Answer:
v = 300.6 m/s
Explanation:
given,
mass of bullet(m) = 10 g = 0.01 kg
mass of block of wood (M)= 5 Kg
speed of the block plus bullet after collision(V) = 0.6 m/s
speed of wood(u) = 0 m/s
original speed of the bullet(v) = ?
using conservation of momentum
m v + Mu = (M + m)V
0.01 x v + 5 x 0 = (5 + 0.01) x 0.6
0.01 x v = (5.01) x 0.6
[tex]v = \dfrac{3.006}{0.01}[/tex]
v = 300.6 m/s
the original speed of the bullet before collision is equal to v = 300.6 m/s
