| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.5 This is a standard projectiles question requiring routine application of kinematic equations. Part (i) involves simple trigonometry (u cos 30° = 40) and using y = ut sin θ - ½gt². Part (ii) requires eliminating t to get the trajectory equation y = x tan θ - gx²/(2u²cos²θ), which is a bookwork derivation. The question is slightly easier than average because the horizontal displacement equation is given directly, removing one step of working. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V\cos30 = 40\) | M1 | Note \(V\) is the velocity of projection |
| \(V = 46.2\) m s\(^{-1}\) | A1 | Allow \(\frac{80}{\sqrt{3}}\) or \(\frac{80\sqrt{3}}{3}\) |
| \(y = 23.1t - 5t^2\) | B1FT | Use \(s = ut + \frac{at^2}{2}\) vertically. FT candidates half \(V\) but not \(V = 40\) used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \frac{23.1x}{40} - \frac{5x^2}{1600}\) | M1 | Attempt to eliminate \(t\) by substituting \(t = \frac{x}{40}\) into answer to part (i) |
| \(y = 0.577x - \frac{x^2}{320}\) or \(y = 0.577x - 0.003125x^2\) | A1 |
## Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V\cos30 = 40$ | M1 | Note $V$ is the velocity of projection |
| $V = 46.2$ m s$^{-1}$ | A1 | Allow $\frac{80}{\sqrt{3}}$ or $\frac{80\sqrt{3}}{3}$ |
| $y = 23.1t - 5t^2$ | B1FT | Use $s = ut + \frac{at^2}{2}$ vertically. FT candidates half $V$ but not $V = 40$ used |
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## Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{23.1x}{40} - \frac{5x^2}{1600}$ | M1 | Attempt to eliminate $t$ by substituting $t = \frac{x}{40}$ into answer to part (i) |
| $y = 0.577x - \frac{x^2}{320}$ or $y = 0.577x - 0.003125x^2$ | A1 | |
---2 A small ball is projected from a point $O$ on horizontal ground at an angle of $30 ^ { \circ }$ above the horizontal. At time $t \mathrm {~s}$ after projection the horizontal and vertically upwards displacements of the ball from $O$ are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively. It is given that $x = 40 t$.\\
(i) Calculate the initial speed of the ball, and express $y$ in terms of $t$.\\
(ii) Hence find the equation of the trajectory of the ball.\\
\hfill \mbox{\textit{CAIE M2 2019 Q2 [5]}}