Two identical particles `A` & `B` of mass `m=2kg` is connected at the end of the spring `(K=50N//m)` and placed on a smooth horizontal surface. Now, force `F=10N` and `2F` are applied on the particles in opposite direction as shown in figure. Initially spring is in natural length.
Simultaneously particle `A` receives an impulse towards `B` which impats speed `v_(0)` to it, find `v_(0)` in m/s for which length of elongation is two times the length of compression in the spring during subsequent motion?

Answer

6

Explanation

`1/2.m/2 v_(0)^(2)=1/2K((3F)/2)^(2)+(3F)/2.(3F)/K`
`v_(0)^(2)=(4xx9F^(2))/(Km)`
`v_(0)=6m//s`