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?

Options

A. `sqrt((2)/(k)(m_(1)+m_(2))).g sin theta`
B. `sqrt((3)/(k)(m_(1)+m_(2))).g sin theta`
C. `sqrt(((m_(1)+m_(2))/(k))).g sin theta`
D. `sqrt((5)/(k)(m_(1)+m_(2))).g sin theta`

Answer

B

Explanation

At the equilibrium condition
`x = ((m_(1)+m_(2))g sin theta)/(k) rarr (1)`
Let `x_(1) = (2)/(k) (m_(1)+m_(2)) g sin theta rarr (2)`
Total compression in spring `=x +x_(1)`
Let `x_(2)` be the displacement of `m_(1)` at the time of loosing the contant with `m_(2)` from mean position `F.B.D.` at the time of loosing contact.
`:. m_(1)a = m_(1)g sin theta - k (x-x_(2))`
`rArr m_(1) omega^(2)x_(2) - m_(1)g sin theta -kx +kx_(2)`
`rArr (m_(1)k)/((m_(1)+m_(2))) x_(2) -kx_(2)+k. ((m_(1)+m_(2))gsin theta)/(k) =m_(1)g sin theta`
(`:' omega = sqrt((k)/(m_(1)+m_(2)))) rArr (-m_(2)k)/((m_(1)+m_(2)))x_(2) =- m_(2)g sin theta`
`rArr x_(2) = ((m_(1)+m_(2))gsin theta)/(k) =x rarr (3)`
`:.` At the time of loosing contact, no compression in spring
`rArr` from law of conservation of energy, `E_("initia") = E_("final")`
`rArr (1)/(2)k(x_(1)+x)^(2) -(m_(1)+m_(2)) g sin theta.(x+x_(1)) =(1)/(2)(m_(1)+m_(2))v^(2)`
`v = sqrt((3)/(k)(m_(1)+m_(2))).gsin theta`