# 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?

1
by Diet

2015-10-09T18:53:52+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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