IIT JEE , AIEEE Forum - Physics , Mechanics - The ultimate challenge in Mechanics from Feynmann

feynmann

Two particles A & B start from positions ( 0, 0 ) & ( 0, -d ) and move with constant speeds v & u respectively . A moves along x - axis and B moves such that its velocity is always aimed at A . Let r be the distance between them and be the angle made by the velocity of B with X - axis , at some time t .

Prove that ,

r ( 1 - cos )^ ( u/v )

--- = ------------------------------

d ( sin ) ^ ( u/v + 1 )

I have solved the problem & will publish it on next sunday .

Let me see ,by then , who among u can solve it !!!

All are welcome .

bhuwanaroracorroded

the path of d particle will be such that at

any time tangent drawn to d curve wil pass

through d particle

let at any time t d coordinates of particle B

be b ai+bj

eq of d curve on which d particle is moving is

y=f(x)

eq. of tangent (x-a)f¹(a)=(y-b)

this will satisfy (vt,0)

(vt-a)f¹(a)=-b

f¹(a)=-b/(vt-a)

also b=f(a)

r²=(vt-a)²+b²

f¹(a)=tan

da/dt=ucos

d(d-b)/dt=usin

getting these equations

feynmann

The above procedure does not solve the problem at all ..

feynmann

C'mon guys . Solve it !!!!!!!!!!!!!!!

feynmann

Hope that at least one goiitian should be able to solve it !!!!

elessar_iitkgp

The problem is best solved in plane polar coordinates. (For an analysis of motion in plane polar coordinates please refer to "Fundamentals of Mechanics" by I E Irodov)
Velocity of B in the frame of A,
(dr/dt)e_r + (r(d

/dt)e_th = [u - vcos

]e_r - (- vsin

)e_th

(dr/dt) = u-vcos

------------------(1)
And
r(d

/dt) = vsin

----------------(2)

Dividing (1) by (2) and rearranging,
(1/r)dr = (u/v)cosec

d

- cot

d

_d

^r (1/r)dr = _pi/2

^theta(u/v)cosec

d

- _pi/2

^thetacot

d

ln(r/d) = ln[(cosec

- cot

)^u/v/sin

]
r/d = (1 - cos

)^u/v]/(sin

)^(u/v)+1

Was quite an enjoyable problem ... can you tell me its source??

feynmann

Excellent answer !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Here is ur five pointer .

Anyway ,I got the problem in a very old magazine .

sid.shah.90

Lol
IE Irodov is the source;) coz i remember I had done this problem last yr

feynmann

Hey , I have solved Irodov three times !!!!!!!!!!!!!!!

Could u plz give me the problem no in Irodov where the same problem with the SAME ANS SOUGHT is posed ?

If u can ,hurry .

If u can't ,plz think twice before posting any comment .

Thank u !!!

feynmann

Anyway, I can provide u a solution in cartesian co-ordinates .

see, the controlling differential eqns are clearly.........

dx/dt = u cos

..................... ( 1 )

dy/dt = usin

.......................( 2 )

y/( x - vt) = tan

....................... ( 3 )

y = - r sin

.................. ( 4 )

Now the task is to solve these coupled differential eqn .

From ( 3 ) we have

x - vt = y cot

dwrt 't'

dx/dt -v = dy/dt cot

- y cosec^2(

) d

/dt

combining ( 1 ) & ( 2 ) with this eqn we have

u cos

- v = u sin

cos

/sin

+ r cosec

d

/dt

or , d

/dt = -v/r sin

.................. ( A )

differentiating ( 4 ) wrt 't ' , we get

u sin

= - d/d

( r sin

) d

/dt

combining with ( A ) we get

u/v = cos

+ sin

1/r dr/d

or , u/v cosec

- cot

= 1/r dr/d

Now integrating it and putting the boundary condition that r= d at

=

/2

we get

r/d = ( 1- cos

) ^u/v / ( sin

) ^ ( u/v + 1 )

Werewolf

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