The diagram represents a smooth track lying in a vertical plane. Section $AB$ is a quarter circle with radius $1.8\,\text{m}$ and centre $O$. Section $BC$ is a horizontal straight segment of length $7.0\,\text{m}$, with $OB$ perpendicular to $BC$. Section $CFE$ is a straight line that is inclined at an angle of $\theta^\circ$ above the horizontal. A particle $P$ of mass $0.5\,\text{kg}$ is released from rest at $A$. Particle $P$ then collides with a particle $Q$ of mass $0.1\,\text{kg}$ that is at rest at $B$. Immediately after the collision, the speed of $P$ is $4\,\text{m s}^{-1}$ in the direction of $BC$. Assume that $P$ is moving horizontally at the moment it collides with $Q$.
(a)[4]
Show that the speed of $Q$ immediately after the collision is $10\,\text{m s}^{-1}$.
(b)[4]
When $Q$ arrives at $C$, it collides with a particle $R$ of mass $0.4\text{ kg}$ which is at rest at $C$. The two particles coalesce. The combined particle comes to rest instantaneously at $F$. Assume that there is no instantaneous change in speed as the combined particle leaves $C$, or when it passes through $C$ again as it moves back down the slope. If the distance $CF$ is $0.4\text{ m}$, find the value of $\theta$.
(c)[5]
Determine the distance from $B$ at which $P$ collides with the combined particle.
Worked solution & mark scheme
This 13-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Conservation of energy used to obtain $v=6$” …