Three particles $A$, $B$ and $C$ have masses $0.3\,\text{kg}$, $0.4\,\text{kg}$ and $m\,\text{kg}$ respectively and are at rest on a smooth horizontal plane in a straight line. The separation of $B$ and $C$ is $2.1\,\text{m}$. $A$ is launched directly at $B$ with speed $2\,\text{m s}^{-1}$. Once $A$ collides with $B$, the speed of $A$ becomes $0.6\,\text{m s}^{-1}$, still travelling in the same direction.
(a)[2]
Demonstrate that the speed of $B$ immediately after the impact is $1.05\,\text{m s}^{-1}$.
(b)[2]
Following the collision between $A$ and $B$, $B$ travels directly towards $C$. Particle $B$ then strikes $C$. After that collision, the two particles join together and move with a common speed of $0.5\,\text{m s}^{-1}$. Determine $m$.
(c)[5]
Determine the time from the moment when $B$ and $C$ collide until $A$ collides with the combined particle.
Worked solution & mark scheme
This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Apply conservation of momentum: $0.3\times2=0.3\times0.6+0.4v$” …