Edexcel M2 2010 January — Question 5 11 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2010
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeCyclist or runner: find resistance or speed
DifficultyStandard +0.3 This is a standard M2 power-force-velocity question requiring the formula P=Fv and resolving forces parallel to motion. Part (a) is direct substitution (490=F×3.5), while part (b) involves resolving forces on an incline with variable resistance, leading to a quadratic equation. The steps are routine for M2 students who have practiced this topic, making it slightly easier than average overall.
Spec3.03f Weight: W=mg3.03m Equilibrium: sum of resolved forces = 06.02l Power and velocity: P = Fv

A cyclist and her bicycle have a total mass of \(70\) kg. She cycles along a straight horizontal road with constant speed \(3.5 \text{ ms}^{-1}\). She is working at a constant rate of \(490\) W.
  1. Find the magnitude of the resistance to motion. [4]
The cyclist now cycles down a straight road which is inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac{1}{14}\), at a constant speed \(U \text{ ms}^{-1}\). The magnitude of the non-gravitational resistance to motion is modelled as \(40U\) newtons. She is now working at a constant rate of \(24\) W.
  1. Find the value of \(U\). [7]

(a)
AnswerMarks
\(\frac{490}{3.5} - R = 0\)B1 M1 A1
\(R = 140\) NA1 (4)
(b)
AnswerMarks
\(\frac{24}{u} + 70g \cdot \frac{1}{14} - 40u = 0\)B1
\(40u^2 - 49u - 24 = 0\)M1 A2, 1, 0
\((5u - 8)(8u + 3) = 0\)DM1
\(u = 1.6\)DM1 A1 (7)
Total: [11]
## (a)

$\frac{490}{3.5} - R = 0$ | B1 M1 A1 |
$R = 140$ N | A1 (4) |

## (b)

$\frac{24}{u} + 70g \cdot \frac{1}{14} - 40u = 0$ | B1 |
$40u^2 - 49u - 24 = 0$ | M1 A2, 1, 0 |
$(5u - 8)(8u + 3) = 0$ | DM1 |
$u = 1.6$ | DM1 A1 (7) |

**Total: [11]**

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A cyclist and her bicycle have a total mass of $70$ kg. She cycles along a straight horizontal road with constant speed $3.5 \text{ ms}^{-1}$. She is working at a constant rate of $490$ W.

\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the resistance to motion.
[4]
\end{enumerate}

The cyclist now cycles down a straight road which is inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac{1}{14}$, at a constant speed $U \text{ ms}^{-1}$. The magnitude of the non-gravitational resistance to motion is modelled as $40U$ newtons. She is now working at a constant rate of $24$ W.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $U$.
[7]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2010 Q5 [11]}}