| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two projectiles meeting - 2D flight |
| Difficulty | Standard +0.3 This is a straightforward two-projectile problem requiring standard kinematic equations. Part (i) uses simple vertical motion under gravity, part (ii) applies the same to projectile B with given angle, and part (iii) requires calculating horizontal separation. All steps are routine applications of SUVAT equations with no novel insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}gt^2 = 7.2\) | B1 | |
| Time taken is 1.2s | B1 | Subtotal: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left[y_B = 7.2 + 5\sin60° \times 1.2 - \frac{1}{2}g(1.2)^2\right]\) | M1 | For using \(y_B = 7.2 + 5t\sin60° - \frac{1}{2}gt^2\) |
| B is approximately 5.2m above the ground | A1 | AG. Subtotal: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Horizontal distance \(= (8 + 5\cos60°) \times 1.2\) \((= 12.6)\) | B1ft | |
| \([AB^2 = 12.6^2 + (3\sqrt{3})^2]\) | M1 | For using \(AB^2 = (\text{HorD})^2 + (\text{VerD})^2\) |
| Distance is 13.6m | A1 | Subtotal: 3. Total: 7 |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}gt^2 = 7.2$ | B1 | |
| Time taken is 1.2s | B1 | **Subtotal: 2** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[y_B = 7.2 + 5\sin60° \times 1.2 - \frac{1}{2}g(1.2)^2\right]$ | M1 | For using $y_B = 7.2 + 5t\sin60° - \frac{1}{2}gt^2$ |
| B is approximately 5.2m above the ground | A1 | AG. **Subtotal: 2** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal distance $= (8 + 5\cos60°) \times 1.2$ $(= 12.6)$ | B1ft | |
| $[AB^2 = 12.6^2 + (3\sqrt{3})^2]$ | M1 | For using $AB^2 = (\text{HorD})^2 + (\text{VerD})^2$ |
| Distance is 13.6m | A1 | **Subtotal: 3. Total: 7** |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{36259e2a-aa9b-4655-b0c2-891f96c3f5a4-4_547_933_269_607}
Particles $A$ and $B$ are projected simultaneously from the top $T$ of a vertical tower, and move in the same vertical plane. $T$ is 7.2 m above horizontal ground. $A$ is projected horizontally with speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $B$ is projected at an angle of $60 ^ { \circ }$ above the horizontal with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 } . A$ and $B$ move away from each other (see diagram).\\
(i) Find the time taken for $A$ to reach the ground.
At the instant when $A$ hits the ground,\\
(ii) show that $B$ is approximately 5.2 m above the ground,\\
(iii) find the distance $A B$.
\hfill \mbox{\textit{CAIE M2 2008 Q5 [7]}}