Year: 2018
Paper: 3
Question Number: 9
Course: UFM Mechanics
Section: Momentum and Collisions 1
The total entry was a record number, an increase of over 6% on 2017. Only question 1 was attempted by more than 90%, although question 2 was attempted by very nearly 90%. Every question was attempted by a significant number of candidates with even the two least popular questions being attempted by 9%. More than 92% restricted themselves to attempting no more than 7 questions with very few indeed attempting more than 8. As has been normal in the past, apart from a handful of very strong candidates, those attempting six questions scored better than those attempting more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
A particle $P$ of mass $m$ is projected with speed $u_0$ along a smooth horizontal floor directly towards a wall. It collides with a particle $Q$ of mass $km$ which is moving directly away from the wall with speed $v_0$.
In the subsequent motion, $Q$ collides alternately with the wall and with $P$. The coefficient of restitution between $Q$ and $P$ is $e$, and the coefficient of restitution between $Q$ and the wall is 1.
Let $u_n$ and $v_n$ be the velocities of $P$ and $Q$, respectively, towards the wall after the $n$th
collision between $P$ and $Q$.
\begin{questionparts}
\item
Show that, for $n\ge2$,
\[
(1+k)u_{n} - (1-k)(1+e)u_{n-1} + e(1+k)u_{n-2} =0\,.
\tag{$*$}
\]
\item
You are now given that $e=\frac12$ and $k = \frac1{34}$, and that the solution of $(*)$ is of the form
\[
\phantom{(n\ge0)}
u_n=
A\left( \tfrac 7{10}\right)^n
+
B\left( \tfrac 5{7 }\right)^n
\ \ \ \ \ \
(n\ge0)
\,,
\]
where $A$ and $B$ are independent of $n$.
Find expressions for $A$ and $B$ in terms of $u_0$ and $v_0$.
Show that, if $0 < 6u_0 < v_0$, then $u_n$ will be negative for large $n$.
\end{questionparts}\begin{questionparts}
\item
Just before collision $n-1$:
Velocity of $P$ is $u_{n-2}$
Velocity of $Q$ is $-v_{n-2}$
\begin{align*}
COM: && mu_{n-2}+km(-v_{n-2}) &= mu_{n-1}+kmv_{n-1} \\
\Rightarrow && u_{n-2}-kv_{n-2} &= u_{n-1}+kv_{n-1} \\
NEL: && v_{n-1}-u_{n-1} &= -e((-v_{n-2})-u_{n-2}) \\
\Rightarrow && v_{n-1}-u_{n-1} &= e(v_{n-2}+u_{n-2})
\end{align*}
\begin{align*}
&&kv_{n-1} &= u_{n-2} - kv_{n-2}-u_{n-1} \\
&&kv_{n-1}&= ku_{n-1}+kev_{n-2}+keu_{n-2} \\
\Rightarrow && kv_{n-2}(1+e) &= u_{n-2}(1-ke)-u_{n-1}(1+k) \\
\Rightarrow && kv_{n-1}(1+e) &= u_{n-1}(1-ke)-u_{n}(1+k) \\
&& k(1+e)v_{n-1}-k(1+e)u_{n-1} &= k(1+e)e(v_{n-2}+u_{n-2}) \\
\Rightarrow && u_{n-1}(1-ke)-u_{n}(1+k)-k(1+e)u_{n-1} &= e(u_{n-2}(1-ke)-u_{n-1}(1+k))+k(1+e)eu_{n-2} \\
\Rightarrow && 0 &= (1+k)u_n + ((ke-1)+k(1+e)-e(1+k))u_{n-1} \\
&&& \quad \quad + (e(1-ke)+k(1+e)e)u_{n-2} \\
\Rightarrow && 0 &= (1+k)u_n- (1-k)(1+e)u_{n-1} +e(1+k)u_{n-2}
\end{align*}
\item $u_0 = A + B$
\begin{align*}
&&& \begin{cases}u_0 - kv_0 &= kv_1 + u_1 \\
\frac12 (u_0+v_0) &= v_1 - u_1 \\
\end{cases} \\
\Rightarrow && (1+k)u_1 &= u_0 - kv_0 - \frac{k}{2}(u_0 + v_0) \\
\Rightarrow && u_1 &= \frac{1}{k+1} \l u_0 (1-\frac{k}{2}) - \frac32 k v_0 \r \\
&&&= \frac{67}{70} u_0 - \frac{3}{70} v_0
\end{align*}
Therefore $A+B = u_0, \frac{49A+50B}{70} = \frac{67}{70} u_0 - \frac{3}{70} v_0$
\begin{align*}
&& A+B &= u_0 \\
&& 49A+50B &= 67u_0 - 3v_0 \\
\Rightarrow && 50u_0 - A &= 67u_0 - 3v_0 \\
\Rightarrow && A &= -17u_0 + 3v_0 \\
&& B &= 18u_0 - 3v_0
\end{align*}
If $0 < 6u_0 < v_0$, then $B < 0$ and as $n \to \infty$ we will find that $\l \frac57 \r^n$ dominates $\l \frac7{10} \r^n$ and so our velocity will be negative and the particle will change direction
\end{questionparts}
Whilst being the most popular of the applied questions, it was much less popular than any of the Pure questions being attempted by only just over a quarter of the candidates. It was fairly weakly attempted with a mean score of 5.5/20, only slightly better than question 6. The low scores were mainly caused by sign errors in the velocities of P and Q, and candidates finding it difficult to know which variables to eliminate. However, most did attempt to express the momentum conservation and Newton's Experimental Law of Impact equations which were the starting points to the question. Most students considered the very first and second collisions, rather than the nth and n+1th; algebraically, these expressions were the same, but they did then need to justify the generalisation. The few candidates that got as far as part (ii), scored higher overall. In this case, often students showed that the given formula was a solution, which was not asked for, rather than finding and. Those who found in terms of and generally managed to solve the question well. Very few attempted the last part, but those that did scored well, taking the limit correctly at the end.