Year: 2017
Paper: 3
Question Number: 11
Course: UFM Mechanics
Section: Momentum and Collisions 1
The total entry was only very slightly smaller than that of 2016, which was a record entry, but was still over 10% more than 2015. No question was attempted by in excess of 90%, although two were very popular and also five others were attempted by 60% or more. No question was generally avoided with even the least popular one attracting more than 10% of the entry. Less than 10% of candidates attempted more than 7 questions, and, apart from 18 exceptions, those doing so did not achieve very good totals and seemed to be 'casting around' to find things they could do: the 18 exceptions were very strong candidates who were generally achieving close to full marks on all the questions they attempted. The general trend was that those with six attempts fared better than those with more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
A railway truck, initially at rest, can move forwards without friction on a long straight horizontal track. On the truck, $n$ guns are mounted parallel to the track and facing backwards, where $n>1$. Each of the guns is loaded with a single projectile of mass $m$. The mass of the truck and guns (but not including the projectiles) is $M$.
When a gun is fired, the projectile leaves its muzzle horizontally with a speed $v-V$ relative to the ground, where $V$ is the speed of the truck immediately before the gun is fired.
\begin{questionparts}
\item All $n$ guns are fired simultaneously.
Find the speed, $u$, with which the truck moves, and show that the kinetic energy, $K$, which is gained by the system (truck, guns and projectiles) is given by
\[
K= \tfrac{1}{2}nmv^2\left(1 +\frac{nm}{M} \right)
.
\]
\item Instead, the guns are fired one at a time.
Let $u_r$ be the speed of the truck when $r$ guns have been fired, so that $u_0= 0$.
Show that, for $1\le r \le n\,$,
\[
u_r - u_{r-1} = \frac{mv}{M+(n-r)m}
\tag{$*$}
\]
and hence that $u_n < u\,$.
\item
Let $K_r$ be the total kinetic energy of the system when $r$ guns have been fired (one at a time), so that $K_0 = 0$. Using $(*)$, show that, for $1\le r\le n\,$,
\[
K_r -K_{r-1} = \tfrac 12 mv^2 + \tfrac12 mv (u_r-u_{r-1})
\]
and hence show that
\[
K_n = \tfrac{1}{2}nmv^2 +\tfrac{1}{2}mvu_n
\,.
\]
Deduce that $K_n < K$.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
\text{COM}: && 0 &= nm(v-0) - Mu \\
\Rightarrow && u &= \frac{nm}{M}v \\
\\
\Rightarrow && \text{K.E.} &= \underbrace{\tfrac12 nmv^2}_{\text{projectiles}} + \underbrace{\tfrac12Mu^2}_{\text{guns and truck}} \\
&&&= \tfrac12nmv^2 + \tfrac12M \frac{n^2m^2}{M^2}v^2 \\
&&&= \tfrac12 nmv^2 \left (1 + \frac{nm}{M} \right)
\end{align*}
\item $\,$ \begin{align*}
\text{COM}: && ((n-r+1)m+M)u_{r-1} &= -m(v-u_{r-1})+ ((n-r)m+M)u_r \\
\Rightarrow && mv &= -((n-r+1)m+M-m)u_{r-1}+((n-r)m+M)u_r \\
\Rightarrow && u_r-u_{r-1} &= \frac{mv}{M+(n-r)m} \\
\\
&& u_n &= \frac{mv}{M+(n-1)m} + \frac{mv}{M+(n-2)m} + \cdots + \frac{mv}{M} \\
&&&< \frac{mv}M + \cdots + \frac{mv}{M} \\
&&&= \frac{mn}{M}v = u
\end{align*}
\item $\,$ \begin{align*}
&& K_r &= \underbrace{K_{r-1}-\frac12(m(n-r+1)+M)u_{r-1}^2}_{\text{energy of already dispersed projectiles}} + \frac12m(v-u_{r-1})^2 + \frac12(m(n-r)+M)u_r^2 \\
\Rightarrow && K_r-K_{r-1} &= \tfrac12(u_r^2-u_{r-1}^2)(M+m(n-r))+\tfrac12mv^2-mvu_{r-1} \\
&&&=\tfrac12mv^2+ \tfrac12(u_r+u_{r-1})mv-mvu_{r-1} \\
&&&= \tfrac12mv^2 + \tfrac12mv(u_r-u_{r-1}) \\
\\
&& K_n &= \frac12nmv^2 + \tfrac12mv(u_n - u_0) \\
&&&= \tfrac12nmv^2 + \tfrac12mvu_n \\
&&&< \tfrac12nmv^2 + \tfrac12mvu \\
&&&= \tfrac12nmv^2 + \tfrac12mv \frac{nm}{M}v \\
&&&= \tfrac12nmv^2 \left (1 +\frac{m}{M} \right) \\
&&&\leq K
\end{align*}
\end{questionparts}
A fifth of the candidates attempted this with just a little less success than that for question 5. Only a minority drew a sketch of the problem; had more done so, some errors might have been precluded. In part (i), a large minority ignored the condition 'initially at rest', a handful gave a negative answer for , and a few attempted to conserve energy, but that was rare. In the second part, some candidates attempted to just write down the given answer employing conserved momentum with verbal justification. The inequality generally followed if a telescoping argument was used although the care shown in dealing with the strict inequality and the case was poor. Attempts at (iii) were generally sound though some took the projectile speed as .