Edexcel M2 2018 June — Question 3 8 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2018
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeCyclist or runner: find resistance or speed
DifficultyStandard +0.3 This is a standard M2 work-energy-power question requiring the power equation P=Fv and Newton's second law. Part (a) involves finding resistance using F=ma and P=Fv with straightforward substitution. Part (b) adds an incline component but remains routine application of resolving forces and the power equation. Slightly above average due to two-part structure and combining multiple mechanics concepts, but follows standard M2 patterns with no novel insight required.
Spec6.02l Power and velocity: P = Fv

3. A cyclist and his bicycle, with a combined mass of 75 kg , move along a straight horizontal road. The cyclist is working at a constant rate of 180 W . There is a constant resistance to the motion of the cyclist and his bicycle of magnitude \(R\) newtons. At the instant when the speed of the cyclist is \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), his acceleration is \(0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
  1. Find the value of \(R\). Later, the cyclist moves up a straight road with a constant speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The road is inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac { 1 } { 21 }\). The cyclist is working at a rate of 180 W and the resistance to the motion of the cyclist and his bicycle from non-gravitational forces is again the same constant force of magnitude \(R\) newtons.
  2. Find the value of \(v\).

3a.
AnswerMarks
Use of \(P = Fv: F = \frac{180}{4}\)B1
Equation of motion: \(F - R = 75 \times 0.2\)M1
Equation in \(R\): \(\frac{180}{4} - R = 75 \times 0.2(45 - R = 15)\)DM1
\(R = 30\)A1
(4)
3b.
AnswerMarks
Equation of motion: \(D - 75g\sin\theta - R = 0\)M1
\(\frac{180}{v} - 75 \times g \times \frac{1}{21}\) – their \(R = 0\)A2 ft
\(v = 2.77\) or \(2.8\)A1
(4)
[8]
Notes for Q3:
3(a)
- B1 for \(F = \frac{180}{4}\) seen
- First M1 for equation of motion with usual rules, \(F\) does not need to be substituted
- Second M1, dependent on first M1, for an equation in \(R\) only with usual rules
- A1 for \(R = 30\)
3(b)
- M1 for equation of motion with usual rules but none of \(D\), \(\sin\theta\) nor \(R\) need to be substituted
- A2 ft for a correct equation, in \(v\) only, ft on their \(R\)
- A1 A0 if one error
- Third A1 for \(2.77\) or \(2.8\) (Only answers)
## 3a.
Use of $P = Fv: F = \frac{180}{4}$ | B1 |
Equation of motion: $F - R = 75 \times 0.2$ | M1 |
Equation in $R$: $\frac{180}{4} - R = 75 \times 0.2(45 - R = 15)$ | DM1 |
$R = 30$ | A1 |
(4) | |

## 3b.
Equation of motion: $D - 75g\sin\theta - R = 0$ | M1 |
$\frac{180}{v} - 75 \times g \times \frac{1}{21}$ – their $R = 0$ | A2 ft |
$v = 2.77$ or $2.8$ | A1 |
(4) | |
[8] | |

**Notes for Q3:**

3(a)
- B1 for $F = \frac{180}{4}$ seen
- First M1 for equation of motion with usual rules, $F$ does not need to be substituted
- Second M1, dependent on first M1, for an equation in $R$ only with usual rules
- A1 for $R = 30$

3(b)
- M1 for equation of motion with usual rules but none of $D$, $\sin\theta$ nor $R$ need to be substituted
- A2 ft for a correct equation, in $v$ only, ft on their $R$
- A1 A0 if one error
- Third A1 for $2.77$ or $2.8$ (**Only answers**)

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3. A cyclist and his bicycle, with a combined mass of 75 kg , move along a straight horizontal road. The cyclist is working at a constant rate of 180 W . There is a constant resistance to the motion of the cyclist and his bicycle of magnitude $R$ newtons. At the instant when the speed of the cyclist is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, his acceleration is $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $R$.

Later, the cyclist moves up a straight road with a constant speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The road is inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 21 }$. The cyclist is working at a rate of 180 W and the resistance to the motion of the cyclist and his bicycle from non-gravitational forces is again the same constant force of magnitude $R$ newtons.
\item Find the value of $v$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2018 Q3 [8]}}