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 μ. When A has travelled 0.45m down the wire, its speed is 0.6 m s−1.
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.
(a) Question
Show that – = 0.25.
▶️Answer/Explanation
(a) Using Work-Energy Principle:
The work done by the friction force is equal to the change in kinetic energy:
\( W_{\text{friction}} = \mu mgh \)
The change in kinetic energy is given by:
\( \Delta KE = \frac{1}{2}mv^2 \)
Equating the work done by friction to the change in kinetic energy:
\( \mu mgh = \frac{1}{2}mv^2 \)
Given data:
\( m = 0.1 \, \text{kg}, \quad \sin^{-1}\left(\frac{7}{25}\right) \approx 16.3^\circ, \quad h = 0.45 \, \text{m}, \quad v = 0.6 \, \text{m/s} \)
Acceleration (\(a\)):
\( 0.6^2 = 0 + 2a \times 0.45 \)
Solving for \(a\):
\( a = 0.4 \, \text{m/s}^2 \)
Normal Reaction (\(R\)):
\( R = 0.1g \cos \alpha = 0.1g \times \cos 16.3^\circ \approx 0.96 \, \text{N} \)
Equating Forces along the Inclined Plane:
\( 0.1g \times \frac{7}{25} – F = 0.1g \times 0.4 \)
Solving for \(F\):
\( F = 0.24 \, \text{N} \)
Expressing Friction Force (\(F\)) in terms of \(\mu\):
\( F = \mu \times 0.1g \times \frac{24}{25} \)
Solving for \(\mu\):
\( \mu = \frac{0.24 \times 25}{24} = 0.25 \)
Therefore, \(\mu = 0.25\).
(b) Question
Find the time from when the collision occurs until A collides with B again.
▶️Answer/Explanation
(b)Using Conservation of Energy for Bead A:
\( 0.1 \times 0.6 = 0.5 \times v \)
\( v = 0.12 \, \text{m/s} \)
Equating Forces for Bead B:
\( 0.5g \times \frac{7}{25} – 0.275 \times 0.5g \times \frac{24}{25} = 0.5a \)
\( a = 0.16 \, \text{m/s}^2 \)
Equations of Motion for Beads A and B:
– For Bead A:
\( s_A = \frac{1}{2} \times 0.4t^2 \)
– For Bead B:
\( s_B = 0.12t + \frac{1}{2} \times 0.16t^2 \)
Equating Positions of Beads A and B:
\( \frac{1}{2} \times 0.4t^2 = 0.12t + \frac{1}{2} \times 0.16t^2 \)
Solving for \(t\):
Combine like terms and solve for \(t\):
\( 0.1t^2 – 0.12t – 0.12 = 0 \)
Solve this quadratic equation to find \(t\).
\(t = 1 \, \text{s}\), which is the required time.
A particle of mass 12 kg is stationary on a rough plane inclined at an angle of 250 to the horizontal. A force of magnitude P N acting parallel to a line of greatest slope of the plane is used to prevent the particle sliding down the plane. The coefficient of friction between the particle and the plane is 0.35.
(a) Question
Draw a sketch showing the forces acting on the particle.
▶️Answer/Explanation
(b) Question
Find the least possible value of P.
▶️Answer/Explanation
P + F = 12g sin 25 =50.7
R =12g cos 25 = 108.8
F = 120 cos 25 × 0.35 = 38.1
P + 38.1 = 120 sin 25
P = 12.6