| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Speed at specific time or position |
| Difficulty | Moderate -0.8 This is a straightforward projectiles question requiring standard application of kinematic equations to find velocity components at a given time, then calculating speed and angle. The calculations are routine with no problem-solving insight needed—simpler than average A-level questions. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v_y = 50\sin40 - 2.5g\) | B1 | Vertical component speed \((=7.139...)\) |
| \(v^2 = (50\sin40 - 2.5g)^2 + (50\cos40)^2\) | M1 | Uses Pythagoras with correct horizontal component |
| \(v = 39(.0) \text{ ms}^{-1}\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = 50\cos40 \times 2.5\) | B1 | Horizontal displacement at 2.5s \((=95.75...)\) |
| \(y = 50\sin40 \times 2.5 - 2.5^2\,g/2\) | B1 | \((=49.09...)\) |
| \(\tan\theta = 49.09/95.75\) | M1 | Appropriate ratio to find angle |
| \(\theta = 27.1°\) | A1 [4] |
## Question 5:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $v_y = 50\sin40 - 2.5g$ | B1 | Vertical component speed $(=7.139...)$ |
| $v^2 = (50\sin40 - 2.5g)^2 + (50\cos40)^2$ | M1 | Uses Pythagoras with correct horizontal component |
| $v = 39(.0) \text{ ms}^{-1}$ | A1 [3] | |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 50\cos40 \times 2.5$ | B1 | Horizontal displacement at 2.5s $(=95.75...)$ |
| $y = 50\sin40 \times 2.5 - 2.5^2\,g/2$ | B1 | $(=49.09...)$ |
| $\tan\theta = 49.09/95.75$ | M1 | Appropriate ratio to find angle |
| $\theta = 27.1°$ | A1 [4] | |
**Total: [7]**
---5 A particle $P$ is projected with speed $50 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $40 ^ { \circ }$ above the horizontal from a point $O$. For the instant 2.5 s after projection, calculate\\
(i) the speed of $P$,\\
(ii) the angle between $O P$ and the horizontal.
\hfill \mbox{\textit{CAIE M2 2013 Q5 [7]}}