Physics ch 9 ex?????

Question

A 8.0 g bullet is fired into a 11 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0 {\rm cm} across the table. The coefficient of kinetic friction for wood sliding on wood is 0.20.

What was the speed of the bullet?

Answer 1

Let:

m be the mass of the bullet,

M be the mass of the block,

v1 be the speed of the bullet before impact,

v2 be the speed of the block and bullet after impact,

a be the acceleration of the block and bullet,

u be the coefficient of friction,

g be the acceleration due to gravity,

s be the distance the block slides.

Conserving momentum:

mv1 = (M + m)v2 ...(1)

Equations of motion for sliding:

0 = v2^2 + 2as ...(2)

u(M + m)g = - (M + m)a

ug = - a ...(3)

Eliminating a from (2) and (3):

v2^2 = 2ugs

Substituting for v2 in (1):

mv1 = (M + m) sqrt(2ugs)

v1 = (M + m) sqrt(2ugs) / m

= 11.008 sqrt(2 * 0.20 * 9.81 * 0.05) / 0.008

= 609 m/s.