Question

A 8.65- g bullet from a 9-mm pistol has a velocity of 333.0 m/s. It strikes the 0.665- kg block of a ballistic pendulum and passes completely through the block. If the block rises through a distance h = 9.79 cm, what was the velocity of the bullet as it emerged from the block?

Answer 1

te velocity imparted to the 0.665 kg block can be determined using energy conservation of the block

0.665*9.8*0.0979=0.5*0.665*v²

v=1.385 m/s

velocity imparted to 0.665 kg block just after collision is 1.385 m/s

velocity at which the bullet emerrged from the block can be calculated using conservation of momentum

Initial momentum = 0.00865*333=2.88 kg-m/s

let velocity at which the bullet emerrged from the block be v

From momentum conservation

2.88=0.665*1.385 + 0.00865*v

v=226.5 m/s