The sphere $A$ of mass $m$ is at the bottom most position in a smooth wedge $B$ of mass $M$ and moves in a radius R, as shown. Wedge $B$ is placed on a smooth horizontal surface. The minimum value of velocity with which sphere is projected so that it just reaches the top of the wedge is:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

${[2 g R]}^{\frac{1}{2}}$

${[\frac{2 g R (M + m)}{m}]}^{\frac{1}{2}}$

${[\frac{2 g R (M + m)}{M}]}^{\frac{1}{2}}$

${[\frac{2 g R M}{(M + m)}]}^{\frac{1}{2}}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Sphere just reaches the top of the wedge, It means relative velocity of sphere w.r.t wedge is zero $m u = (M + m) V$
So, V=horizontal velocity of sphere and wedge
$= \frac{m u}{(M + m)}$
From conservation of mechanical energy,
$\frac{1}{2} m u^{2} = \frac{1}{2} (M + m) V^{2} + m g R$
$u^{2} = \frac{(M + m)}{m} (\frac{m^{2} u^{2}}{{(M + m)}^{2}}) + 2 g R$
$u = {[\frac{2 g R (M + m)}{M}]}^{\frac{1}{2}}$