Saved Bookmarks
| 1. |
Block A has a mass of 5kg and is placed on the top of smooth triangular block B having amass of 25kg. If the system is released fromrest, determine the distance B moves, inmetre, when A reaches the bottom. Neglectthe size of block A.h=10/3mK30°30° |
|
Answer» lock A has a mass of 5 kg and is placed on the top of smooth triangular block B having a mass of 25 kg. To find:If the system is released from rest, determine the distance B moves, in metre, when A reaches the BOTTOM.Solution:Let mA and VA be the mass and VELOCITY of block A and mB and vB be the mass and velocity of block B.Applying the conservation of momentum, we get,mA(vA)₀ + mB(vB)₀ = mAvA + mBvBusing the given data in above equation, we get,5 × 0 + 25 × 0 = 5 (-vA) + 25vB5vA = 25vB∴ vA = 5vB ...........(1)Now, using relative velocity RELATION, we get,vB/A = vB - (-vA)vB/A = vB + vAusing equation (1) in above equation, we get,vB/A = vB + 5vB ∴ vB/A = 6vBintegrating this equation w.r.t TIME, we get,sB/A = 6sBsin 30° = 1/2 = 0.5sB = sB/A/6sB = 0.5/6 = 0.0833Therefore, the distance B moves, in metre, when A reaches the bottom is 0.083 m. |
|