The block of mass `m_1` shown in figure is fastened to the spring and the block of mass `m_2` is placed asgainst ilt. A. Find the compression of the spring in the equilibrium position. b.The blocks are pushed a further distance `(2/k)(m_1+m_2)gsintheta` against the spring and released. Find the position where the two blocks separate. c. What is the common speed of blocks at the time of separation?

Explanation

(a) In equilibrium
`(m_1+m_2)gsintheta=kx_0`
`x_0=((m_1+m_2)gsintheta)/(k)`
(b)
A: initial position of blocks
O: equilibrium position
Block `m_1` is attached to spring. When blocks are released from B, blocks move with same acceleration up to A. After A, when spring extends, `m_1` is under retardation due to spring and gravitational force and `m_2` is under retardation due to gravitational force. Retardation of `m_1` is greater than `m_2`, hence, blocks separate when springs attains natural length.
After O:

(c) Applying energy conservation between B and A
`1/2k[3/k(m_1+m_2)gsmtheta]^2`
`=(m_1+m_2)g3/k(m_1+m_2)gsinthetasintheta+1/2(m_1+m_2)v^2`
`((m_1+m_2)^2g^2sin^2theta)/(k)[9/2-3]=1/2(m_1+m_2)v^2`
`v=sqrt((3(m_1+m_2)g^2sin^2theta)/(k))`
`v:` common speed of blocks