Answer:
The speed of bullet before impact is 480.95 m/s.
Explanation:
mass of block, M = 762 g = 0.762 kg
mass of bullet, m = 0.0158 kg
distance, s = 6.5 m
coefficient of friction = 0.75
Let the velocity of bullet before impact is u and after impact is v.
Use third equation of motion
[tex]v'^2 = v^2 - 2 \mu g s\\\\0 = v^2 - 2 \times 0.75\times 9.8\times 6.5\\\\v =9.77 m/s[/tex]
Use conservation of momentum
m x u + M x 0 = (M + m) v
0.0158 x u = (0.0158 + 0.762) x 9.77
u = 480.95 m/s
