NEED ANSWER ASAP!!!

Question

A 0.01 kg bullet moves at a constant speed of 500 m/s and collides with a 1.5 kg wooden block initially at rest. The surface of the table is friction-less and 0.7 m above the floor level. After the collision the bullet becomes embedded into the block. The bullet-block system slides off the top of the table and strikes the floor. What is the resulting kinetic energy of the bullet and block system just before impact with the ground?

a. 0 J

b. 3.31 J

c. 4.1 J

d. 5 J

e. 8.3 J

f. 12.7 J

g 13.4 J

h. 1250 J

i. don't be silly, it's none of these!!

Answer 1

First use conservation of momentum to find the velocity of the system on the table. 0.01*500=1.51v, so v=5/1.51=3.311 m/s.

After the collision, the total potential and kinetic energy is conserved. Mv^2/2+Mgh=MVf^2/2, so the final kinetic energy is 1.51*(3.311^2/2+9.8*.7)=18.63 J. This falls under none of these.