A disc of mass m and radius R is rolling on a rough surface and encounters a movable wedge of sufficient height as shown in figure. There is no friction between the wedge and the ground and also between the disc and the wedge. Then :-

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No friction at the wedge, so the disc keeps its spin and its rotational KE does not change

Horizontal momentum (wedge mass m) at maximum height, both move with vx

$m v = 2 m v_{x} \Rightarrow v_{x} = \frac{v}{2}$

Translational energy

$\frac{1}{2} m v^{2} = \frac{1}{2} (2 m) (\frac{v}{2})^{2} + m g h$

$h = \frac{v ^{2}}{4 g}$

ω is unchanged and the disc moves with the wedge, so C and D are wrong

Answer: A, B

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