A sphere of mass `m` and radius `R` rolls without sliding on a horizontal surface. It collides with a light spring of stiffness `K` with a kinetic energy `E`. If the surface (`AB`) under the spring is smooth, find the maximum compression of the spring.

Explanation

The total kinetic energy of the sphere `=E=E_("tran")+E_("rot")`
`=1/2mv^(2)+1/2 2/5mR^(2) (v^(2))/(R^(2))=7/10mv^(2)implies v=sqrt((10E)/(7m))`
Since the spring force `F` passes through the centre `O` of the sphere, it cause no torque about `O`. therefore, the angular momentum and hence angular velocity of the sphere remains constant. Since the surface `AB` is smooth no frictional loss take place. Therefore, conserve in the energy of the system (sphere spring) between the given (initial) position and the final position (maximum compression of the spring), we obtain
`/_\KE_("sphere")+/_\E_("spring")=0`
`implies/_\KE_("tran")+/_\KE_("rot")+/_\PE_("spring")=0`
Since the initial angular velocity of the sphere remains constant due to the absence of friction.

`/_\KE_("rotation")=0implies/_\KE_("trans")+/_\PE_("spring")=0`
`=(0-1/2mv^(2))+(1/2kx^(2)-0)=0`
where `v=sqrt((10E)/(7m))` and `x=` maximum compression of the spring
`implies1/2 Kx^(2)=1/2m(sqrt((10E)/(7m))^(2))impliesx^(2)=(10E)/(7k)`
`implies x=sqrt((10E)/(7k))`