A hemispherical bowl is placed in inverted position on frictionless horizontal floor and a small disk is placed on the top of the bowl as shown in the figure. The disk and the bowl are of equal mass and there is no friction between them. The disk is slightly pushed horizontally so that it starts sliding on the bowl with negligible speed.

A) Only vertical force present in the system. So vertical momentum does not change.
B) Also horizontal velocity will remain same.
Let’s work from CM frame, in this frame disk moves in circular path. For circular motion 
$mg\cos \theta -m{a}_0\sin \theta -N=\frac{m{V}_1^2}{R}$    ……..(i)
$\text{'}{V}_1\text{'}$ is velocity of disk relative to the hemisphere
According to conservation of linear momentum  $m(V_1\cos \theta -V_2)-m{V}_2$
$\Rightarrow V_2=\frac{V_1\cos \theta }{2}$                 ………(ii)
WE theorem (for whole system)
$mgR(1-\cos \theta )=\frac{1}{2}m{V}_2^2+\frac{1}{2}m\{{(V_1\cos -V_2)}^2+V_1^2{\sin }^2\theta \}$            ……(iii)
At the time of breakoff  $N\to 0\Rightarrow a_0\to 0(BecauseN\sin \theta =m{a}_0)$
So from (i)
$V_1^2=gR\cos \theta$                      ……..(iv)
Now from (i) and (iii)
$2gR(1-\cos \theta )=\frac{{\left(V_1\cos \theta \right)}^2}{4}+\frac{{\left(V_1\cos \theta \right)}^2}{4}+V_1^2{\sin }^2\theta$
$=\frac{V_1^2{\cos }^2\theta }{2}+V_1^2(1-{\cos }^2\theta )$
$2gR(1-\cos \theta )=V_1^2-V_1^2\frac{{\cos }^2\theta }{2}$
From (iv),
$2gR(1-\cos \theta )=gR\cos \theta -\frac{gR\cos \theta .{\cos }^2\theta }{2}$
$2-2\cos \theta =\cos \theta -\frac{{\cos }^3\theta }{2};2(2-2\cos \theta )=2\cos \theta -\cos 5\theta$
  $\Rightarrow {\cos }^3\theta -6\cos \theta +4=0\Rightarrow (\cos \theta =\sqrt{3}-1)$
$(\cos \theta -2)({\cos }^2\theta +2\cos \theta -2)=0$
$\cos \theta =\frac{-2\pm \sqrt{4+8}}{2}$
$=-1\pm \sqrt{3}$
D) Disc will be accelerating only not bowl.