Edexcel M2 — Question 3 10 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeMaximum speed on incline vs horizontal
DifficultyStandard +0.3 This is a straightforward M2 work-energy-power question requiring standard application of P=Fv at maximum speed (where acceleration=0). Part (a) involves simple algebraic manipulation to find k, while part (b) adds a component of weight down the slope but follows the same method. The question is slightly above average difficulty due to the two-part structure and inclusion of an inclined plane, but requires no novel insight—just systematic application of familiar mechanics principles.
Spec6.02l Power and velocity: P = Fv

3. A car of mass 1200 kg experiences a resistance to motion, \(R\) newtons, which is proportional to its speed, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). When the power output of the car engine is 90 kW and the car is travelling along a horizontal road, its maximum speed is \(50 \mathrm {~ms} ^ { - 1 }\).
  1. Show that \(R = 36 v\). The car ascends a hill inclined at an angle \(\theta\) to the horizontal where \(\sin \theta = \frac { 1 } { 14 }\).
  2. Find, correct to 3 significant figures, the maximum speed of the car up the hill assuming that the power output of the engine is unchanged.
    (6 marks)

AnswerMarks Guidance
(a)\(R \propto v\) \(\therefore R = kv\) where \(k\) is a constant M1
\(\frac{dv}{dt} - R = 0\) \(\therefore \frac{90000}{m} - 50k = 0\)M1 A1
\(k = 36\) \(\therefore R = 36v\)A1
(b)\(\frac{dv}{dt} - R - mg\sin\theta = 0\) \(\therefore \frac{90000}{m} - 36v - 1200(9.8)\frac{1}{3} = 0\) M1 A1
\(90000 - 36v^2 - 840v = 0\) \(\therefore 3v^2 + 70v - 7500 = 0\)M1 A1
Quad. form. giving \(v = 39.7\text{ ms}^{-1}\) (3sf) (clearly 63.0 not suitable)M1 A1 (10 marks total)
(a) | $R \propto v$ $\therefore R = kv$ where $k$ is a constant | M1 |
| $\frac{dv}{dt} - R = 0$ $\therefore \frac{90000}{m} - 50k = 0$ | M1 A1 |
| $k = 36$ $\therefore R = 36v$ | A1 |

(b) | $\frac{dv}{dt} - R - mg\sin\theta = 0$ $\therefore \frac{90000}{m} - 36v - 1200(9.8)\frac{1}{3} = 0$ | M1 A1 |
| $90000 - 36v^2 - 840v = 0$ $\therefore 3v^2 + 70v - 7500 = 0$ | M1 A1 |
| Quad. form. giving $v = 39.7\text{ ms}^{-1}$ (3sf) (clearly 63.0 not suitable) | M1 A1 | (10 marks total)

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3. A car of mass 1200 kg experiences a resistance to motion, $R$ newtons, which is proportional to its speed, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. When the power output of the car engine is 90 kW and the car is travelling along a horizontal road, its maximum speed is $50 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $R = 36 v$.

The car ascends a hill inclined at an angle $\theta$ to the horizontal where $\sin \theta = \frac { 1 } { 14 }$.
\item Find, correct to 3 significant figures, the maximum speed of the car up the hill assuming that the power output of the engine is unchanged.\\
(6 marks)
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2  Q3 [10]}}