Question

A 12g bullet is fired into a 12kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0cm across the table. The coefficient of kinetic friction for wood sliding on wood is 0.20.

awnser should be in m/s

Answer 1

m1 = 12g = 0.012kg
m2 = 12kg
delta(x) = 5.0cm = 0.05m
Fnet = ma = Ff = coefficient of kinetic friction(N) = coefficient of kinetic friction(mg)
= 12.012(a) = 0.20(12.012)(9.8)
a = -1.96m/s^2 {negative because we assume that the intial velocity of the bullet to be positive}
vf^2 = vo^2 + 2a(delta(x))
0 = vo^2 + 2(-1.96)(0.05)
vo = 0.443m/s

m1v1 + m2v2 = m1+m2(vf)
0.012(vb) = 12.012(0.443)
vb = 443m/s

Answer 2

Since fs is (μ)n=(μ)mg, fs=(.2)(12)(9.8). From there, we use the conservation of energy equations. Since (1/2)*mv² = -W_{Other Forces} , and the only force acting upon it is the work due to friction, (1/2)*mv² = -W_Friction. Since that is true, (1/2)*(m₁+m₂)v² = F_s*d or (1/2)*(.012+12)*(v_{Bullet + Block}²) = (.2)(12)(9.8)(.005). After you solve for v_{Bullet + Block}, use the conservation of energy equation again as (1/2)*(.012)*v_Bullet² = (1/2)*(.012+12)*(v_{Bullet + Block}²). Then all you have to do is solve for v_Bullet.