Question

A 5.00-g bullet is fired horizontally into a 1.50-kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.0250 m along the surface before stopping. What was the initial speed of the bullet?

Answer 1

m = 0.005 kg          M = 1.500 kg
mu = 0.20
v = velocity of bullet before collision with the block

V = velocity of bullet + wooden block after bullet is embedded inside the block
Frictional force on the block = mu (m+M) g
Deceleration of the block = force / mass = mu * g = 0.20 * 9.8 m/sec²
       a = - 1.96 m/sec²
v² - u² = 2 a s
     =>    0² - V² = - 2 * 1.96 * 0.025
    => V = 0.313 m/s

Let apply conservation of linear momentum:
      m v = (m + M) V
       0.005 * v = 1.505 * 0.313
           v = 94.22 m/sec