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Home / Doubts and Solutions/JEE/Physics

. A light spring of force constant 'K' is held
between two blocks of masses 'm' and '2m'
The two blocks and the spring system rests
on a smooth horizontal floor. Now the blocks
are moved towards each other compressing
the spring by 'x' and suddenly released. The
relative velocity between the blocks when the
spring attains its natural length will be

Asked by chandanbr6004 2nd November 2017, 5:43 PM
Answered by Expert
Answer:
When the spring of force constant 'K' is pressed to a distance x, the workdone
which is stored in the spring as potential energy is begin mathsize 12px style 1 half K x squared end style. This stored energy 
is utilized to give kinetic energy of masses m and 2m when the pressed spring is 
released.
 
begin mathsize 12px style 1 half m v squared plus 1 half left parenthesis 2 m right parenthesis V squared space equals space 1 half K x squared space left parenthesis space E n e r g y space C o n s e r v a t i o n right parenthesis m v space plus space left parenthesis 2 m right parenthesis V space equals space 0 space left parenthesis M o m e n t u m space c o n s e r v a t i o n right parenthesis end style
 
here v is velocity of block of mass m and V is that of block of mass 2m
 
begin mathsize 12px style w e space w i l l space g e t space V space equals space minus v over 2 end style from momentum conservation. Substitute this V in
energy conservation equation and solve for v. we will get begin mathsize 12px style v space equals space square root of fraction numerator 2 K over denominator 3 m end fraction end root x end style 
 
Relative velocity with respect to block of mass m is begin mathsize 12px style v minus V space equals space v plus v over 2 space equals space 3 over 2 v space equals space 3 over 2 square root of fraction numerator 2 K over denominator 3 m end fraction end root x end style
Answered by Expert 13th December 2017, 7:45 PM
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