| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Speed at specific time or position |
| Difficulty | Standard +0.3 This is a standard projectiles question requiring resolution of velocity components and use of kinematic equations. Part (i) uses the angle condition to find V through tan(angle) = v_y/v_x, and part (ii) applies standard displacement formulas. The multi-step nature and angle interpretation elevate it slightly above routine exercises, but it follows a well-practiced method with no novel insight required. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Velocity component vertically \(= \pm(V\sin60 - 3g)\) | B1 | Use \(v = u + at\) |
| \(\tan30 = \frac{30 - V\sin60}{V\cos60}\) | M1 | Use trigonometry of a right angled triangle |
| \(V = 15\sqrt{3} = 26(.0)\) m s\(^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = 26\sin60 \times 3 - \frac{g \times 3^2}{2}\) | B1FT | Use \(s = ut + \frac{at^2}{2}\) vertically. Their \(V\) from part (i) |
| \(D^2 = (26\sin60 \times 3 - g \times 3^2)^2 + (26\cos60 \times 3)^2\) | M1 | Use Pythagoras's Theorem |
| \(D = 45(.0)\) m | A1 | |
| Total: 3 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Velocity component vertically $= \pm(V\sin60 - 3g)$ | B1 | Use $v = u + at$ |
| $\tan30 = \frac{30 - V\sin60}{V\cos60}$ | M1 | Use trigonometry of a right angled triangle |
| $V = 15\sqrt{3} = 26(.0)$ m s$^{-1}$ | A1 | |
## Question 4(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = 26\sin60 \times 3 - \frac{g \times 3^2}{2}$ | B1FT | Use $s = ut + \frac{at^2}{2}$ vertically. Their $V$ from part (i) |
| $D^2 = (26\sin60 \times 3 - g \times 3^2)^2 + (26\cos60 \times 3)^2$ | M1 | Use Pythagoras's Theorem |
| $D = 45(.0)$ m | A1 | |
| **Total: 3** | | |
---4 A particle is projected from a point $O$ on horizontal ground with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $60 ^ { \circ }$ above the horizontal. At the instant 3 s after projection the direction of motion of the particle is $30 ^ { \circ }$ below the horizontal.\\
(i) Find $V$.\\
..................................................................................................................................\\
(ii) Calculate the distance of the particle from $O$ at the instant 3 s after projection.\\
\hfill \mbox{\textit{CAIE M2 2019 Q4 [6]}}