A(n) 9.9 g bullet is fired into a(n) 138 g block of wood at rest on a horizontal surface and stays inside. after impact, the block slides 13 m before coming to rest. the acceleration of gravity is 9.8 m/s^2 .
if the coefficient of friction between the surface and the block is 0.6, find the speed of the bullet before impact. answer in units of m/s.

185.25 m/s

Explanation:

consider the motion of the combination of bullet and block after the collision

v₀ = initial speed just after the collision

v' = final speed = 0 m/s

μ = Coefficient of friction = 0.6

g = acceleration due to gravity = 9.8 m/s²

a = acceleration of the combination = - μ g = - (0.6) (9.8) = - 5.88 m/s²

d = stopping distance = 13 m

using the kinematics equation

v'² = v₀² + 2 a d

0² = v₀² + 2 (- 5.88) (13)

v₀ = 12.4 m/s

m = mass of the bullet = 9.9 g = 0.0099 kg

M = mass of the wood = 138 g = 0.138 kg

v = speed of bullet before collision

v₀ = speed of combination after the collision = 12.4 m/s

Using conservation of momentum

m v = (m + M) v₀

(0.0099) v = (0.0099 + 0.138) (12.4)

v = 185.25 m/s

the answer would be 32*f