| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Session | Specimen |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Difficulty | Standard +0.3 This is a standard Further Mechanics question involving power-force-velocity relationships and energy methods. Part (i) is straightforward P=Fv. Part (ii) uses work-energy principle with given answer to show, requiring careful accounting of energy terms but following a standard template. Part (iii) requires checking if available power exceeds required power during motion, involving forces on an incline and kinematics. All techniques are routine for Further Mechanics students, though the multi-step nature and careful bookkeeping elevate it slightly above average difficulty. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (i) | 5000(cid:32)F(cid:117)2.5 |
| F = 2000. Now R = F so R is 2000 N | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 1.1 | M |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (ii) | (A) |
| Answer | Marks |
|---|---|
| so 71750 J AG | M1 |
| Answer | Marks |
|---|---|
| [4] | I |
| Answer | Marks |
|---|---|
| 1.1 | Use of work-energy equation with |
| Answer | Marks | Guidance |
|---|---|---|
| correct | KE is kinetic energy | |
| 8 | (ii) | (B) |
| Answer | Marks |
|---|---|
| F(cid:32)2609.09...so resistance is 2610N (3 s.f.) | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| 1.1 | Work done is Fd | |
| 8 | (iii) | 3.252 (cid:32)32 (cid:14)200a |
| Answer | Marks |
|---|---|
| not achieve16.2 kW | M1 |
| Answer | Marks |
|---|---|
| [7] | 3.1b |
| Answer | Marks |
|---|---|
| 3.2a | Use suvat to find acceleration |
Question 8:
8 | (i) | 5000(cid:32)F(cid:117)2.5
F = 2000. Now R = F so R is 2000 N | M1
A1
[2] | 3.4
1.1 | M
Use of P(cid:32)Fv
8 | (ii) | (A) | 1 1
(cid:117)6000(cid:117)32 (cid:16) (cid:117)6000(cid:117)2.52
2 2
(cid:32)8000(cid:117)10(cid:16)W
P
W (cid:32)80000(cid:16)27000(cid:14)18750 = 71750,
so 71750 J AG | M1
EB1
B1
E1
[4] | I
C
3.3
1.1
3.4
1.1 | Use of work-energy equation with
all terms
One KE term correct
Work done by driving force
correct | KE is kinetic energy
8 | (ii) | (B) | S
2.5(cid:14)3
Distance travelled = (cid:117)10(cid:32)27.5m
2
Hence F(cid:117)27.5(cid:32)71750
F(cid:32)2609.09...so resistance is 2610N (3 s.f.) | B1
M1
A1
[3] | 3.1b
2.2a
1.1 | Work done is Fd
8 | (iii) | 3.252 (cid:32)32 (cid:14)200a
3.252 (cid:16)32
Acceleration = = 0.0078125
200
N2L up slope: D(cid:16)2000(cid:16)6000gsin(cid:68)(cid:32)6000a
D = 4986.875
Max power required = 4986.875 × 3.25
= 16.2 kW
Max power available is now 16 kW so could
not achieve16.2 kW | M1
A1
M1
A1
M1
A1
E1
[7] | 3.1b
1.1
3.4
1.1
3.2a
1.1
3.2a | Use suvat to find acceleration
N
N2L used
E
Use of max speed
MA tractor has a mass of 6000 kg. When developing a power of 5 kW, the tractor is travelling at a steady speed of 2.5 m s$^{-1}$ across a horizontal field.
\begin{enumerate}[label=(\roman*)]
\item Calculate the magnitude of the resistance to the motion of the tractor. [2]
\end{enumerate}
The tractor comes to horizontal ground where the resistance to motion is different. The power developed by the tractor during the next 10 s has an average value of 8 kW. During this time, the tractor accelerates uniformly from 2.5 m s$^{-1}$ to 3 m s$^{-1}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item \begin{enumerate}[label=(\Alph*)]
\item Show that the work done against the resistance to motion during the 10 s is 71 750 J. [4]
\item Assuming that the resistance to motion is constant, calculate its value. [3]
\end{enumerate}
\end{enumerate}
The tractor can usually travel up a straight track inclined at an angle $\alpha$ to the horizontal, where $\sin\alpha = \frac{1}{20}$, while accelerating uniformly from 3 m s$^{-1}$ to 3.25 m s$^{-1}$ over a distance of 100 m against a resistance to motion of constant magnitude of 2000 N.
The tractor develops a fault which limits its maximum power to 16kW.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Determine whether the tractor could now perform the same motion up the track. [You should assume that the mass of the tractor and the resistance to motion remain the same.] [7]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major Q8 [16]}}