A block of mass `m=0.1 kg` is connceted to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest. After approaching half the distance `((x)/(2))` from the euilibrium position, it hits another block and comes to rest momentarily, while the other block moves with velocity `3ms^(-1)`. The total initial energy of the spring is :

Options

A. 0.3 J
B. 0.6 J
C. 0.8 J
D. 1.5 J

Answer

C

Explanation

Conservation of linear momentum
`m_(1)u_(1) +m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)`
`m_(1)u+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)`
`m_(1)u+m(0) = m_(1)(0)+m(3)`
`(0.1)u = m(3), " "…(i)`
`1/2 (0.1)u^(2) =1/2m(3)^(2)" " …(ii)`
Solving equation (i) & (ii) we get , u = 3
`1/2 kx^(2) = 1/2 k(x/2)^(2) +1/2(0.1)3^(2)`
`rArr " " 3/4 kx^(2) = 0.9 rArr 3/4 xx1/2kx^(2)=0.9`
Intial energy of spring , `1/2Kx^(2) = 0.6 J `