If u = 10, the bead comes to rest at a point D between B and C. FindA m-kg bead is projected along the wire from A with speed u ms-1 so that it has enough energy to reach B but not C. Prove that 8.16<u<10.95 and that the speed at B >/= 2.58 ms-1. The vertical distance from (0, 0) to (3, 3) is so the increase in potential energy is Joules. The vertical distance from (0, 0) to (6, 6) is so the increase in potential energy is Joules. Since the bead has enough kinetic energy, initially, to go between B and C, what bounds must its speed be between? (I presume you know that kinetic energy is .)
(a) The greatest speed of the bead between B and D.[/quote]
Obviously, the bead is losing speed as it goes up so its greatest speed 'between B and D' is at B. You are given the bead's speed at A so you can calculate its kinetic energy there. It will stop when its potential energy is equal to that kinetic energy (all kinetic energy has converted to potential energy). At B, what is the potential energy? So what is the remaining kinetic energy? And, from that, is the bead's speed at B?
At what height will all of the initial kinetic energy be converted to potential energy?(b) The coordinates of D.
Well, that should be obvious! It has come to a stop at D and now gravity is pulling it back down!What happens after the bead reaches D?
The answer for (a) is 6.83ms-1 and for (b) is (5.7 , 5)