AP Physics B question?

Question

This chapter is driving me crazy...help with this problem please:

A pitcher claims he can throw a 0.142 kg
baseball with as much momentum as a 2.33 g
bullet moving with a speed of 52570 m/s.
What must be its speed if the pitcher's
claim is valid?
Answer in units of m/s.

What is the kinetic energy of the bullet? An-
swer in units of J.

What is the kinetic energy of the ball? Answer
in units of J.

Answer 1

Momentum = mass * velocity
p = m * v

if the momentums are the same, that would be that
p(ball) = p(bullet)
m(ball) * v(ball) = m(bullet) * v(bullet)
0.142 * v(ball) = 0.00233 * 52570 (mass of the bullet must be in Kg)
v(ball) = 863 m/s

Kinetic Energy = 1/2 * m * v^2
K(bullet) = 1/2 * 0.00233 * (52570)^2 = 3,220,000 J

K(ball) = 1/2 * 0.142 * 863^2 = 52,900 J