- #1

steejk

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*A smooth wire forms a circular hoop of radius 1m. It is fixed in a vertical plane. Two beads, A and B, of masses m and 2m respectively, are threaded onto the wire. The coefficient of restitution between the beads is 0.5. Bead B rests at the bottom of the hoop. Bead A is projected from the topmost point with speed u, and subsequently collides with B.*

**Q. Show that Bead A is brought to rest by the collision.**

My working so far:

Using the energy change of GPE --> KE

GPE = 2mg = 0.5mv^2

Increase in velocity is therefore 2sqrt(g)

At the point A collides with B its velocity is therefore u + 2sqrt(g)

Using conservation of momentum,

u + 2sqrt(g) = 2x + y (where x is the speed of B and y is speed of A after collision)

Using Newton's law of restituiton:

0.5 = (2x-y)/(u + 2sqrt(g))

And this is where I get stuck :( ...