Question

A solid ball is placed on a rough horizontal surface with initial velocity of its $COM$ as $v_0$ and the initial angular velocity ${\omega }_0$ as given in column-I

Answer 1

(A) Since angular velocity is in opposite direction and external force acting on the system is friction which passes through point of contact therefore angular momentum will be conserved around point of contact.
Initially ball moves in right direction and angular velocity is in anticlockwise therefore, if $v^{\prime }$ is final velocity and ${\omega }^{\prime }$ is final angular velocity
Now initial angular momentum around point of contact
$L_i=m{v}_{o}R-I{\omega }_o$
$L_i=m{v}_{o}R-3m{R}^{2}3v_o/2R$
$L_i=m{v}_{o}R-9m{v}_{o}R/2=-7m{v}_{o}R/2$
Therefore, final angular momentum will also be negative therefore, ball will return in its journey.
A->3

(B) Initial angular momentum $L_i=m{v}_{o}R+3m{R}^{2}2v_o/2R=4m{v}_{o}R$
Let final linear velocity be $v^{\prime }$
$L_f=m{v}^{\prime }R+3m{v}^{\prime }R/2=5m{v}^{\prime }R/2=L_i$
Therefore, $v^{\prime }=8v_0/5$
Initial$K{E}_i=m{v}_o^2/2+I{\omega }_o^2/2=m{v}_o^2/2+m{R}^2{v}_o^2/4R^2=3m{v}_o^2/4$
final $K{E}_f=3m{v}^{\prime 2}/4=3m64v_o^2/100=192m{v}_o^2/100$
Clearly Final $KE>1/2\left(initialKE\right)$
B->2,4

(C) Initial $KE=m{v}_o^2/2$
Angular momentum around point of contact is conserved.
$L_i=m{v}_{o}R$
$L_f=m{v}^{\prime }R+3m{v}^{\prime }R/2=5m{v}^{\prime }R/2=L_i$
this gives $v^{\prime }=2v_o/5$
final $KE=m{v}^{\prime 2}/2+3m{R}^2{v}^{\prime 2}/4R=5m{v}^{\prime 2}/4=5m4v_o^2/100=20m{v}_o^2/100$
clearly, $finalKE\ne \left(initialKE\right)/2$
C->4

(D) No further information is given initial angular momentum is +ve therefore, final will also be +ve.
D->4