| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Standard +0.3 Part (i) is standard bookwork requiring recall of projectile equations and elimination of parameter t. Part (ii) is a routine range equation problem with two solutions from a quadratic. Part (iii) requires minimal physical reasoning. This is slightly easier than average as it's mostly standard technique with no novel problem-solving required. |
| Spec | 3.02i Projectile motion: constant acceleration model |
**Question 8**
**(i)**
$x = Ut\cos\theta$ — B1
$y = Ut\sin\theta - \frac{1}{2}gt^2$ — B1 (Allow $g = 9.8$)
$t = \frac{x}{U\cos\theta}$ — M1 (Make $t$ the subject of $x$ equation and substitute)
$\therefore y = U\left(\frac{x}{U\cos\theta}\right)\sin\theta - \frac{1}{2}g\left(\frac{x}{U\cos\theta}\right)^2$
$= x\tan\theta - \frac{gx^2}{2U^2\cos^2\theta}$ — A1 **[4]**
Accept any correct form/unsimplified.
**(ii)** $y = 0$ and $x \neq 0$ gives $x = \frac{U^2}{g}\sin 2\theta$ — M1
(Set $y = 0$ and attempt to make $x$ or $\sin 2\theta$ the subject. Allow other equivalent methods e.g. by solving a quadratic $(t^2 - 4t + 1 = 0)$ in $\tan\theta$ $(= 2 \pm \sqrt{3})$.)
$\therefore \sin 2\theta = \frac{gx}{U^2} = \frac{10 \times 45}{30^2} = 0.5$ — A1
(Substitute and obtain 0.5 (or $\tan\theta$) correctly.)
This has 2 solutions so there are 2 trajectories. — B1
(Require an explicit statement to this effect.)
$\therefore \theta = 15°$ or $75°$ — A1 **[4]** (Both correct.)
**(iii)** $\theta = 15°$ is fast (and low). — B1 ("Advantage" of one. (ft (ii)))
$\theta = 75°$ is high (more likely to clear any obstacles). — B1 **[2]**
("Advantage" of the other. (ft (ii)))
SC B1 only for just "high" and "low". Allow other reasonable "advantages".
**Total: [10]**8 A particle is projected from a point $O$ with initial speed $U$ at an angle $\theta$ above the horizontal. At time $t$ after projection the position of the particle is $( x , y )$ relative to horizontal and vertical axes through $O$.\\
(i) Write down expressions for $x$ and $y$ at time $t$. Hence derive the cartesian equation of the trajectory of the particle.\\
(ii) A player in a cricket match throws the ball with speed $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to another player who is 45 metres away. Assume that the players throw and catch the ball at the same height above the ground. Show that there are two possible trajectories and find their respective angles of projection. [4]\\
(iii) Describe briefly one advantage of each trajectory.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2013 Q8 [10]}}