| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.3 This is a projectile motion on an inclined plane problem requiring coordinate transformation and multi-step calculations. Part (a) involves resolving velocities in rotated coordinates and finding range on slope (standard Further Maths technique, ~6-7 steps). Part (b) adds collision analysis to find maximum rebound angle. While systematic, it requires careful geometric reasoning and is more involved than typical A-level mechanics, but follows established Further Maths patterns without requiring novel insight. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05b Sine and cosine rules: including ambiguous case3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (a) | x g s i n , y g c o s = − = − |
| y ( 2 0 s i n ) t 0 .5 ( y ) t 2 = + | M1* | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 0 ( 35 ) t − 12 g ( 11 23 ) t 2 = 0 t = ... | M1dep* | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| t=2.6 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| y 2 0 s i n ( g c o s ) t 2 0 35 1 0 11 23 t = + − = − | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| x = 6 , y = − 1 2 | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | M1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| OA = 28.6 (m) | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | Maximum value of e 1 = | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| x 1 2 | M1 | 3.4 |
| 8 6 = | A1 | 3.2a |
Question 9:
9 | (a) | x g s i n , y g c o s = − = − | B1 | 1.1 | Possibly implied by later working
y ( 2 0 s i n ) t 0 .5 ( y ) t 2 = + | M1* | 3.3 | Applying s=ut+0.5at2perpendicular to the plane –
allow sin/cos confusion with the component of 20
(but not in terms of )
2 0 ( 35 ) t − 12 g ( 11 23 ) t 2 = 0 t = ... | M1dep* | 3.4 | Sets y = 0 and solves for t (acceleration must have
been a component of either s i n or c o s )
t=2.6 | A1 | 1.1 | 2 6
Allow (provided g not later used as 9.8). Could be
g
implied by correct values for the velocity components
(provided g = 10) or correct value of OA
( ) ( )
x 2 0 c o s ( g s i n ) t 2 0 45 1 0 51 t = + − = −
3 | M1 | 3.3 | Applying v=u+at(oe) parallel to the plane – allow
sin/cos confusion (but correct use of theta/alpha for
those components) – allow with their t or just t but
must have substituted trigonometric values
( ) ( )
y 2 0 s i n ( g c o s ) t 2 0 35 1 0 11 23 t = + − = − | M1 | 3.3 | Applying v=u+at(oe) perpendicular to the plane –
allow sin/cos confusion (but correct use of theta/alpha
for those components) – allow with their t or just t but
must have substituted trigonometric values
x = 6 , y = − 1 2 | A1 | 2.2a | Allow 12
O A ( 2 0 c o s ) t 12 ( g s i n ) t 2 2 0 ( 45 ) t 5 ( 51 ) t 2 = + − = −
3 | M1 | 3.4 | Applying s=ut+0.5at2parallel to the plane – allow
with their t or just t (or equivalent e.g.
x2 =(20cos)2 +2(−gsin)(OA)) but must have
substituted trigonometric values but allow sign errors
OA = 28.6 (m) | A1 | 2.2a | If g missing from acceleration terms can score first
two M marks only
If using g = 9.8 then first and last A marks cannot
be awarded (so max. 7/9 – look out for t = 2.653…
and OA = 29.183…)
[9]
(b) | Maximum value of e 1 = | B1 | 3.1b | Indication that the greatest angle occurs when the
collision is elastic (award if using the values for x,y
found in part (a))
y 5
t a n 1 t a n 1 = − + −
x 1 2 | M1 | 3.4 | Correct method for finding - all values substituted
8 6 = | A1 | 3.2a | For reference: 86.05481… - allow awrt 86 (but not
from incorrect working) or awrt 1.50 (radians)
[3]In this question take $g = 10$.
A small ball P is projected with speed $20 \text{ m s}^{-1}$ at an angle of elevation of $(\alpha + \theta)$ from a point O at the bottom of a smooth plane inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac{5}{12}$ and $\tan \theta = \frac{3}{4}$. The ball subsequently hits the plane at a point A, where OA is a line of greatest slope of the plane, as shown in the diagram.
\includegraphics{figure_9}
\begin{enumerate}[label=(\alph*)]
\item Determine the following, in either order.
\begin{itemize}
\item The components of the velocity of P, parallel and perpendicular to the plane, immediately before P hits the plane at A.
\item The distance OA.
\end{itemize}
[9]
\end{enumerate}
After P hits the plane at A it continues to move away from O. Immediately after hitting the plane at A the direction of motion of P makes an angle $\beta$ with the horizontal.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Determine the maximum possible value of $\beta$, giving your answer to the nearest degree. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2023 Q9 [12]}}