CAIE M1 2016 March — Question 1 3 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2016
SessionMarch
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeEnergy method - driving force on horizontal road
DifficultyStandard +0.3 This is a straightforward work-energy problem requiring application of the work-energy principle with two components: change in kinetic energy and work against resistance. The calculation involves standard formulas (KE = ½mv², work = force × distance) with no conceptual subtleties or multi-step reasoning beyond combining these two contributions. Slightly above average difficulty due to requiring recognition that total work = ΔKE + work against resistance, but still a routine mechanics question.
Spec6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts
1 A cyclist has mass 85 kg and rides a bicycle of mass 20 kg . The cyclist rides along a horizontal road against a total resistance force of 40 N . Find the total work done by the cyclist in increasing his speed from \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) while travelling a distance of 50 m .
Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Attempt KE gain or WD against ResistanceM1
KE gain \(= \frac{1}{2} \times 105 \times (10^2 - 5^2)\); WD against Resistance \(= 50 \times 40\)A1 Both correct (unsimplified); KE gain \(= 3937.5\) J, WD \(= 2000\) J
Total WD \(= 5937.5\) JB1 3
Alternative method:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(10^2 = 5^2 + 2 \times 50 \times a \Rightarrow DF - 40 = 105a\)M1 Using \(v^2 = u^2 + 2as\) and applying Newton's 2nd law to the system
\(DF = 40 + 105 \times 0.75 = 118.75\)A1
Total WD \(= 118.75 \times 50 = 5937.5\) JB1 3 
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt KE gain **or** WD against Resistance | M1 | |
| KE gain $= \frac{1}{2} \times 105 \times (10^2 - 5^2)$; WD against Resistance $= 50 \times 40$ | A1 | Both correct (unsimplified); KE gain $= 3937.5$ J, WD $= 2000$ J |
| Total WD $= 5937.5$ J | B1 | 3 | $\text{WD} = \text{KE gain} + \text{WD against Res}$ |

**Alternative method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $10^2 = 5^2 + 2 \times 50 \times a \Rightarrow DF - 40 = 105a$ | M1 | Using $v^2 = u^2 + 2as$ and applying Newton's 2nd law to the system |
| $DF = 40 + 105 \times 0.75 = 118.75$ | A1 | |
| Total WD $= 118.75 \times 50 = 5937.5$ J | B1 | 3 | $\text{WD} = DF \times 50$ |

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1 A cyclist has mass 85 kg and rides a bicycle of mass 20 kg . The cyclist rides along a horizontal road against a total resistance force of 40 N . Find the total work done by the cyclist in increasing his speed from $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ while travelling a distance of 50 m .

\hfill \mbox{\textit{CAIE M1 2016 Q1 [3]}}
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