AP physics help pleeaseee

A(n) 5.76 g bullet moving with an initial speed of 482 m/s is fired into and passes through a 0.86 kg block. The block, initially at rest on a frictionless horizontal surface, is connected to a spring with a spring constant
of 766 N/m. The block moves 4.73 cm to the right after impact.
a)Find the speed at which the bullet emerges from the block. Answer in units of m/s
b)How much mechanical energy was lost in the collision? Answer in units of J

we will use conservation of energy and momentum to solve this problem

we know the momentum before collision = momentum after collision

0.00576kg x 482m/s = 0.00576 v1 + 0.86kg v2

where v1 and v2 are the speeds of the bullet and block after collision

we find the speed of the block after collision using conservation of energy; the initial KE of the block will equal the final PE of the compressed spring, so we have

1/2 x 0.86kg v2^2 = 1/2 x 766N/m (0.0473m)^2

solve for v2: v2 = 1.41m/s

knowing this, we can solve for v1:

0.00576kg x 482m/s = 0.00576 v1 + 0.86kg v2

if v2 = 1.42m/s, then v1 = 271.5m/s

compare initial and final KE:

1/2 x 0.00576 x (482m/s)^2 - 1/2(0.00576 x (271.5m/s)^2 +0.86 x (1.41m/s)^2)

the initial KE = 669J
the final KE = 213.1J

loss of KE = - 456J