Year: 2016
Paper: 3
Question Number: 11
Course: UFM Mechanics
Section: Variable Force
A substantially larger number of candidates took the paper this year: 14% more than in 2015. However, the mean score was virtually identical to that in 2015. Five questions were very popular, with two being attempted by in excess of 90% of the candidates, but once again, all questions were attempted by significant numbers, with only one dipping under 10% attempting it, and every question was answered perfectly by at least one candidate. Most candidates kept to six sensible attempts, although some did several more scoring weakly overall, except in six outstanding cases that earned very high marks.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
A car of mass $m$ travels along a straight horizontal road with its engine working at a constant rate $P$. The resistance to its motion is such that the acceleration of the car is zero when it is moving with speed $4U$.
\begin{questionparts}
\item
Given that the resistance is proportional to the car's speed, show that the distance $X_1$ travelled by the car while it accelerates from speed $U$ to speed $2U$, is given by
\[
\lambda X_1 = 2\ln \tfrac 9 5 - 1
\,,
\]
where $\lambda= P/(16mU^3)$.
\item
Given instead that the resistance is proportional to the square of the car's speed, show that the distance $X_2$ travelled by the car while it accelerates from speed $U$ to speed $2U$ is given by
\[
\lambda X_2 = \tfrac43 \ln \tfrac 98
\,.
\]
\item
Given that $3.17<\ln 24 < 3.18$ and $1.60<\ln 5 < 1.61$, determine which is the larger of $X_1$ and $X_2$.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& F_{res} &= kv \\
&& P &= Fv \\
v = 4U: && 0 &= F-F_{res} \\
\Rightarrow && 0 &= \frac{P}{4U} - 4Uk \\
\Rightarrow && k &= \frac{P}{16U^2} \\
\\
&&m v \frac{\d v}{\d x}&= \frac{P}{v} - \frac{P}{16U^2}v \\
\Rightarrow && X_1 &= \int_{v=U}^{v=2U} \frac{16U^2mv^2}{P(16U^2-v^2)} \d v \\
v = Ut&& &= \frac{16mU^2}{P} \int_{t=1}^{t=2}\left ( \frac{t^2}{16-t^2} \right)U\d t \\
&&&= \frac{16mU^3}{P} \int_1^2 \left ( -1 + \frac{16}{16-t^2} \right) \d t \\
&&&= \frac{16mU^3}{P} \int_1^2 \left ( -1 +\frac{2}{4+t} +\frac{2}{4-t} \right) \d t \\
&&&= \frac{1}{\lambda}\left (-1 + 2\ln(6)-2\ln(2)-2\ln(5)+2\ln(3) \right) \\
\Rightarrow && \lambda X_1 &= 2\ln \tfrac95-1
\end{align*}
\item $\,$ \begin{align*}
&& F_{res} = kv^2 \\
v = 4U: && 0 &= \frac{P}{4U} - 16U^2k \\
\Rightarrow && k &= \frac{P}{64U^3} \\
\\
&& mv \frac{\d v}{\d x} &= \frac{P}{v} - \frac{P}{64U^3}v^2 \\
\Rightarrow && X_2 &= \int_{v=U}^{v=2U} \frac{64U^3mv^2}{P(64U^3-v^3)} \d v \\
&&&= \frac{64U^3m}{P} \int_{v=U}^{v=2U} \frac{v^2}{64U^3-v^3} \d v\\
v = Ut &&&= \frac{64U^3m}{P} \int_{t=1}^{t=2} \frac{U^2t^2}{64U^3-U^3v^3} U \d t\\
&&&= \frac{4}{\lambda} \int_1^2 \frac{t^2}{64-t^3} \d t \\
&&&= \frac{4}{\lambda} \left [ -\frac13\ln(64-t^3) \right]_1^2 \\
&&&= \frac{4}{3\lambda} \ln (63/56) \\
\Rightarrow && \lambda X_2 &= \tfrac43 \ln \tfrac98
\end{align*}
\item $\,$ \begin{align*}
&& 2\ln \tfrac95 - 1 &\overset{?}{>} \frac43 \ln \frac98 \\
\Leftrightarrow && 4 \ln 3 - 2\ln 5 - 1 &\overset{?}{>} \frac83\ln 3 -4 \ln 2 \\
\Leftrightarrow && \frac43(3\ln 3 + 3\ln 2 - 2 \ln 3) &\overset{?}{>} 2 \ln 5 + 1\\
\Leftrightarrow && \frac43\ln 24 &\overset{?}{>} 2 \ln 5 + 1\\
\end{align*}
The $LHS$ is $>4.22$. The $RHS$ is $< 4.22$, and therefore our inequality holds, in particular, $X_1 > X_2$.
\end{questionparts}
The most popular of the non-Pure questions, it was attempted by a third of the candidates, but generally, only about a quarter of the marks were scored. A large number attempted to use constant acceleration formulae, or if they realised that calculus was required, failed to appreciate that they needed to use v = dx/dt. For those that separated variables, the integrations caused few problems. Part (iii) caused difficulties as candidates were not comfortable using the bounds; had they considered λ₁ - λ₂ they might have encountered fewer problems.