A block A slides on the surface of wedge B. When the block reaches at bottom most point, the velocity of 10 kg wedge is 4 m/s, the velocity of block with respect to wedge is - (Assume all the surface in contact are smooth)
Text Solution
Verified by ExpertsThe correct answer is:
D
Apply conservation of linear momentum along horizontal direction.
10 × 4 = 2(u cos 60º – 4)
20 = 
– 4 ⇒ 24 = 
⇒ u = 48 m/s
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