Question

If a 0.75kg bullet made a 3.0kg block swing to change its height 0.5 Meters.

How fast was the bullet traveling?

Answer 1

The mass of the system block + bullet = 3.75 kg

The initial speed of the system bullet + block is given by

Vf² = Vi² +2*g*d...........


Vf  = 0   (Final speed equals zero)

d is the height traveled in the vertical component (0.5m)

Vi = sqrt(-2*(-9.8[m/s²])* (0.5m)) = 3.13 m/s

The momentum is conserved, so 

m*V1 (before impact) = M*Vi  (after impact)

0.75kg*V1 = 3.75kg*(3.13m/s)

>> V1 = 15.65 [m/s]   (Initial speed of the bullet)