| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.3 This is a standard projectile motion problem requiring routine application of SUVAT equations with given angle and time to maximum height. Part (i) uses v=u+at to find U and then standard height formula; part (ii) applies range formula. The sin θ = 12/13 setup is straightforward (giving cos θ = 5/13), and all steps follow directly from memorized formulas without requiring problem-solving insight or novel approaches—slightly easier than average due to its formulaic nature. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| At max height \(v_y = 0\) i.e. \(U \sin \theta - 10t = 0\) | M1 | Use of condition for max height. |
| \(\therefore U \times \frac{12}{13} - 10 \times 2.4 = 0\) | A1 | c.a.o. |
| \(\therefore U = 26\) (ms\(^{-1}\)) | A1 | c.a.o. |
| Max \(y = Ut \sin \theta - \frac{1}{2}gt^2\) | M1 | Use of formula for \(y\). |
| \(y = 26 \times 2.4 \times \frac{12}{13} - 5 \times 2.4^2\) | M1 | Ft c's \(U\). |
| \(\therefore y = 28.8\) (m) | A1 | [4] |
| Answer | Marks |
|---|---|
| \(\cos \theta = \frac{5}{13}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| At horizontal range \(t = 4.8\) (sec) | B1 | |
| \(R = Ut \cos \theta\) | M1 | Use of formula for \(x\) |
| \(\therefore R = 26 \times 4.8 \times \frac{5}{13}\) | A1 | Ft c's \(U\). |
| \(\therefore R = 48\) (m) | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin 2\theta = 2 \sin \theta \cos \theta = 2 \times \frac{12}{13} \times \frac{5}{13}\) | B1 | |
| \(\therefore R = \frac{U^2}{g} \sin 2\theta\) | M1 | Use of formula for \(R\). |
| \(\therefore R = \frac{26^2}{10} \times \frac{2 \times 12 \times 5}{13^2}\) | M1 | Use of formula for \(R\). |
| \(\therefore R = 48\) (m) | A1 | Ft c's \(U\). |
### (i)
At max height $v_y = 0$ i.e. $U \sin \theta - 10t = 0$ | M1 | Use of condition for max height.
$\therefore U \times \frac{12}{13} - 10 \times 2.4 = 0$ | A1 | c.a.o.
$\therefore U = 26$ (ms$^{-1}$) | A1 | c.a.o.
Max $y = Ut \sin \theta - \frac{1}{2}gt^2$ | M1 | Use of formula for $y$.
$y = 26 \times 2.4 \times \frac{12}{13} - 5 \times 2.4^2$ | M1 | Ft c's $U$.
$\therefore y = 28.8$ (m) | A1 | [4]
Allow M1A0 for sin/cos error here only; but absence is M0.
### (ii)
$\cos \theta = \frac{5}{13}$ | B1 |
**ALT version 1:**
At horizontal range $t = 4.8$ (sec) | B1 |
$R = Ut \cos \theta$ | M1 | Use of formula for $x$
$\therefore R = 26 \times 4.8 \times \frac{5}{13}$ | A1 | Ft c's $U$.
$\therefore R = 48$ (m) | [4]
**ALT version 2:**
$\sin 2\theta = 2 \sin \theta \cos \theta = 2 \times \frac{12}{13} \times \frac{5}{13}$ | B1 |
$\therefore R = \frac{U^2}{g} \sin 2\theta$ | M1 | Use of formula for $R$.
$\therefore R = \frac{26^2}{10} \times \frac{2 \times 12 \times 5}{13^2}$ | M1 | Use of formula for $R$.
$\therefore R = 48$ (m) | A1 | Ft c's $U$. | [4]
---A particle is projected with speed $U \text{ m s}^{-1}$ at an angle $\theta$ above the horizontal, where $\sin \theta = \frac{12}{13}$, and reaches its maximum height after $2.4$ seconds.
\begin{enumerate}[label=(\roman*)]
\item Find $U$ and the maximum height reached by the particle. [4]
\item Find the horizontal range of the particle. [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2016 Q8 [8]}}