A smooth block of mass m slides on a wedge of mass M with a speed V at the lowest position. The wedge is moving towards left with speed V. Coefficient of friction between the wedge & ground is $μ$ . Radius of wedge is R. A constant force F starts acting horizontally on the wedge. At this instant :-

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V c m ​ = m + M ∣ mV − M V ∣ ​ = m + M ∣ m − M ∣ V ​ - (A) true. Block's speed relative to the wedge = 2 V , so the wedge must supply N = m g + R m ( 2 V ) 2 ​ = m g + R 4 m V 2 ​ . Ground normal = M g + N ; wedge moves left, so ground friction μ ( M g + m g + R 4 m V 2 ​ ) acts right, same as F. a c m ​ = M + m F + μ ( M g + m g + R 4 m V 2 ​ ) ​ - (D) true. Answer: (A), (D)

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