AL app math~circular motion~

by conservation of enegry
0.5mu² + mg(2l) = 0.5mv²
u² + 4gl = v²-----------(1)

by conservation of momentum
mv = (m + 2m)v'
v = 3v'--------------(2)

by conservation of energy again, the minimum v'
0.5mv'² => mg(2l)
v' => sqrt(4gl)

using (2)
v => 3 sqrt(4gl)

using (1)
u² => (36 - 4) gl
u => 4sqrt(2gl)

To complete a whole circle, by the law of conservation of energy,

Loss of G.P.E. at the highest point <= Gain of K.E. at the lowest point

mg(2l) <= 1/2 mv2

v <= 2sqrt(gl)

2009-05-14 21:11:06 補充：
At the lowest position where the two particles collide, by the law of conservation of momentum,

mu' = (m + 2m)v

u' <= 6sqrt(gl)

This is the speed of the particle with mass m just before the collision.

2009-05-14 21:11:16 補充：
By the law of conservation of energy,

Loss of G.P.E. = Gain of K.E.

mg(2l) = 1/2 mu'2 - 1/2 mu2

4gl <= 36gl - u2

u <= 4sqrt(2gl)