Question

A bullet of mass 6.00 g is fired horizontally into a wooden block of mass 1.21 kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.210. The bullet remains embedded in the block, which is observed to slide a distance 0.260 m along the surface before stopping. What was the initial speed of the bullet?

I know that all the kinetic energy is turned into potential energy but I know I also need to take momentum into account. so 1/2mv^2 = mgh?

Answer 1

The energy absorved by friction is 1.216*.230*9.8 J. That means that amount of energy must have been the initial KE of the block and bullet. From that, you can solve the initial Velocity of the block and bullet.
conservation of momentum:
momentumblock/bullet=initail momentum bullet
1.216*V=.06*velocitybullet
solve for velocity bullet