| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Up and down hill: two equations |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring application of standard formulas (P=Fv, F=ma, resolving forces on an incline) across two parts. The steps are routine: find resistance force from horizontal motion, use it with acceleration down the slope to verify consistency, then apply power equation going uphill. No novel insight required, just systematic application of A-level mechanics principles. |
| Spec | 3.02d Constant acceleration: SUVAT formulae6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | Driving force = resistance = P1 |
| 2 | B1 | 3.3 |
| Answer | Marks |
|---|---|
| 2 | M1 |
| B1 | 3.4 |
| 1.1 | Attempt at N2L, correct number of |
| Answer | Marks | Guidance |
|---|---|---|
| P=6624.568 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | 1 6 0v 0 0 − 1 4 0 0 g s i n 7 − 6 6 2 4 .5 6 8 = 0 | |
| 1 2 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| v = 7 .1 9 3 9 | A1 | 1.1 |
Question 2:
2 | (a) | Driving force = resistance = P1
2 | B1 | 3.3 | soi
1 4 0 0 g s i n 7 − P1 = 1 4 0 0 0 .8
2 | M1
B1 | 3.4
1.1 | Attempt at N2L, correct number of
terms and weight resolved (but
condone sign errors and sin/cos
confusion)
Correct weight component if seen
in (a) or (b)
P=6624.568 | A1 | 1.1
[4]
(b) | 1 6 0v 0 0 − 1 4 0 0 g s i n 7 − 6 6 2 4 .5 6 8 = 0
1 2 | M1 | 3.3 | Attempt at N2L with zero
acceleration – correct number of
terms (condone sign errors)
FT their value of P or resistive
force from (a)
v = 7 .1 9 3 9 | A1 | 1.1
[2]2 A car of mass 1400 kg , travels along a straight horizontal road AB , after which it descends a hill BC inclined at a constant angle of $7 ^ { \circ }$ to the horizontal (see diagram). $\mathrm { A } , \mathrm { B }$ and C all lie in the same vertical plane. Throughout the entire journey, the total resistance to the car's motion is constant.\\
\includegraphics[max width=\textwidth, alt={}, center]{cad8805d-59f6-4ed2-81f4-9e8c749461f5-3_232_1227_392_251}
Between A and B, the car moves at a constant speed of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and the power developed by the car is a constant $P \mathrm {~W}$. When the car reaches B , the engine is switched off and the car travels down a line of greatest slope from $B$ to $C$ with an acceleration of $0.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. The resistance to motion is unchanged.
\begin{enumerate}[label=(\alph*)]
\item Determine the value of $P$.
When the car reaches C it turns round and travels back up the hill towards B at a constant speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The power developed by the car between C and B is a constant 16 kW . The resistance to motion is unchanged.
\item Determine the value of $v$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2023 Q2 [6]}}