Chimpoo and McZee - II

Our little MacZee …When Chimpoo is not here.

He likes to take a little ride with balloons

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Now see what he has done…

The balloon, the light rope and McZee are at rest in the air. If McZee reaches the top of the rope, by what distance does the balloon descend? Mass of the balloon = M, mass of McZee = m and the length of the rope ascended by the monkey = L

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Oh come on chose one balloon Mczee wud ya?

Promise me we both stay

together balloon

You got it maczee

What an idiot !!

Now as balloonHas promised to stay together

With McZee and McZee since he has

No choice , they shall be now pronouncedAs A SYSTEM

-A proclamation-Signed by

His highnessIssac Newton

Thanks balloon. Wait I

am coming over to kiss you

as a token of appreciation

Are you sure? Well than I will also be pulled down

anyways towards you if you come towards me.

Its ok don’t come. I don’t want to pulled downwards

Man this is not cool. I wish I could deflate myself fully

to avoid any embarrassment

Ok , my friend, I am Acceleration at A m/s2

now upwards.

Doing this I will have to

pull you downwards

which means that as I go up, you come

down.

Fine. What would be my acceleration

downwards McZee? I want to see how much I

move downwards for you?

I have no idea. Let meAsk his highness Mr.

newtonDamn it you monkey!! You never rest. Do you? Promise me this is the last time you are doing

your stupid climbing..Ok balloon and monkey , since you both a

system together any forces inside the system need not to be considered if I see your overall motion of the system. Now the net force acting

on the system is the weights of you two i.e (M+m) g and if I am considering a system the

internal forces shall be equal and opposite which is tension force T. So coming to your situation, when you are ascending upwards

you are applying a pulling force on rope which in reaction pulls you upwards

T

mg

TMg

Remember these forces are drawn w.ref. t the ground and

here in this case if I also consider ground plus you two as a system, there is no external force which is acting on you. So that would help me use the Law of Conservation of Momentum which says “ In absence of any external force acting on the system, the net

momentum of the system remains unchanged which is true

in deed for your case.Now just check out your momentums,

balance them and find out what you are looking for

I was initially at rest and not travelling. So my initial momentum was zero. Let us say that I

gained a velocity v when I climbed L on the rope , my final momentum w.r.t ground is then mv..What

about you balloon?

Well, my momentum initially was zero and as you moved

upwards, I decended downwards and reached a velocity V when

you climbed L upwards w.r.t rope. So my final momentum is

MV

What do we do now Mr. Newton.

Does that means our total momentum should be equated

to Zero which means in this case MV-mv = 0 ( taking in to considerations signs of velocities

too..Yea this monkey is getting

smarter day by day ;) )

Let me check out with the rope now.

Rope you might know how much I

moved on over you and with what velocity . Any

comments where shall we proceed?

Yea McZee you moved with a velocity of v+V upwards assuming that M moves down with a velocity V

(opposite direction. So L = (v+V)t.And balloon you , you moved down with a velocity V

w.r.t ground and the length descended by you is Vt i.e. L = vt + L1 or L1 = vt – L . Put t = L/ (v+V)

L1 = v L/ (v+V) – L. Use MV= mv or v /V= M/mL1 = M/m (L) ( M/m +1) – L

L1 = mL / (M+m) So, baloon descended mL/ ( M+m) distance

downwards the w.r.t ground.I hope that’s clear guys….

Well indeed it makes sense, SO I go upwards a distance of L w.r.t balloon and the balloon comes down with a

distance of mL/ ( M+m) w.r.t ground .Right.And we meet each

other pretty soon as I expected…Love you balloon.

I am never holding with this fool again.. And I seriously pity for this rope..Again stuck with

this monkey