| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.2 This is a projectile motion problem on an inclined plane requiring coordinate system setup and resolution of motion in two directions. While it involves multiple steps (resolving initial velocity, finding height above inclined plane, and range on slope), the techniques are standard for Further Maths mechanics. The incline angle is given explicitly, making this more straightforward than problems requiring geometric insight or optimization. |
| Spec | 3.02i Projectile motion: constant acceleration model |
(i) Solution 1: **V, H**
$y = 30t\sin50° - 5t^2$; $x = 30t\cos50°$
$[y = 26.2]\ \ x = 48.2$
Height of slope $= x\tan10° = 8.5$
Difference $= \mathbf{17.7}$
- M1A1: Both equations attempted; both correct
- A1: Correct value of $x$; SC: If M0, $y = 26.2$ gets B1
- M1: Find height
- A1: Correct answer $[17.703]$
**[5]**
Solution 2: $\parallel, \perp$
$y = 30t\sin40° - 5t^2\cos10° = 17.43$
Height above field $= y \div \cos10° = \mathbf{17.7}$
- M1A1, A1, M1, A1
(ii) Solution 1: **V, H**
$30t\sin50° - 5t^2 = 30t\cos50°\tan10°$
$t = 3.916$
$PX = 30t\cos50° \div \cos10°$
$X = \mathbf{76.7}$
- M1: Put $y = x\tan10°$ and solve
- A1: Correct value of $t$, can be implied
- M1: Calculate $x \div \cos10°$ or $y \div \sin10°$
- A1: $X$, a.r.t. 76.7 $[76.684]$
**[4]**
Solution 2: $\parallel, \perp$
$Y = 30t\sin40° - 5t^2\cos10°$; $Y = 0$ at $t = 3.916$
$X = 30t\cos40° - 5t^2\sin10°$; $PX = \mathbf{76.7}$
- M1, A1, A1, A1
Solution 3: **traj**
$Y = X\tan50 - \dfrac{10X^2\sec^250}{2\times30^2} = x\tan10°$
$PX = 180(\tan50 - \tan10)\cos^250 = \mathbf{76.7}$
- M1: Use trajectory equation; A1: All correct
- M1: Solve for $X$; A1: $X$, a.r.t. 76.7 $[76.684]$13 A cricket ball is hit from a point $P$ on a sloping field. The initial velocity of the ball is $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $40 ^ { \circ }$ above the field, which under the path of the ball slopes upwards at $10 ^ { \circ }$ to the horizontal. Air resistance is to be ignored.\\
(i) Find the vertical height of the ball above the field after 2.5 seconds.\\
(ii) The ball lands on the field at the point $X$. Find the distance $P X$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q13 [9]}}