Question

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)

• A. 24 m/s
• B. 36 m/s
• C. 42 m/s
• D. 48 m/s

Answer 1

Answer: (D)

48 m/s

Answer 2

Apply conservation of linear momentum along horizontal direction.

10 × 4 = 2(u cos 60º – 4)

20 = $\frac { \mathrm { u } } { 2 }$ – 4 ̃ 24 = $\frac { \mathrm { u } } { 2 }$ ̃ u = 48 m/s