OCR MEI Further Mechanics A AS 2019 June — Question 5 10 marks

Exam BoardOCR MEI
ModuleFurther Mechanics A AS (Further Mechanics A AS)
Year2019
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeWork-energy over time interval
DifficultyStandard +0.3 This is a straightforward application of standard mechanics formulas (P=Fv for part a, then work-energy principle for part b). Part (a) requires resolving forces and using P=Fv, while part (b) uses energy methods with given time. Both parts follow standard textbook procedures with no novel problem-solving required, making it slightly easier than average.
Spec6.02k Power: rate of doing work6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product
5 A car of mass 4000 kg travels up a line of greatest slope of a straight road inclined at an angle of \(\theta\) to the horizontal, where \(\sin \theta = 0.1\).
The power developed by the car's engine is constant and the resistance to the motion of the car is constant and equal to 850 N . The car passes through a point A on the road with speed \(18 \mathrm {~ms} ^ { - 1 }\) and acceleration \(0.75 \mathrm {~ms} ^ { - 2 }\).
  1. Calculate the power developed by the car. The car later passes through a point B on the road with speed \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The car takes 17.8 s to travel from A to B .
  2. Find the distance AB .
Question 5:
AnswerMarks Guidance
5(a) M1
slope with all terms and weight
AnswerMarks
component resolvedAllow with sign errors,
sin/cos confusion and
errors with g
AnswerMarks Guidance
D8504000gsin4000  0.75 A1 1.1
D7770A1 1.1
P7770  18 M1 3.4
(velocity)Used at some point
139860 (W)A1 2.2a
[5]
AnswerMarks
(b)1  
KE change 4000 252182
AnswerMarks Guidance
2B1 1.1
602000 (J)
AnswerMarks Guidance
PE gained = 4000𝑔(ABsin𝜃)B1 1.1
WD by car’s engine 13986017.8B1ft 1.1
power from (a)
12500003920  AB 648000
 850  
AnswerMarks Guidance
139860 17.8 ABM1 3.3
present
AnswerMarks Guidance
Distance AB = 400 mA1 1.1
[5]
Question 5:
5 | (a) | M1 | 3.3 | Attempt at N2L parallel to the
slope with all terms and weight
component resolved | Allow with sign errors,
sin/cos confusion and
errors with g
D8504000gsin4000  0.75  | A1 | 1.1 | Correct eqn – allowsin or 0.1
D7770 | A1 | 1.1 | Could be implied by later working
P7770  18  | M1 | 3.4 | Use of P = (tractive force) ×
(velocity) | Used at some point
139860 (W) | A1 | 2.2a | Cao (2 s.f. or better) | 140kW
[5]
(b) | 1  
KE change 4000 252182
2 | B1 | 1.1 | 1250000 – 648000 =
602000 (J)
PE gained = 4000𝑔(ABsin𝜃) | B1 | 1.1 | In terms of AB
WD by car’s engine 13986017.8 | B1ft | 1.1 | 2489508 J − follow through their
power from (a)
12500003920  AB 648000
 850  
139860 17.8 AB | M1 | 3.3 | Work-energy principle – all terms
present
Distance AB = 400 m | A1 | 1.1 | Cao 2 s.f. or better | 395.70398
[5]
5 A car of mass 4000 kg travels up a line of greatest slope of a straight road inclined at an angle of $\theta$ to the horizontal, where $\sin \theta = 0.1$.\\
The power developed by the car's engine is constant and the resistance to the motion of the car is constant and equal to 850 N . The car passes through a point A on the road with speed $18 \mathrm {~ms} ^ { - 1 }$ and acceleration $0.75 \mathrm {~ms} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the power developed by the car.

The car later passes through a point B on the road with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car takes 17.8 s to travel from A to B .
\item Find the distance AB .
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2019 Q5 [10]}}
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