| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done against friction/resistance - inclined plane or slope |
| Difficulty | Moderate -0.3 This is a straightforward application of work-energy principles with clearly stated forces and standard mechanics formulas. Part (i) requires simple subtraction after calculating work done against rolling friction; part (ii) is a direct energy conservation calculation; part (iii) combines P=Fv with F=ma and resolving forces. All steps are routine for M2 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(455 = 0.01 \times 80 \times 9.8 \times \cos4° \times 12 + \text{WD}\) | M1 | Use of \(Fx\) |
| Rolling friction force correct (7.82); 12 not needed | B1 | |
| All correct terms in equation (allow sign errors) | A1 | |
| \(\text{WD} = 361\) J (3 s.f.) | A1 | cao; SC B1B1 for final answer 30.1 seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.5 \times 80 \times v^2 - 0.5 \times 80 \times 2^2\) | M1 | Use of W-E equation. Must include GPE, at least one KE and the WD |
| Either KE term | B1 | |
| \(= 80 \times 9.8 \times 12 \times \sin4° - 455\) | B1 | GPE term (656.27) |
| All correct terms in equation (allow sign errors) | A1 | |
| \(v = 3.01\) ms\(^{-1}\) (3 s.f.) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using N2L with driving force \(S\): \(S - (15 + 0.01 \times 80 \times 9.8 \times \cos5°) - 80 \times 9.8 \times \sin5° = 80 \times 1.5\) | M1 | N2L with at most one force term missing |
| Both resistance terms seen (15 and 7.81) | B1 | |
| Condone wrong sign (68.33) | B1 | |
| All correct terms present; allow sign errors | A1 | May be implicit |
| \(S = 211.1402...\) | A1 | |
| \(405 = Sv\); \(v = 1.92\) ms\(^{-1}\) (3 s.f.) | M1, A1 | Use of Power \(= Sv\) with any \(S\) calculated using N2L; FT their \(S\); Note: missing one term in N2L can earn 4/7 |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $455 = 0.01 \times 80 \times 9.8 \times \cos4° \times 12 + \text{WD}$ | M1 | Use of $Fx$ |
| Rolling friction force correct (7.82); 12 not needed | B1 | |
| All correct terms in equation (allow sign errors) | A1 | |
| $\text{WD} = 361$ J (3 s.f.) | A1 | cao; SC B1B1 for final answer 30.1 seen |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.5 \times 80 \times v^2 - 0.5 \times 80 \times 2^2$ | M1 | Use of W-E equation. Must include GPE, at least one KE and the WD |
| Either KE term | B1 | |
| $= 80 \times 9.8 \times 12 \times \sin4° - 455$ | B1 | GPE term (656.27) |
| All correct terms in equation (allow sign errors) | A1 | |
| $v = 3.01$ ms$^{-1}$ (3 s.f.) | A1 | cao |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using N2L with driving force $S$: $S - (15 + 0.01 \times 80 \times 9.8 \times \cos5°) - 80 \times 9.8 \times \sin5° = 80 \times 1.5$ | M1 | N2L with at most one force term missing |
| Both resistance terms seen (15 and 7.81) | B1 | |
| Condone wrong sign (68.33) | B1 | |
| All correct terms present; allow sign errors | A1 | May be implicit |
| $S = 211.1402...$ | A1 | |
| $405 = Sv$; $v = 1.92$ ms$^{-1}$ (3 s.f.) | M1, A1 | Use of Power $= Sv$ with any $S$ calculated using N2L; FT their $S$; Note: missing one term in N2L can earn 4/7 |
---2 This question is about 'kart gravity racing' in which, after an initial push, unpowered home-made karts race down a sloping track.
The moving karts have only the following resistive forces and these both act in the direction opposite to the motion.
\begin{itemize}
\item A force $R$, called rolling friction, with magnitude $0.01 M g \cos \theta \mathrm {~N}$ where $M \mathrm {~kg}$ is the mass of the kart and driver and $\theta$ is the angle of the track with the horizontal
\item A force $F$ of varying magnitude, due to air resistance
\end{itemize}
A kart with its driver has a mass of 80 kg .\\
One stretch of track slopes uniformly downwards at $4 ^ { \circ }$ to the horizontal. The kart travels 12 m down this stretch of track. The total work done by the kart against both rolling friction and air resistance is 455 J .\\
(i) Calculate the work done against air resistance.\\
(ii) During this motion, the kart's speed increases from $2 \mathrm {~ms} ^ { - 1 }$ to $v \mathrm {~ms} ^ { - 1 }$. Use an energy method to calculate $v$.
To reach the starting line, the kart (with the driver seated) is pushed up a slope against rolling friction and air resistance.
At one point the slope is at $5 ^ { \circ }$ to the horizontal, the air resistance is 15 N , the acceleration of the kart is $1.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ up the slope and the power of the pushing force is 405 W .\\
(iii) Calculate the speed of the kart at this point.
\hfill \mbox{\textit{OCR MEI M2 2013 Q2 [16]}}