Edexcel M4 2005 January — Question 4 9 marks

Exam BoardEdexcel
ModuleM4 (Mechanics 4)
Year2005
SessionJanuary
Marks9
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Mark schemeDownload PDF ↗
TopicPower and driving force
TypeFind acceleration given power
DifficultyChallenging +1.2 This is a standard M4 variable acceleration problem requiring integration of F=ma with power P=Fv. Students must set up the equation of motion (RU/v - R = M dv/dt), separate variables, and integrate between given limits. While it involves multiple steps and careful algebraic manipulation, it follows a well-practiced template for this module with no conceptual surprises or novel insights required.
Spec6.02k Power: rate of doing work6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods
A car of mass \(M\) moves along a straight horizontal road. The total resistance to motion of the car is modelled as having constant magnitude \(R\). The engine of the car works at a constant rate \(RU\). Find the time taken for the car to accelerate from a speed of \(\frac{1}{4}U\) to a speed of \(\frac{1}{2}U\). [9]
AnswerMarks Guidance
Part (a): Apply \(F = ma\): \(m\frac{dv}{dt} = \frac{RU}{v} - R\)M1 Application of Newton's second law
Part (b): Separate variables: \(\int_u^v \frac{mvdv}{R(U-v)} = \int_0^t dt\)M1 Variable separation
Part (c): \(\frac{m}{R}\int_u^v \left(-1 + \frac{U}{(U-v)}\right)dv = [t]_0^T\)M1 Integration setup
Part (d): \(\frac{m}{R}\left[v - U\lnU-v \right]_u^v = T\)
Part (e): \(T = \frac{m}{R}\left(\left(-\frac{1}{4}U - U\ln\frac{1}{4}U\right) - \left(-\frac{1}{4}U - U\ln\frac{3}{4}U\right)\right)\)M1 Substitution of limits
Part (f): \(T = \frac{mU}{R}\left(-\frac{1}{4} + \ln\frac{3}{2}\right)\)A1 Final answer 
**Part (a):** Apply $F = ma$: $m\frac{dv}{dt} = \frac{RU}{v} - R$ | M1 | Application of Newton's second law |

**Part (b):** Separate variables: $\int_u^v \frac{mvdv}{R(U-v)} = \int_0^t dt$ | M1 | Variable separation |

**Part (c):** $\frac{m}{R}\int_u^v \left(-1 + \frac{U}{(U-v)}\right)dv = [t]_0^T$ | M1 | Integration setup |

**Part (d):** $\frac{m}{R}\left[v - U\ln|U-v|\right]_u^v = T$ | A1 | Correct integration |

**Part (e):** $T = \frac{m}{R}\left(\left(-\frac{1}{4}U - U\ln\frac{1}{4}U\right) - \left(-\frac{1}{4}U - U\ln\frac{3}{4}U\right)\right)$ | M1 | Substitution of limits |

**Part (f):** $T = \frac{mU}{R}\left(-\frac{1}{4} + \ln\frac{3}{2}\right)$ | A1 | Final answer |

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A car of mass $M$ moves along a straight horizontal road. The total resistance to motion of the car is modelled as having constant magnitude $R$. The engine of the car works at a constant rate $RU$.

Find the time taken for the car to accelerate from a speed of $\frac{1}{4}U$ to a speed of $\frac{1}{2}U$. [9]

\hfill \mbox{\textit{Edexcel M4 2005 Q4 [9]}}
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