Two blocks, of masses `M` and `2 M`, are connected to a light spring of spring constant `K` that has one end fixed, as shown in figure. The horizontal surface and the pulley are frictionless. The blocks are released from when the spring is non deformed. The string is light.
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Options

A. Maximum extension in the spring is `(4 Mg)/(K)`
B. Maximum kinetic energy of the stystem is `(2M^(2)g^(2))/(K)`
C. Maximum energy stored in the spring is four times that of maximum kinetic energy of the system.
D. When kinetic energy of the system is maximum, energy stored in the spring is `(4 M^(2)g^(2))/(K)`

Answer

A, B, C

Explanation

Maximum extension will be at the moment when both masses stop momentarily after going down Applying W-E theorem from starting to that instant.
`k_(f)-k_(i)=W_(gr)+W_(sp)+W_(ten)`
`0-0=2 M.g x+(-(1)/(2) kx^(2))+0 x=(4 Mg)/(K)`
System will have maximum KE when net force on the system becomes zero. Therefore
`2 Mg = T and T = kx rArr x = (2Mg)/(K)`
Hence KE will be maximum when 2M mass has gone down by `(2 Mg)/(K)`
Applying W.E theorem
`k_(f)-0=2Mg.(2Mg)/(K)-(1)/(2)K(4 M^(2)g^(2))/(K^(2))`
`K_(f) =(2 M^(2)g^(2))/(K^(2))`
Maximum energy of strong
`=(1)/(2)K.((4 Mg)/(K))^(2)=(8 M^(2)g^(2))/(K)`
Therefore Maximum spring energy
`= 4 xx` maximum K.E
When K.E is maximum `x = (2 Mg)/(K)`
Spring energy `= (1)/(2). K. (4 M^(2) g^(2))/(K^(2))`
`=(2 M^(2)g^(2))/(K^(2))` i.e. (D) is wrong.