A horizontal spring block system executes SHM with ampitude A = 10 cm initial phase `phi =0` and angular frequency `omega`. The mass of block is M = 13kg and there is no friction between the block and the horizontal surface. The spring constant being 2500 N/m
At `t = t_(1)` sec [for which `omega t_(1) = phi_(1) = 30^(@)`). A mass m = 12kg is gently put on the block. [Assume that collision between the block and the mass is perfectly inelastic and mass m remains stationary w.r.t the block M always]

The new amplitude of the system will be

Options

A. 5 cm
B. 8 cm
C. 10 cm
D. none of these

Answer

B

Explanation

`omega. = omega sqrt((M)/(M + m))`
`x_(1) = A sin phi_(1)`
Potential energy stored in the spring at `(t = t_(1)) = (1)/(2) kx_(1)^(2) = (1)/(2) M omega^(2) sin^(2) phi_(1)`
Let v and v. be the velocity of system just before collsion and just after collision, so using COLM
`v. = (Mv)/((M + m)) = (MA omega cos theta)/((M + m))`
Total energy after collision `= PE + KE = (1)/(2) kx_(1)^(2) + (1)/(2) (M + m)v^(2)`
`=(1)/(2) MA^(2) omega^(2) [(M + m sin^(2) theta)/(M + m)] = (1)/(2) kA^(2) [(M + m sin^(2) theta)/(M + m)] = 8J`
`(1)/(2) (M + m) A^(2) omega^(2) = (1)/(2) MA^(2) omega^(2) [(M + m sin^(2) theta)/(M + m)]`
`A. = A sqrt((M + m sin^(2) theta)/(M + m)) = 8 cm`