Question

Two small identical discs, each of mass `m`, lie on a smooth horizontal plane. The discs are interconnected by a light non-deformed spring of length `l_0` and stiffness `ϰ`. At a certain moment one of the disc is set in motion in a horizontal direction perpendicular to the spring with velocity `v_0`. Find the maximum elongation of the spring in the process of motion, if it is known to be considerably less than unity.

Answer 1

In the C.M. frame of the system (both the discs+spring), the linear momentum of the discs are related by the relation, `overset~vecp_1=-overset~vecp_2`, at all the moments of time.
where, `overset~p_1=overset~p_2=overset~p=muv_(rel)`
And the total kinetic energy of the system,
`T=1/2muv^2_(rel)`
Bearing in mind that at the moment of maximum deformation of the spring, the projection of `vecv_(rel)` along the length of the spring becomes zero, i.e `v_(rel(x))=0`.
The conservation of mechanical energy of the considered system in the C.M. frame gives.
`1/2(m/2)v_0^2=1/2kx^2+1/2(m/2)v_(rel(y))^2` (1)
Now from the conservation of angular momentum of the system about the C.M.,
`1/2(l_0/2)(m/2v_0)=2((l_0+x)/(2))m/2v_(rel(y))`
or, `v_(rel(y))=(v_0l_0)/((l_0+x))=v_0(1+x/l_0)^-1~~v_0(1-x/l_0)`, as `xlt ltl_0` (2)
Using (2) in (1), `1/2mv_0^2[1-(1-x/l_0)^2]=kx^2`
or, `1/2mv_0^2[1-(1-(2x)/(l_0)+(x^2)/(l_0^2))^2]=kx^2`
or, `(mv_0^2x)/(l_0)~~kx^2`, [neglecting `x^2//l_0^2`]
As `x!=0`, thus `x=(mv_0^2)/(kl_0)`