A bullet of mass 20 g moving with 600 ms-1 collides with a block of mass 4 kg hanging with the string of length 0.4 m. What is velocity of bullet when it comes out of block, if block rises to height 0.2 m after collision? (Take, g = 10 ms-2)

Velocity of block just after collision =2gh=2×10×0.2                                                        =2 ms-1Now, applying conservation of linear momentum just before and just after collision,                         0.02 × 600 = 4 × 2 + 0.02 × v∴                                     v = 200 ms-1