | A1 | [2] |
---4 A particle $P$ is projected with speed $50 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ above the horizontal from a point $O$ on a horizontal plane.\\
(i) Calculate the speed of $P$ when it has been in motion for 4 s , and calculate another time at which $P$ has this speed.\\
(ii) Find the distance $O P$ when $P$ has been in motion for 4 s .
\hfill \mbox{\textit{CAIE M2 2014 Q4 [7]}}