Mass `m_(1)` hits & sticks with `m_(2)` while sliding horizontally with velocity `v` along the common line of centres of the three equal masses `(m_(1)=m_(2)=m_(3)=m)`. Initially masses `m_(2)` and `m_(3)` are stationary and the spring is unstretched. Find
(a) the velocities of `m_(1),m_(2)` and `m_(3)` immediately after impact.
(b) the maximum kinetic energy of `m_(3)`.
(c) the minimum kinetic energy of `m_(2)`.
(d) the maximum compression of the spring.

Answer

(a) u/2,mu/2; (b) `2mv^(2)//9;(c)mv^(2)//72;(d)x=sqrt(m//6k)v`

Explanation

Just before collision


(a) We can use conservation of momentum as spring force is not an impulsive force.
`rArrmv+0=(2m)v'rArr=(v)/(2)`
`rArrv_(m_(1))=(v)/(2),v_(m_(2))=(v)/(2),v_(m_(3))=0`
(b) Looking from CM frome of refernce
`v_(CM)=((2m)v'+m(0))/(2m+m)=(2)/(3)v'=(v)/(3)`

(Both the blocks will execute SHM of same time period in CM frome of reference)
`rArr(v_(m_(3)))_(max)=(v)/(3)+v_(CM)=(v)/(3)+(v)/(3)=(2v)/(3)`
`rArr(KE_(m_(3)))_(max)=(1)/(2)m(v_(3))_(max)^(2)=(1)/(2)m((2v)/(3))^(2)=(2mv^(2))/(9)`
(c) `(v_(m_(2)))_(min)=(v)/(3)-(v)/(6)=(v)/(6)`
`rArr(KE_(m_(2)))_(min)=(1)/(2)m(v_(2))_(min)^(2)=(1)/(2)m((v)/(6))^(2)=(mv^(2))/(72)`
(d) At the time of maximum compression, both blacks will be at rest in CM frame of reference
`rArr(1)/(2)kx_(max)^(2)=(1)/(2)(2m)((v)/(6))^(2)+(1)/(2)m((v)/(3))^(2)`
`rArr(1)/(2)Kx_(max)^(2)=(1)/(12)mv^(2)`
`rArrx_(max)=vsqrt((m)/(6k))`