AQA M2 2015 June — Question 6 9 marks

Exam BoardAQA
ModuleM2 (Mechanics 2)
Year2015
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAdvanced work-energy problems
TypeEngine power on road constant/variable speed
DifficultyStandard +0.3 This is a standard M2 power-resistance problem requiring application of F=ma, resolving forces on an incline, and P=Fv. While it involves multiple steps (finding driving force from power, resolving parallel to slope, using Newton's second law), these are routine techniques that follow a well-practiced procedure with no novel insight required. The 9 marks reflect the working steps rather than conceptual difficulty, making it slightly easier than average.
Spec6.02k Power: rate of doing work6.02l Power and velocity: P = Fv

6 A van, of mass 1400 kg , is accelerating at a constant rate of \(0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) as it travels up a slope inclined at an angle \(\theta\) to the horizontal. The van experiences total resistance forces of 4000 N .
When the van is travelling at a speed of \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the power output of the van's engine is 91.1 kW . Find \(\theta\).
[0pt] [9 marks]

Question 6:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Power = Force × velocity: \(F = \frac{91100}{20} = 4555\) NM1 Finding driving force from P = Fv
\(F = 4555\) NA1
Newton's second law along slope: \(F - R - mg\sin\theta = ma\)M1 Equation of motion along slope with all relevant forces
\(4555 - 4000 - 1400g\sin\theta = 1400 \times 0.2\)A1 Correct substitution (g = 9.8)
\(4555 - 4000 - 13720\sin\theta = 280\)
\(13720\sin\theta = 275\)DM1 Solving for \(\sin\theta\)
\(\sin\theta = \frac{275}{13720}\)A1
\(\theta = 1.15°\)A1 Accept \(\theta \approx 1.1°\) or \(1.2°\)
# Question 6:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Power = Force × velocity: $F = \frac{91100}{20} = 4555$ N | M1 | Finding driving force from P = Fv |
| $F = 4555$ N | A1 | |
| Newton's second law along slope: $F - R - mg\sin\theta = ma$ | M1 | Equation of motion along slope with all relevant forces |
| $4555 - 4000 - 1400g\sin\theta = 1400 \times 0.2$ | A1 | Correct substitution (g = 9.8) |
| $4555 - 4000 - 13720\sin\theta = 280$ | | |
| $13720\sin\theta = 275$ | DM1 | Solving for $\sin\theta$ |
| $\sin\theta = \frac{275}{13720}$ | A1 | |
| $\theta = 1.15°$ | A1 | Accept $\theta \approx 1.1°$ or $1.2°$ |

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6 A van, of mass 1400 kg , is accelerating at a constant rate of $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ as it travels up a slope inclined at an angle $\theta$ to the horizontal.

The van experiences total resistance forces of 4000 N .\\
When the van is travelling at a speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the power output of the van's engine is 91.1 kW .

Find $\theta$.\\[0pt]
[9 marks]

\hfill \mbox{\textit{AQA M2 2015 Q6 [9]}}