| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Maximum speed on incline vs horizontal |
| Difficulty | Standard +0.3 This is a straightforward application of the power-force-velocity relationship (P=Fv) with standard mechanics principles. Part (a) uses P=Rv at maximum speed, (b) applies F=ma with power equation, and (c) involves resolving forces on an incline. All steps are routine for Further Mechanics students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03f Weight: W=mg6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | D = 90000/25 = 3600 |
| Answer | Marks |
|---|---|
| 3600 (N) | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | Use of P = Fv to find ‘driving |
| Answer | Marks |
|---|---|
| (b) | 6000a = 60000/10 – “3600” |
| a = 2400/6000 = 0.4 (ms–2 ) | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 1.1 | Using NII with correct number of |
| Answer | Marks |
|---|---|
| (c) | 6000g sinθ |
| Answer | Marks |
|---|---|
| => θ = awrt 1.3° or 0.0230 rads | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | DR |
Question 3:
3 | (a) | D = 90000/25 = 3600
At constant speed, R = D so resistance is
3600 (N) | M1
A1
[2] | 3.3
1.1 | Use of P = Fv to find ‘driving
force’
Using fact that a = 0 so the two
forces balance. Must indicate
they’ve calculated a different
force to D.
(b) | 6000a = 60000/10 – “3600”
a = 2400/6000 = 0.4 (ms–2 ) | M1
A1
[2] | 3.4
1.1 | Using NII with correct number of
terms, dimensionally consistent.
Correct power of 60000 soi.
(c) | 6000g sinθ
6000g sinθ + 90000/40 – 3600 = 0
sinθ = 9/392 (0.022959...)
=> θ = awrt 1.3° or 0.0230 rads | B1
M1
A1
[3] | 1.1
3.4
1.1 | DR
Correct weight component
Three terms, condone sign errors
and m seen. Do not condone
60000 in place of 90000.
Condone 178.7° or 3.12 rads.
Note 1dp for 1.3 degrees is good
enough (typically 1.32 or 1.31)3 The mass of a truck is 6000 kg and the maximum power that its engine can generate is 90 kW . In a model of the motion of the truck it is assumed that while it is moving the total resistance to its motion is constant.
At first the truck is driven along a straight horizontal road. The greatest constant speed that it can be driven at when it is using maximum power is $25 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of the resistance to motion.
The truck is being driven along the horizontal road with the engine working at 60 kW .
\item Find the acceleration of the truck at the instant when its speed is $10 \mathrm {~ms} ^ { - 1 }$.
The truck is now driven down a straight road which is inclined at an angle $\theta$ below the horizontal. The greatest constant speed that the truck can be driven at maximum power is $40 \mathrm {~ms} ^ { - 1 }$.
\item Determine the value of $\theta$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2024 Q3 [7]}}