2019-03-01_Physics_Times

The two particles are shown in the figure

From previous problem we already know that

3

'

8 8

T T

t   t

The particles are represented on a circular path on

which they are in uniform circular motion.

Here the shadows of the two particles are moving

in same direction. We reverse particle-1 and it is

represented by

Method-1 The time of meeting is

Method-

Method-

3 / 2 3

2 8

r

r

T

t

 

 

  

r-relative angular displacement

r-relative angular velocity

The equations of the particles are

1

3

sin( )

2

x A t



 

x A t 2  sin( )

when they meet

x x 1  2

3

sin( ) sin

2

A t A t



  

3

( 1)

2

  t n    n t

For n 0

3

2

t t



   (not possible)

For n 1

3

2

t t



    

3

2

2 2

t

 

   

8

T

t



 (not possible)

For n 2

3

2

2

t t



     (not possible)

For n 3

3

3

2

t t



    

3 3

2

2 8

T

t t



   

For n 5

3

5

2

t t



    