| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Topic | Projectiles |
| Type | Projection from elevated point - angle above horizontal |
| Difficulty | Standard +0.3 This is a two-part mechanics problem combining motion on an inclined plane with projectile motion. Part (i) involves standard SUVAT equations on a slope with constant acceleration (g sin 30°). Part (ii) requires resolving the velocity at A into horizontal and vertical components, then applying projectile motion equations. While it requires multiple steps and careful coordinate work, the techniques are all standard A-level mechanics procedures with no novel insight required. The problem is slightly easier than average because the calculations are straightforward (nice numbers like 30° and speeds involving √3) and the method is routine for students who have practiced combined mechanics problems. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
Total horizontal distance travelled from O is given by *their* horizontal distance \(OA\) + (*their* horizontal velocity at \(A\)) \(\times\) (*their* time of flight) — M1
**(i)** Acceleration parallel to the slope is $-g\cos60 = -5$ ms$^{-2}$ — B1
Use 'suvat': $10 = 20t + 0.5(-5)t^2$ — M1 A1 *(Any appropriate 'suvat' used. Correct equation.)*
Solve quadratic $t^2 - 8t + 4 = 0$
Obtain $4 - \sqrt{12}$ $(= 4 - 2\sqrt{3} = 0.536$ seconds$)$ — A1 *(Correct outcome.)*
Initial speed of projectile is $20 - 5(4 - 2\sqrt{3}) = 10\sqrt{3}$ $(= 17.32$ ms$^{-1})$ — M1 A1 [6] *(2nd appropriate 'suvat' used. Correct outcome.)*
**(ii)** For the vertical motion, the particle strikes the ground when
$-5 = 10\sqrt{3}\sin30\, t + 0.5(-10)t^2$ — M1 *(Condone sin/cos confusion.)*
$t^2 - \sqrt{3}\,t - 1 = 0$ — M1
Solve quadratic $t = \frac{\sqrt{3} + \sqrt{7}}{2}$
Obtain positive solution $= 2.189$ s — A1
Total horizontal distance travelled from O is given by *their* horizontal distance $OA$ + (*their* horizontal velocity at $A$) $\times$ (*their* time of flight) — M1
$= 10\cos30$ — B1
$+ 10\sqrt{3}\cos30 \times 2.189$ — B1 *(ft their $t$)*
$= 41.5$ metres — A1 [7] *(c.a.o.)*
**Total: [13]**11 A particle $P$ of mass 2 kg can move along a line of greatest slope on the smooth surface of a wedge which is fixed to the ground. The sloping face $O A$ of the wedge has length 10 metres and is inclined at $30 ^ { \circ }$ to the horizontal (see Fig. 1). $P$ is fired up the slope from the lowest point $O$, with an initial speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f0c32e07-f3a0-4d58-bd00-c266177ceaac-5_295_1529_484_310}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
(i) Find the time taken for $P$ to reach $A$ and show that the speed of $P$ at $A$ is $10 \sqrt { 3 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
After $P$ has reached $A$ it becomes a projectile (see Fig. 2).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f0c32e07-f3a0-4d58-bd00-c266177ceaac-5_424_1533_1123_306}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
(ii) Find the total horizontal distance travelled by $P$ from $O$ when it hits the ground.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2012 Q11 [13]}}