Answer:
1.52m/s
Explanation:
Using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity
Substitute the given values into the formula
0.013(270)+2(130) = (270+130)v
3.51+260 = 400v
263.51 = 400v
v = 400/263.51
v = 1.52m/s
Hence the velocity after the bullet emerges is 1.52m/s