The block of mass `m_1` shown in figure is fastened to the spring and the block of mass `m_2` is placed as against it. Find the compression of the spring in the equilibrium position.

Answer

`rarr x=((m_1+m_2)gsintheta)/k`
`x_1=2/k(m_1+m_2)gsintheta`
`v=sqrt({3/k(m_1+m_2)})gsintheta`

Explanation

`a`. At the equilibrium condition
`kx=(m_1+m_2)gsin theta`
`rarr x=((m_1+m_2)gsintheta)/k`
`b`. `x_1=2/k(m_1+m_2)gsintheta`
given When the system is released it will start to make SHM.
where , `omega=sqrt(k/(m_1+m_2))`
when the blocks lose contact
`p=0`
So `m_2gsintheta=m_2x_2xxk/(m_1+m_2)`
`rarr x_2=((m_1+m_2)gsintheta)/k`
so the blocks will lose contact with each other when the springs attain its natural length.
`c`.Let the common speed attained you both the block be v
`1/2(m_1+m_2)v^2-0`
`=1/1k(x_1+x_2)^2-(m_1+m_2)gsintheta(x+x_1)`
`[x+x_1=`total compression]
`rarr 1/2(m_1+m_2)v^2`
`=1/2k(3/k)(m_1+m_2)gsintheta-(m_1+m_2)gsintheta-(x_1+x_2)`
`rarr 1/2(m_1_m_2)v^2=`
`1/2(m_1+m_2)gsinthetaxx(3/k)(m_1+m_2)gsintheta
`rarr `v=sqrt({3/k(m_1+m_2)})gsintheta`