Question

A bullet with mass 0.04000 kg, traveling at a speed of 310.0 m/s, strikes a stationary block of wood having mass 6.960 kg. The bullet embeds into the block. The speed, in m/s, of the bullet-plus-wood combination immediately after the collision is

Answer 1

1.77 m/s

Step-by-step 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