| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - projected vertically upward |
| Difficulty | Challenging +1.8 This Further Mechanics question requires setting up and manipulating differential equations for variable force motion, completing the square, integrating with arctan, and critically evaluating a model against experimental data. The conceptual demand of understanding why the falling motion integral differs (sign conventions, force directions) and the multi-step nature (deriving the DE, integrating, comparing with bounds, analyzing the falling case) place it well above average difficulty, though the individual techniques are standard for Further Maths students. |
| Spec | 1.08d Evaluate definite integrals: between limits6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) | (a) |
| Answer | Marks |
|---|---|
| (cid:16)9(cid:14)9.8 (cid:14)0.8 | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 2.1 | Divide by m and rearrange |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) | (b) |
| Answer | Marks |
|---|---|
| time recorded by A. | M1 |
| Answer | Marks |
|---|---|
| [5] | 3.1a |
| Answer | Marks |
|---|---|
| 2.2b | n |
| Answer | Marks |
|---|---|
| likely to be correct | Or use |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (ii) | e.g. student C’s claim is incorrect as they have |
| Answer | Marks |
|---|---|
| 0 | B1 |
| Answer | Marks |
|---|---|
| [3] | 2.2a |
| Answer | Marks |
|---|---|
| 3.3 | For explaining that the differential |
Question 5:
5 | (i) | (a) | dv (cid:11) (cid:16)6mv(cid:14)mv2(cid:12)
m (cid:32)(cid:16)mg(cid:16)
dt
dt 1
(cid:32)(cid:16)
dv v2 (cid:16)6v(cid:14)9.8
1 1
(cid:32)(cid:16) (cid:32)(cid:16)
(cid:11)v(cid:16)3(cid:12)2 (cid:11)v(cid:16)3(cid:12)2
(cid:16)9(cid:14)9.8 (cid:14)0.8 | B1
M1
E1
[3] | 3.3
1.1
2.1 | Divide by m and rearrange
AG Completing the square in the
denominator
5 | (i) | (b) | dv
t(cid:32)(cid:16)(cid:179)
(cid:11)v(cid:16)3(cid:12)2
(cid:14)0.8
0
(cid:180) dv
t (cid:32)(cid:16) , and
1 (cid:181) (cid:11)v(cid:16)3(cid:12)2
(cid:182) (cid:14)0.8
8.25
0
(cid:180) dv
t (cid:32)(cid:16)
2 (cid:181) (cid:11)v(cid:16)3(cid:12)2
(cid:182) (cid:14)0.8
8.75
t (cid:32)2.9997903...
1 p
S
t (cid:32)3.0159247...
2
2.9997...(cid:31)t(cid:31)3.0159... so B’s model for the
vast majority of speeds in the interval
(cid:11)8.5(cid:114)0.25(cid:12)ms-1 does produce a value of t
greater than 3 so it is extremely likely (but not
certain) that B’s model is consistent with the
time recorded by A. | M1
B1
e
A1
A1
E1
[5] | 3.1a
3.4
i
c
1.1
1.1
2.2b | n
e
Separation of variables
m
Using the limits correctly at least once
BC
BC
Indication that the non-exact nature of
the initial speed leads to a result that is
likely to be correct | Or use
1 1 (cid:167)x(cid:183)
(cid:179) dx(cid:32) tan(cid:16)1
(cid:168) (cid:184)
a2 (cid:14)x2 a (cid:169)a(cid:185)
5 | (ii) | e.g. student C’s claim is incorrect as they have
assumed that the differential equation for the
motion of the particle is the same in both
directions
In fact the differential equation for when the
particle is falling is given by
dv
mg(cid:16)((cid:16)6mv(cid:14)mv2)(cid:32)m
dt
1
(cid:180) 1
(cid:16) dv
(cid:181)
(cid:182)(cid:11)v(cid:16)3(cid:12)2
(cid:16)18.8
0 | B1
B1
B1
[3] | 2.2a
2.3
3.3 | For explaining that the differential
equation of motion is not the same in
each direction
For eithenr this corrected model or
stating that the two forces in the
veertical direction are now acting in
opposite directions
m5 A particle $P$ of mass $m \mathrm {~kg}$ is projected vertically upwards through a liquid. Student $A$ measures $P$ 's initial speed as $( 8.5 \pm 0.25 ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and they also record the time for $P$ to attain its greatest height above the initial point of projection as over 3 seconds.\\
In an attempt to model the motion of $P$ student $B$ determines that $t$ seconds after projection the only forces acting on $P$ are its weight and the resistance from the liquid. Student $B$ models the resistance from the liquid to be of magnitude $m v ^ { 2 } - 6 m v$, where $v$ is the speed of the particle.
\begin{enumerate}[label=(\roman*)]
\item (a) Show that $\frac { \mathrm { d } t } { \mathrm {~d} v } = - \frac { 1 } { ( v - 3 ) ^ { 2 } + 0.8 }$.\\
(b) Determine whether student $B$ 's model is consistent with the time recorded by $A$ for $P$ to attain its greatest height.
After attaining its greatest height $P$ now falls through the liquid. Student $C$ claims that the time taken for $P$ to achieve a speed of $1 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when falling through the liquid is given by
$$- \int _ { 0 } ^ { 1 } \frac { 1 } { ( v - 3 ) ^ { 2 } + 0.8 } \mathrm {~d} v$$
\item Explain why student $C$ 's claim is incorrect and write down the integral which would give the correct time for $P$ to achieve a speed of $1 \mathrm {~ms} ^ { - 1 }$ when falling through the liquid.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics Q5 [11]}}