| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.3 This is a straightforward two-part projectile motion question requiring standard SUVAT equations. Part (i) uses time of flight to find the angle (routine application of vertical motion equation), and part (ii) finds when velocity makes a specific angle (standard use of tan and velocity components). Both parts are textbook exercises 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 | Marks | Guidance |
| \(-15\sin\theta = 15\sin\theta - 2g\) | M1 | Use \(v = u + at\) vertically |
| \((\theta =)\ 41.8\) | A1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertically: \(\frac{v}{15\cos\theta} = \pm\tan 20\) | M1 | \(v =\) vertical velocity |
| \(v = (\pm)\ 4.07\) | A1 | |
| \(-4.07 = 15\sin 41.8 - gt\) | M1 | Use \(v = u + at\) vertically |
| \((t =)\ 1.41\) s | A1 | |
| Total | 4 |
**Question 2(i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-15\sin\theta = 15\sin\theta - 2g$ | M1 | Use $v = u + at$ vertically |
| $(\theta =)\ 41.8$ | A1 | |
| **Total** | **2** | |
**Question 2(ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertically: $\frac{v}{15\cos\theta} = \pm\tan 20$ | M1 | $v =$ vertical velocity |
| $v = (\pm)\ 4.07$ | A1 | |
| $-4.07 = 15\sin 41.8 - gt$ | M1 | Use $v = u + at$ vertically |
| $(t =)\ 1.41$ s | A1 | |
| **Total** | **4** | |
---2 A particle is projected from a point on horizontal ground with speed $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $\theta ^ { \circ }$ above the horizontal. The particle strikes the ground 2 s after projection.\\
(i) Find $\theta$.\\
\includegraphics[max width=\textwidth, alt={}, center]{4cd525d5-d59b-4ab9-85a3-fc3d97fd09fc-03_67_1571_438_328}\\
(ii) Calculate the time after projection at which the direction of motion of the particle is $20 ^ { \circ }$ below the horizontal.\\
\hfill \mbox{\textit{CAIE M2 2019 Q2 [6]}}