0.05 kg bullet collides and sticks to a 2.5 kg stationary block suspended from a string. the bullet and block swings to a maximum height of 12 cm. what was the initial speed of the bullet?

First calculate the combine velocity of the bullet and stick, using the formula

kinetic energy = potential energy

0.5mv^2 = mgh

where m is the combined mass

v is the velocity

g is the acceleration due to gravity (9.81 m/s^2

h is the heigth

0.5 (0.05 + 2.5 kg) v^2 = (9.81 m/s^2) (0.05 + 2.5 kg) (0.12 m)

solve for v

v = 1.53 m/s

using momentum balance

M = mv

m1v1 + m2v2 = mv

where m1 is the mass of the bullet

v1 is the initial velocity of the bullet

m2 mass of the stick

v2 initial velocity of stick

(0.05 kg) (v1) + (2.5 kg) (0) = (0.05 + 2.5 kg) (1.53 m/s)

v1 = 78.25 m/s