OCR M2 2015 June — Question 1 7 marks

Exam BoardOCR
ModuleM2 (Mechanics 2)
Year2015
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAdvanced work-energy problems
TypeEngine power on road constant/variable speed
DifficultyModerate -0.3 This is a straightforward M2 power-resistance question requiring standard formulas (P=Fv, work-energy principle). Part (i) is direct substitution; part (ii) uses work-energy theorem with given values. Both parts follow textbook methods with no novel problem-solving required, making it slightly easier than average.
Spec6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv

1 A cyclist travels along a straight horizontal road. The total mass of the cyclist and her bicycle is 80 kg and the resistance to motion is a constant 60 N .
  1. The cyclist travels at a constant speed working at a constant rate of 480 W . Find the speed at which she travels.
  2. The cyclist now instantaneously increases her power to 600 W . After travelling at this power for 14.2 s her speed reaches \(9.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find the distance travelled at this power.

Question 1:
Part (i):
AnswerMarks Guidance
AnswerMarks Guidance
\(D = 480/v\)B1 Use of \(D = P/v\)
\(480/v - 60 = 0\)M1 Use of N2L with 2 terms to find \(v\)
\(v = 8 \text{ m s}^{-1}\)A1
Part (ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(600 \times 14.2\)B1 WD by cyclist (8520 J)
\(\frac{1}{2}(80)(9.4^2 - 8^2)\)B1ft ft \(v\) from (i); KE gained (974.4 J); may be implied in energy equation
\(8520 = 974.4 + 60d\)M1 Attempt at energy equation with all terms
\(d = 126 \text{ m}\)A1 Exact 125.76
## Question 1:

### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $D = 480/v$ | B1 | Use of $D = P/v$ |
| $480/v - 60 = 0$ | M1 | Use of N2L with 2 terms to find $v$ |
| $v = 8 \text{ m s}^{-1}$ | A1 | |

### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $600 \times 14.2$ | B1 | WD by cyclist (8520 J) |
| $\frac{1}{2}(80)(9.4^2 - 8^2)$ | B1ft | ft $v$ from (i); KE gained (974.4 J); may be implied in energy equation |
| $8520 = 974.4 + 60d$ | M1 | Attempt at energy equation with all terms |
| $d = 126 \text{ m}$ | A1 | Exact 125.76 |

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1 A cyclist travels along a straight horizontal road. The total mass of the cyclist and her bicycle is 80 kg and the resistance to motion is a constant 60 N .\\
(i) The cyclist travels at a constant speed working at a constant rate of 480 W . Find the speed at which she travels.\\
(ii) The cyclist now instantaneously increases her power to 600 W . After travelling at this power for 14.2 s her speed reaches $9.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the distance travelled at this power.

\hfill \mbox{\textit{OCR M2 2015 Q1 [7]}}