OCR MEI Further Mechanics A AS 2020 November — Question 1 4 marks

Exam BoardOCR MEI
ModuleFurther Mechanics A AS (Further Mechanics A AS)
Year2020
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeMaximum speed on incline vs horizontal
DifficultyStandard +0.3 This is a straightforward application of P=Fv at maximum speed where driving force equals resistance. First part finds power (P=19.2×6.25=120W), second part requires resolving weight component down slope and solving 120=v(19.2+72g sin3°). Standard two-step mechanics problem with clear method, slightly above average due to requiring power equation and component resolution.
Spec6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product

1 Brent is riding his bicycle along a straight horizontal road.
While riding along this road Brent can attain a maximum speed of \(6.25 \mathrm {~ms} ^ { - 1 }\) and the wind resistance acting on Brent and his bicycle is constant and equal to 19.2 N . Brent and his bicycle have a combined mass of 72 kg . Brent later begins to ride up a hill which is inclined at an angle of \(3 ^ { \circ }\) to the horizontal.
Given that the wind resistance and the maximum power developed by the bicycle is unchanged, determine Brent's maximum speed up the hill.

Question 1:
AnswerMarks Guidance
1Let the driving force be D therefore D =19.2
1 1B1 1.1
the level
(e.g. 19.2 × 6.25 seen implies B1M1 )
AnswerMarks Guidance
Maximum power output P=19.2×6.25=120M1 1.1
1 1
Driving force on slope D =19.2+72gsin3°(=56.128)
AnswerMarks Guidance
2M1 1.1
cos/sin confusion) identified as, or used as, the
driving force up the slope.
Note 56.1 seen does not, by itself, imply M1
120
So maximum speed = =2.14(m s-1)
D
AnswerMarks Guidance
2A1 1.1
[4]
Question 1:
1 | Let the driving force be D therefore D =19.2
1 1 | B1 | 1.1 | 19.2 identified as, or used as, the driving force on
the level
(e.g. 19.2 × 6.25 seen implies B1M1 )
Maximum power output P=19.2×6.25=120 | M1 | 1.1 | Use of P=Dvwith their D and correct v
1 1
Driving force on slope D =19.2+72gsin3°(=56.128)
2 | M1 | 1.1 | 19.2 plus resolved weight (allow sign errors and
cos/sin confusion) identified as, or used as, the
driving force up the slope.
Note 56.1 seen does not, by itself, imply M1
120
So maximum speed = =2.14(m s-1)
D
2 | A1 | 1.1 | For reference: 2.13796080…
[4]
1 Brent is riding his bicycle along a straight horizontal road.\\
While riding along this road Brent can attain a maximum speed of $6.25 \mathrm {~ms} ^ { - 1 }$ and the wind resistance acting on Brent and his bicycle is constant and equal to 19.2 N . Brent and his bicycle have a combined mass of 72 kg .

Brent later begins to ride up a hill which is inclined at an angle of $3 ^ { \circ }$ to the horizontal.\\
Given that the wind resistance and the maximum power developed by the bicycle is unchanged, determine Brent's maximum speed up the hill.

\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2020 Q1 [4]}}