M1 need help in a complicated problem

Points A,B and C on the wire have coordinates (0,0) , (3,3) , (6,6). A 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.

If u = 10, the bead comes to rest at a point D between B and C. Find

(a) The greatest speed of the bead between B and D.

(b) The coordinates of D.

What happens after the bead reaches D?

The answer for (a) is 6.83ms-1 and for (b) is (5.7 , 5)

Re: M1 need help in a complicated problem

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Originally Posted by

**abdulrehmanshah** Points A,B and C on the wire have coordinates (0,0) , (3,3) , (6,6).

So the points are on the line y= x.

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A 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 $\displaystyle 3\sqrt{2}$ so the increase in **potential energy** is $\displaystyle 3\sqrt{2}mg= 3\sqrt{2}(3)(9.8)= 153$ Joules. The vertical distance from (0, 0) to (6, 6) is $\displaystyle 6\sqrt{2}$ so the increase in potential energy is $\displaystyle 6\sqrt{2}mg= 6\sqrt{2}(3)(9.8)= 306$ 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 $\displaystyle (1/2)mv^2$.)

If u = 10, the bead comes to rest at a point D between B and C. Find

(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?

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(b) The coordinates of D.

At what height will all of the initial kinetic energy be converted to potential energy?

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What happens after the bead reaches D?

Well, that should be obvious! It has come to a stop at D and now gravity is pulling it back down!

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The answer for (a) is 6.83ms-1 and for (b) is (5.7 , 5)

Re: M1 need help in a complicated problem

how do u get vertical distance as 6\sqrt{6} . it should be 6

Re: M1 need help in a complicated problem

oh sorry i missed something i forgot to write the starting part of the question i.e a smooth wire is bent into the sahpe of graph of y=(1/6)x^3-(3/2)x^2+4x

Re: M1 need help in a complicated problem