| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find steady/maximum speed given power |
| Difficulty | Standard +0.3 Part (a) is a standard power-speed relationship question requiring P=Fv at terminal velocity (routine 2-mark calculation). Part (b) requires energy conservation with multiple components (KE, PE, work against resistance, work done by engine) but follows a standard framework for Further Mechanics questions. The calculation is multi-step but methodical with no novel insight required, making it slightly easier than average overall. |
| Spec | 6.02a Work done: concept and definition6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks |
|---|---|
| 5(a) | 18000 |
| Answer | Marks |
|---|---|
| v = 30 (m s−1) | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | Use of P = Dv and using N2L horizontally with correct |
| Answer | Marks |
|---|---|
| 5(b) | ±0.5×850×(182 −152) |
| Answer | Marks |
|---|---|
| d =168(m) | B1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Correct expression for the change in KE – need not be |
Question 5:
--- 5(a) ---
5(a) | 18000
−600=0
v
v = 30 (m s−1) | M1
A1
[2] | 3.3
1.1 | Use of P = Dv and using N2L horizontally with correct
number of terms (condone 18 for 18 000) – allow sign errors
--- 5(b) ---
5(b) | ±0.5×850×(182 −152)
Change in KE = (=±42 075)
Work done by the car = 18000×10 (= 180 000)
Change in PE = ±850g× 1 ×d (=±208.25d)
40
0.5(850)(182 −152)=18000×10−850g× 1 ×d −103000
40
d =168(m) | B1
B1
B1
M1
A1
[5] | 1.1
1.1
1.1
3.4
1.1 | Correct expression for the change in KE – need not be
simplified e.g. ±(137700−95625)- both terms could appear
on different sides of their work-energy equation – using 30
(for one of the velocities) should not be treated as a MR
Need not be simplified
Where d is the length of the hill – allow
±850g×d×sin(1.4(325...))
Use of work-energy principle – dimensionally consistent
with the correct number of relevant terms. Allow sin/cos
mix on PE term and sign errors. If using 850gdfor the PE
then M0 but do watch out for those who multiply their final
answer by 40 (oe) as this could then imply the correct
change in PE (and possibly this M mark too). Using 30 for
one of the velocities is M0.
167.7… awrt 168 (but not 170) www - any attempts to use
N2L and SUVAT only scores no marksA car of mass 850 kg is travelling along a straight horizontal road. The power developed by the car is constant and is equal to 18 kW. There is a constant resistance to motion of magnitude 600 N.
\begin{enumerate}[label=(\alph*)]
\item Find the greatest steady speed at which the car can travel. [2]
\end{enumerate}
Later in the journey, while travelling at a speed of $15 \text{ m s}^{-1}$, the car comes to the bottom of a straight hill which is inclined at an angle of $\sin^{-1}\left(\frac{1}{40}\right)$ to the horizontal.
The power developed by the car remains constant at 18 kW. The magnitude of the resistance force is no longer constant but changes such that the total work done against the resistance force in ascending the hill is 103 000 J. The car takes 10 seconds to ascend the hill and at the top of the hill the car is travelling at $18 \text{ m s}^{-1}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Determine the distance the car travels from the bottom to the top of the hill. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2024 Q5 [7]}}