| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 7 |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.2 This is a standard inclined plane projectile problem requiring resolution of motion parallel and perpendicular to the plane, then algebraic manipulation to show the given ratio. While it involves multiple steps and careful coordinate geometry, the technique is well-established in Further Maths syllabi and the result is given to prove, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 3.02i Projectile motion: constant acceleration model |
**Question 4**
With point of projection as origin, equation of trajectory is:
$$y = x\sqrt{3} - \frac{gx^2\sec^260°}{2V^2} = \sqrt{3}x - \frac{2gx^2}{V^2}$$ M1A1
Equation of upward plane is $y = \dfrac{1}{\sqrt{3}}x$ B1
On landing: $\dfrac{1}{\sqrt{3}}x = \sqrt{3}x - \dfrac{2gx^2}{V^2} \Rightarrow x = 3x - \dfrac{2\sqrt{3}gx^2}{V^2}$ M1
$$0 = 2x\left(1 - \frac{\sqrt{3}gx}{V^2}\right) \Rightarrow x = \frac{V^2}{\sqrt{3}g} \quad (x \neq 0)$$ A1
$$R_1 = \frac{V^2}{\sqrt{3}g} \times \frac{2}{\sqrt{3}} = \frac{2V^2}{3g}$$ A1
Equation of downward plane is $y = -\dfrac{1}{\sqrt{3}}x$
On landing: $-\dfrac{1}{\sqrt{3}}x = \sqrt{3}x - \dfrac{2\sqrt{3}gx^2}{V^2} \Rightarrow -x = 3x - \dfrac{2\sqrt{3}gx^2}{V^2}$ M1
$$0 = 2x\left(2 - \frac{\sqrt{3}gx}{V^2}\right) \Rightarrow x = \frac{2V^2}{\sqrt{3}g} \quad (x \neq 0)$$ A1
$$R_2 = \frac{2V^2}{\sqrt{3}g} \times \frac{2}{\sqrt{3}} = \frac{4V^2}{3g} \Rightarrow R_2 = 2R_1$$ (AG) A1 **[9 marks]**4 A particle is projected with velocity $V$, at an angle of elevation of $60 ^ { \circ }$ to the horizontal, from a point on a plane inclined at an angle of $30 ^ { \circ }$ to the horizontal. The path of the particle is in a vertical plane through a line of greatest slope. If $R _ { 1 }$ and $R _ { 2 }$ are the respective ranges when the particle is projected up the plane and down the plane, show that
$$R _ { 2 } = 2 R _ { 1 }$$
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q4 [7]}}