A uniform solid sphere of mass m and radius R is placed on a horizontal surface. A horizontal linear impulse P is applied at a height R2 above its center. Find the velocity of center of mass and angular velocity of sphere just after the impulse is applied.
All you have to recall is
linear impulse = change In linear momentum - - - - - - (1)
angular impulse = change in angular momentum - - - - - - (2)
Here,
linear impulse = P (given)
change in linear momentum = linear momentum final - linear momentum initial
= mVCM−0 [as initially the body was at rest
and let's say it moves with VCM after the impulse]
= mVCM
According to (1)
P =mVCM
⇒VCM=Pm
Angular impulse = linear impulse × perpendicular distance of line of
application of force from the axis of rotation
Change in angular momentum =I(ωf−ωi)
[ωi=0,ωf=ω]
=Iω
According to (2)
PR2=Iω
⇒PR2=25mR2ω⇒ω=5P4mR