See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Momentum conservation along x 0 = mv2  Mv1 Work energy theorem mgR = 2 2 2 1 1 1 mv Mv 2 2  m R M v2 M m v1 x y from (i) and (ii)  2mgR = 2 2 2 2 2 m mv v M  2 2gRM v M m   and 1 m 2gRM v M M m   Making FBD of particle with respect wedge at lowest point 2 1 2 m(v v ) 7mg mg 2 R    Substituting the values, we get : M 4 m  mg N = (7mg/2) 2 2 1 (v v ) R  