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?

Two identical particles A & B of mass m=2kg is connected at the end of

1.	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» SOLUTION :`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`

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