An 8.00g bullet is fired at 220m/s into a 250g wooden block that is initially at rest. The bullet remains in the block and after the collision the two slide up a 30 degree incline.

a) Solve for the velocity of the bullet block after the collision.
b) Determine the distance along the incline the bullet block travel if the incline is frictionless

(a) use conservation of momentum, note that after collision, there is a common velocity v1=v2=v

m1u1+m2u2=m1v1+m2v2=(m1+m2)v

m1,m2=masses of bullet and block
u1,u2=initial velocities of bullet & block
v1,v2=(common) velocity of bullet and block after collision.

substituting values,
8*220+250*0=(8+250)*v
solve for v to get: v=880/129 m/s
Kinetic energy right after impact:
KE=(1/2)(m1+m2)v^2
=6003.1 J (approx.)
After the block has moved up the incline over x metres, potential energy
PE = (m1+m2)gx(sin(30°)
=258*9.81*x*(0.5)
=1265.49x
Equate sum of KE and PE at bottom and top of incline:
1265.49x = 6003.1 J
x=4.74 m (approx.)