A particle of mass m collides elastically with the pan of mass (M = 2m) of a spring balance, as shown in figure. Pan is in equillibrium before collision. Spring constant is k and speed of the particle before collision is `v_(0)` . Answer the following three questions regarding this collision.

Maximum height attained by the particle from the point of collision after collision is

Options

A. `(v_(0)^(2))/(16 g)`
B. `(v_(0)^(2))/(8g)`
C. `(v_(0)^(2))/(36 g)`
D. `(v_(0)^(2))/(72 g)`

Explanation

`v_(1)+v_(2)=V_(0)//2`
`2mv_(2)-mv_(1)=m(v_(0))/2rArr v_(1)=(v_(0))/6 , v_(2)=(v_(0))/3`
Maximum compression =`(v_(2))/(sqrt(K/(2m)))=sqrt((2m)/k)(v_(0))/3`
Maximum height =`(v_(1)^(2))/(2g)=(v_(0)^(2))/(72 g)`
minimum kinetic energy =`1/2 m((sqrt(3)v_(0))/2)^(2)=(3mv_(0)^(2))/8`