| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics B AS (Further Mechanics B AS) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.2 This is a standard projectile-on-inclined-plane problem from Further Mechanics. Part (i) is a 'show that' requiring resolution perpendicular to the plane and applying standard SUVAT equations—straightforward for students who have practiced this topic. Part (ii) requires resolution parallel to the plane to find range. While the inclined plane context adds complexity beyond basic projectiles, this is a textbook example with no novel insight required, making it moderately above average difficulty for A-level but routine for Further Maths students. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (i) | Take x and y horiz and vert; take X and Y parallel and |
| Answer | Marks |
|---|---|
| gcos30 | B1 |
| Answer | Marks |
|---|---|
| A1 | 1.1a |
| Answer | Marks |
|---|---|
| 2.2a | X not needed here |
| Answer | Marks | Guidance |
|---|---|---|
| x(cid:32)14cos70t | B1 | I |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | C | |
| B1 | B1 | |
| y(cid:32)xtan30 | M1 | Including substitution |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | A1 | A1 |
| Answer | Marks |
|---|---|
| g | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (ii) | Either |
| Answer | Marks |
|---|---|
| = 11.725136… so 11.7 m ( to 3 s. f.) | M1 |
| Answer | Marks |
|---|---|
| A1 | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | N |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | M |
| Answer | Marks | Guidance |
|---|---|---|
| gcos30 2 (cid:169) gcos30 (cid:185) | C | |
| M1 | M1 | I |
| (2.120649) | Or substitute numerical value for T |
| Answer | Marks | Guidance |
|---|---|---|
| so CD is 11.725136… so 11.7 m ( to 3 s. f.) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Question | AO1 | AO2 |
| 1i | 2 | 1 |
| 1ii | 4 | 0 |
| 1iii | 5 | 0 |
| 2i | 4 | 0 |
| 2ii | 0 | 0 |
| 3iA | 2 | 0 |
| 3iB | 1 | 0 |
| 3ii | 0 | 0 |
| 4i | 3 | 0 |
| 4ii | 0 | 0 |
| 4iii | 1 | 0 |
| 5i | 0 | 0 |
| 5ii | 2 | 0 |
| 5iii | 0 | 1 |
| 5iv | 0 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 6i | 3 | 0 |
| 6ii | 2 | 2 |
| 0 | 5 | |
| 6iii | 2 | 1 |
| 7i | 3 | 1 |
| 7ii | 3 | 0 |
| 1 | 0 | 4 |
| Totals | 37 | 6 |
Question 7:
7 | (i) | Take x and y horiz and vert; take X and Y parallel and
perp to slope
Method 1
Y (cid:32)(cid:16)gcos30 (and X (cid:32)(cid:16)gsin30)
In Y direction
1
Y (cid:32)14sin40t(cid:16) gcos30t2
2
Y = 0 gives
1
14sin40t(cid:32) gcos30t2
2
28sin40
T (cid:32) AG
gcos30 | B1
M1
A1
M1
A1 | 1.1a
3.4
1.1
1.1
2.2a | X not needed here
N
Must have correct signs and attempt to
Eresolve g
g need not be substituted until final ans
Must use Y not y.
M
Method 2
x(cid:32)14cos70t | B1 | I
1
y(cid:32)14sin70t(cid:16) gt2
2 | C
B1 | B1
y(cid:32)xtan30 | M1 | Including substitution
E
1
14sin70T (cid:16) gT2 (cid:32)14cos70T(cid:117)tan30
2 | A1 | A1 | Substitution correct | Substitution correct
P
28
so T (cid:32) (cid:11)sin70(cid:16)cos70tan30(cid:12)
g | A1
S
28
(cid:32) sin(70(cid:16)30)
gcos30
28sin40
T (cid:32) AG
gcos30
[5]
A1
7 | (ii) | Either
In time T, the particle travels a horizontal distance CQ
and a distance
CQ
CD up the slope where CD(cid:32)
cos30
CQ = 14cos70T
14cos70 28sin40
so CD = (cid:117)
cos30 9.8cos30
= 11.725136… so 11.7 m ( to 3 s. f.) | M1
B1
M1
A1 | 3.1b
1.1
1.1
1.1 | N
Complete method
E
Or substitute numerical value for T
(2.120649)
cao
or
1
Use X (cid:32)14cos40T (cid:16) (cid:117)9.8(cid:117)sin30T2
2 | M1 | M
Must attempt to resolve g in X direction
A1
2
28sin40 1 (cid:167)28sin40(cid:183)
(cid:32)14cos40(cid:117) (cid:16) (cid:117)9.8(cid:117)sin30(cid:168) (cid:184)
gcos30 2 (cid:169) gcos30 (cid:185) | C
M1 | M1 | I | I | Or substitute numerical value for T
(2.120649) | Or substitute numerical value for T
(2.120649)
so CD is 11.725136… so 11.7 m ( to 3 s. f.) | A1 | cao
[4]
Question | AO1 | AO2 | AO3(PS) | AO3(M) | Total
1i | 2 | 1 | 0 | 0 | 3
1ii | 4 | 0 | 0 | 0 | 4
1iii | 5 | 0 | 0 | 0 | 5
2i | 4 | 0 | 0 | 1 | 5
2ii | 0 | 0 | 0 | 1 | 1
3iA | 2 | 0 | 1 | 0 | 3
3iB | 1 | 0 | 1 | 0 | 2
3ii | 0 | 0 | 1 | 0 | 1
4i | 3 | 0 | 0 | 2 | 5
4ii | 0 | 0 | 0 | 1 | 1
4iii | 1 | 0 | 1 | 0 | 2
5i | 0 | 0 | 0 | 1 | 1
5ii | 2 | 0 | 0 | 2 | 4
5iii | 0 | 1 | 0 | 0 | 1
5iv | 0 | 0 | 0 | 1 | N
1
6i | 3 | 0 | 0 | 0 | 3
6ii | 2 | 2 | 1 | E
0 | 5
6iii | 2 | 1 | 1 | 0 | 4
7i | 3 | 1 | 0 | 1 | 5
7ii | 3 | 0 | M
1 | 0 | 4
Totals | 37 | 6 | 7 | 10 | 60
PPMMTT
Y415 Mark Scheme June 20XX
BLANK PAGE
N
E
M
I
C
E
P
S
14
PPMMTT
Y415 Mark Scheme June 20XX
BLANK PAGE
N
E
M
I
C
E
P
S
15
PPMMTT
Y415 Mark Scheme June 20XX
BLANK PAGE
N
E
M
I
C
E
P
S
167 A plane is inclined at $30 ^ { \circ }$ above the horizontal. A particle is projected up the plane from a point C on the plane with a velocity of $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $40 ^ { \circ }$ above a line of greatest slope of the plane. The particle hits the plane at D. See Fig. 7.
\begin{figure}[h]
\begin{center}
\includegraphics[max width=\textwidth]{a01b2e46-e213-4f20-bc2e-5852061d8b91-7_305_766_484_589}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Using the standard model for projectile motion, show that the time of flight, $T$, is given by
$$T = \frac { 28 \sin 40 ^ { \circ } } { g \cos 30 ^ { \circ } }$$
(ii) Calculate the distance CD.
\section*{END OF QUESTION PAPER}
}{www.ocr.org.uk}) after the live examination series.
If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.
For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.\\
OCR is part of the
\hfill \mbox{\textit{OCR MEI Further Mechanics B AS Q7 [9]}}