Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
Feb/Mar 2022 p42 q7
3384
A bead, A, of mass 0.1 kg is threaded on a long straight rigid wire which is inclined at \(\sin^{-1}\left(\frac{7}{25}\right)\) to the horizontal. A is released from rest and moves down the wire. The coefficient of friction between A and the wire is \(\mu\). When A has travelled 0.45 m down the wire, its speed is 0.6 m s\(^{-1}\).
(a) Show that \(\mu = 0.25\).
Another bead, B, of mass 0.5 kg is also threaded on the wire. At the point where A has travelled 0.45 m down the wire, it hits B which is instantaneously 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.
(b) Find the time from when the collision occurs until A collides with B again.
Solution
(a) Using the equation of motion: \(0.6^2 = 0 + 2a \times 0.45\), we find \(a = 0.4\).
The normal reaction \(R = 0.1g \cos \alpha = 0.1g \times \frac{24}{25} = 0.1g \times \cos 16.3^\circ\).
Using Newton's second law: \(0.1g \frac{7}{25} - F = 0.1 \times 0.4\), where \(F = \mu R\).
Substituting \(F = \mu \times 0.1g \times \frac{24}{25}\), we solve for \(\mu\) and find \(\mu = 0.25\).
(b) Using conservation of momentum: \(0.1 \times 0.6 = 0.5v\), we find \(v = 0.12\).
For B, \(0.5g \frac{7}{25} - 0.275 \times 0.5g \times \frac{24}{25} = 0.5a\), leading to \(a = 0.16\).
Using \(s = ut + \frac{1}{2}at^2\), for A: \(s_A = 0 + \frac{1}{2} \times 0.4 \times t^2\), and for B: \(s_B = 0.12t + \frac{1}{2} \times 0.16 \times t^2\).
Setting \(s_A = s_B\) and solving for \(t\), we find \(t = 1\) s.