Particles $A$, $B$ and $C$, with masses $5\text{ kg}$, $1\text{ kg}$ and $2\text{ kg}$ respectively, are initially at rest in that sequence on the straight smooth horizontal track $XYZ$. At the start, $A$ is at $X$, $B$ is at $Y$ and $C$ is at $Z$. $A$ is launched towards $B$ with speed $6\,\text{m s}^{-1}$, and at the same moment $C$ is launched towards $B$ with speed $v\,\text{m s}^{-1}$. In the later motion, $A$ collides with $B$ and they coalesce to form particle $D$. Then $D$ collides with $C$ and they coalesce to form particle $E$, which then moves towards $Z$.
(a)[3]
Show that after the second collision the speed of $E$ is $\frac{15 - v}{4}\,\text{m s}^{-1}$.
(b)[3]
The system loses $63\,\text{J}$ of kinetic energy altogether because of the two collisions. Use the result from (a) to show that $v = 3$.
(c(i))[4]
It is stated that $XY = 36\,\text{m}$ and $YZ = 98\,\text{m}$. Determine the time interval between the two collisions.
(c(ii))[1]
Find the time interval between the instant when $A$ is launched from $X$ and the instant when $E$ reaches $Z$.
Worked solution & mark scheme
This 11-mark question has a full step-by-step worked solution and mark scheme. One marking point: “For the first collision, apply conservation of momentum, for example $5 \times 6 = (5+1)v_D$” …