Question

A 9.05- g bullet from a 9-mm pistol has a velocity of 353.0 m/s. It strikes the 0.625- kg block of a ballistic pendulum and passes completely through the block. If the block rises through a distance h = 23.49 cm, what was the velocity of the bullet as it emerged from the block?

Answer 1

here,

mass of bullet , m1 = 9.05 g = 0.00905 kg

initial velocity of bullet , u1 = 353 m/s

the mass of block , m2 = 0.625 kg

h = 23.49 cm = 0.2349 m

the speed block after the collison , v2 = sqrt(2*g*h)

v2 = sqrt(2*9.81*0.2349) = 2.15 m/s

let the final speed of bullet be v1

using conservation of momentum

m1 * u1 = m1 * v1 + m2 * v2

0.00905 * 353 = 0.00905 * v1 + 0.625 * 2.15

solving for v1

v1 = 204.5 m/s

the velocity of bullet when it emerges out of the block is 204.5 m/s