Question:
A bullet of mass 5g is shot through a wooden block of mass 1kg, that is attached to a 2m string. The center of the block rises .4 cm. Find the velocity of the bullet when it leaves the block if its initial velocity is 300 m/s.

Attempted Solution:
Using the information I have (mB=.005kg, mW=1kg, vB=300m/s, vW=0), and with thr assumption that this is a 1D elastic collision, I set up two equations.

(1) uB - uW = vB - vW --> uB - uW = 300
(2) mBvB + mWvW = mBuB + mWuW -- > 1.5 = .005uB + uT

Solving the system of equations got me uB = 300 m/s

However the answer key says 244 m/s. Where did I go wrong?

Where dod you get that first equation? It seems you think that the difference in velocity between bullet and weight is constant - but it's not. Obviousl the bullet slows fdown some, and some of its original momentum is transferred into the weight, giving it a velocity in the same direction as the bullet. Hence the differences in velocity decreases.

The principle you should be using is conservation of momentum:

u_Bm_B = v_Bm_B + v_W M_W

To calculate v_B you first need to determine v_W. To do that, use the fact that it rises 0.4 cm after being hit. Hence the KE it obtains when struck by the bullet is converted to potential energy in Earth's gravity. From conservation of energy:

\Delta PE = mgh = \Delta KE = \frac 1 2 m v_B^2

Can you take it from here?