A bead, $A$, with mass $0.1\,\text{kg}$ is placed on a long straight rigid wire inclined at $\sin^{-1}\left(\frac{7}{25}\right)$ to the horizontal. $A$ starts from rest and slides down the wire. The coefficient of friction between $A$ and the wire is $\mu$. After $A$ has moved $0.45\,\text{m}$ down the wire, its speed is $0.6\,\text{m s}^{-1}$.
(a)[6]
Show, therefore, that $\mu = 0.25$.
(b)[6]
A second bead, $B$, of mass $0.5\,\text{kg}$ is likewise threaded on the wire. When $A$ has moved $0.45\,\text{m}$ down the wire, it strikes $B$, which is momentarily at rest on the wire. $A$ is brought to instantaneous rest in the collision. The coefficient of friction between $B$ and the wire is $0.275$. Determine the time from when the collision occurs until $A$ collides with $B$ again.
Worked solution & mark scheme
This 12-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Apply a constant acceleration equation, for example $0.6^2 = 2a \times 0.45$.” …