Block A in ?Fig. E8.24? has mass 1.00 kg, and block B has mass 3.00 kg. The blocks are forced together, compressing a spring S between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block B acquires a speed of 1.20 m/s. (a) What is the final speed of block A? (b) How much potential energy was stored in the compressed spring?
Solution 24E Step 1: a) According to the momentum conservation principle, we can write, The momentum acquired by block A = Momentum acquired by block B Momentum acquired by block A, P = m v A A A Momentum acquired by block B, P = m v B B B Provided, mass of block A, m = 1.00 kg A Mass of block B, m = 3.00 kg B Speed of block B, v = 1.20B/s Therefore, m v =A A B B 1.00 kg × v = 3A0 kg × 1.20 m/s vA 3.6 kg m/s / 1 kg = 3.6 m/s Final speed of block A, v = 3.6 m/s A
