Particles $A$ and $B$, with masses $0.4\,\text{kg}$ and $0.2\,\text{kg}$ respectively, travel down the same line of greatest slope on a smooth plane. The plane is inclined at $30^\circ$ to the horizontal, and $A$ is situated above $B$ on the plane. At the moment of collision, the speeds of $A$ and $B$ are $3\,\text{m s}^{-1}$ and $2\,\text{m s}^{-1}$ respectively. During the collision, the speed of $A$ is lowered to $2.5\,\text{m s}^{-1}$.
(a)[2]
Determine the speed of $B$ immediately after the collision.
(b)[7]
After the collision, when $B$ has travelled $1.6\,\text{m}$ down the plane from the collision point, it strikes a barrier and then moves back up the same line of greatest slope. $B$ reaches the barrier $0.4\,\text{s}$ after the collision, and at the barrier its speed is reduced by $90\%$. The two particles meet again $0.44\,\text{s}$ after their earlier collision, and they then coalesce on impact. Show that the speed of $B$ immediately after it strikes the barrier is $0.5\,\text{m s}^{-1}$. Hence find the speed of the combined particle immediately after the second collision between $A$ and $B$.
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: $0.4\times3+0.2\times2=0.4\times2.5+0.2v$.” …