Answer:
1.77 m/s
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m')............................ Equation 1
Where m = mass of the bullet, m' = mass of the block, u = initial velocity of the bullet, u' = initial velocity of the block, V = Velocity of the bullet-wood combination immediately after collision
make V the subject of the equation
V = (mu+m'u')/(m+m').................. Equation 2
Given: m = 0.04 kg, m' = 6.96 kg, u = 310 m/s, u' = 0 m/s (stationary)
Substitute into equation 2
V = (0.04×310+6.96×0)/(0.04+6.96)
V = 12.4/7
V = 1.77 m/s
Hence the speed, in m/s, of the bullet-plus-wood combination immediately after the collision is 1.77 m/s
