Question

spring is compressed between two tickets of mass M1 and M2 when the toy cards are released the spring and results on each equal and opposite average force for the same time T if the coefficient of friction between the ground and the chords are equal then the displacement of the two tickets are in the ratio

Answer 1

Let the displacements be s1 and s2  coefficient of friction=U

Frictional force
=UR
=Umg
or,
ma=Umg
or,
a=Ug

v^2=u^2+2as (u=o)
v=sqrt(2as)

Now since they exert equal but opposite force so 

F1  = - F2

or,

Change in momentum of m1= - change in momentum of m2

or

m1*v1= -m2*v2 (time cancels)

or

m1*sqrt(2as1) = - m2*sqrt(2as2)

Squaring,

m1^2 * 2as1 =  - m2^2 2as2

So

s1/s2=-(m2^2.m1^2)

Answer 2

Answer:

further u can solve to get displacement ratio