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

B

Explanation

Applying momentum conservation
`m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)`
0.1u+m(0)=0.1(0)+m(3)
0.1u=3m
`(1)/(2)0.1u^(2)=(1)/(2)m(3)^(2)`
`(1)/(2)kx^(2)=(1)/(2)K((x)/(2))^(2)+(1)/(2)(0.1)3^(2)`
`implies(3)/(4)kx^(2)=0.9implies(3)/(2)xx(1)/(2)kx^(2)=0.9`
`therefore (1)/(2)kx^(2)=0.6 J` (total initial energy of the spring)