A block of mass `m` is pushed up against a spring, compressing it to a distance `x`, and the block is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed `v`. The same spring projects a second block of mass `4m`, giving it a speed `3v`. What distance was the spring compressed in the second case ?

Options

A. `6x`
B. `x/6`
C. `36x`
D. `12x`

Answer

A

Explanation

`(1)/(2)k(x')^(2)=(1)/(2)xx(4m)xx(3v)^(2)`
`=36((1)/(2)mV^(2))=36((1)/(2)kx^(2))=(1)/(2)k(6x)^(2)`
`rArrx'=6x`