| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Standard +0.3 This is a standard two-part projectiles question requiring application of well-known formulae (max height = u²sin²θ/2g, range = u²sin2θ/g) to find u and θ, then deriving the trajectory equation. While it involves simultaneous equations and algebraic manipulation, the approach is routine for M2 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u^2\sin^2\theta \div 2g = 10\) or \(\frac{1}{2}(u\sin\theta + 0)T = 10\) | B1 | Using maximum height |
| \(2u^2\sin\theta\cos\theta \div g = 40\) or \(u(2T)\cos\theta = 40\) or \(uT\cos\theta = 20\) | B1 | Using range (or half range) |
| \(\left[\dfrac{\sin^2\theta}{\sin\theta\cos\theta} = \dfrac{2g(10)}{g(40)\div2}\right]\) or \(\dfrac{\sin\theta}{\cos\theta} = \dfrac{2\times10}{40\div2}\) | M1 | For eliminating \(u^2\) or \(uT\) |
| \(\theta = 45\) | A1 | |
| \(u^2 = 20\times10 \div \frac{1}{2} \Rightarrow u = 20\) or \(u \div \sqrt{2} = gT\) and \(uT \div 2\sqrt{2} = 10 \Rightarrow u = 20\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = x\tan45° - gx^2 \div (2\times20^2\cos^245°)\) | M1 | For substituting for \(u\) and \(\theta\) in the general equation |
| \(y = x - x^2/40\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = kx(40-x)\) | M1 | For quadratic equation with roots \(x=0\) and \(x=40\) |
| \(10 = 400k\) | M1 | For using \(y=10\) when \(x=20\) |
| \(y = x - x^2/40\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1 = \tan\theta\) | M1 | For equating coefficients of \(x\) with that of general form |
| \(\theta = 45°\) | A1 | |
| \(-\dfrac{1}{40} = -\dfrac{10}{2u^2 \times 1/2}\) | M1 | For equating coefficients of \(x^2\) with general form with \(\theta=45°\) substituted |
| \(u = 20\) | A1 | [4] |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u^2\sin^2\theta \div 2g = 10$ or $\frac{1}{2}(u\sin\theta + 0)T = 10$ | B1 | Using maximum height |
| $2u^2\sin\theta\cos\theta \div g = 40$ or $u(2T)\cos\theta = 40$ or $uT\cos\theta = 20$ | B1 | Using range (or half range) |
| $\left[\dfrac{\sin^2\theta}{\sin\theta\cos\theta} = \dfrac{2g(10)}{g(40)\div2}\right]$ or $\dfrac{\sin\theta}{\cos\theta} = \dfrac{2\times10}{40\div2}$ | M1 | For eliminating $u^2$ or $uT$ |
| $\theta = 45$ | A1 | |
| $u^2 = 20\times10 \div \frac{1}{2} \Rightarrow u = 20$ or $u \div \sqrt{2} = gT$ and $uT \div 2\sqrt{2} = 10 \Rightarrow u = 20$ | A1 | **[5]** |
### Part (ii) — Method 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = x\tan45° - gx^2 \div (2\times20^2\cos^245°)$ | M1 | For substituting for $u$ and $\theta$ in the general equation |
| $y = x - x^2/40$ | A1 | **[2]** |
### Part (ii) — Method 2 (OR):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = kx(40-x)$ | M1 | For quadratic equation with roots $x=0$ and $x=40$ |
| $10 = 400k$ | M1 | For using $y=10$ when $x=20$ |
| $y = x - x^2/40$ | A1 | **[3]** |
### Part (i) — Alternate order method:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 = \tan\theta$ | M1 | For equating coefficients of $x$ with that of general form |
| $\theta = 45°$ | A1 | |
| $-\dfrac{1}{40} = -\dfrac{10}{2u^2 \times 1/2}$ | M1 | For equating coefficients of $x^2$ with general form with $\theta=45°$ substituted |
| $u = 20$ | A1 | **[4]** |
---4 A particle is projected from horizontal ground with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $\theta ^ { \circ }$ above the horizontal. The greatest height reached by the particle is 10 m and the particle hits the ground at a distance of 40 m from the point of projection. In either order,\\
(i) find the values of $u$ and $\theta$,\\
(ii) find the equation of the trajectory, in the form $y = a x - b x ^ { 2 }$, where $x \mathrm {~m}$ and $y \mathrm {~m}$ are the horizontal and vertical displacements of the particle from the point of projection.
\hfill \mbox{\textit{CAIE M2 2005 Q4 [7]}}