The 5kg collar B is at rest and when it is in the position s

The 5kg collar B is at rest and when it is in the position shown, the spring is un-stretched. Another 1 kg collar A strikes it at the velocity of 2m/s and as a result. B slides down the smooth circular path, as shown. The coefficient of restitution between A and B is e = 1.0. Determine, The velocity of collar B when it is at position C. Answer: The force exerted by the guide on collar B, when it is at point C Answer: Motion occurs in the vertical plane and all surfaces are smooth.

Solution

a)

Initial potential energy = mgh = 5*9.81*0.6 = 29.43 J

Initial kinetic energy = 1/2*mv^2 = 1/2 *1*2^2 = 2 J

Total energy = 29.43+2 = 31.43 J

When at C, length of spring = sqrt((0.6+1.4)^2 + 0.6^2) = 2.088 m

When at C, spring extension = 2.088 - 1.4 = 0.688 m

Energy stored in spring = 1/2*k*x^2 = 1/2 * 20 * 0.688^2 = 4.734 J

By energy conservation: 31.43 = 4.734 + 1/2*m*V^2....where m = 5 kg

Solving, V = 3.268 m/s

b)

At C, spring force = kx = 20*0.688 = 13.76 N

Component of force in radial direction = 13.76*(0.6/2.088) = 3.95 N