Question

A 0.0200 kg bullet moving horizontally at 400 m/s embeds itself into an initially stationary 0.500 kg block.

(a) What is their velocity just after the collision? m/s

(b) The bullet-embedded block slides 8.0 m on a horizontal surface with a 0.30 kinetic coefficient of friction. Now what is its velocity? m/s

(c) The bullet-embedded block now strikes and sticks to a stationary 2.00 kg block. How far does this combination travel before stopping? m

Answer 1

a) conservation of momentum to find initial velocity after collision.

0.020(400) + 0.500(0) = (0.500 + 0.020)v
v = 15.4 m/s


b) Initial kinetic energy after collision

KE_i = ½(0.500 + 0.020)15.4²
KE_i = 61.54 J

The work of friction in sliding is

U = Fd
U = μmgd
U = 0.30(0.520)9.81(8.0)
U = 12.24 J

KE_f = KE_i - U
KE_f = 61.54 - 12.24
KE_f = 49.3 J

KE_f = ½m(v_f)²
v_f = √(2(49.3) / 0.520)
v_f = 13.8 m/s


c) Again conservation of momentum

0.520(13.8) + 2.00(0) = (2.00 + 0.520)v_x
v_x = 2.84 m/s

assuming the coefficient of friction is still 0.3

The initial kinetic energy will be converted to the work of friction

½mv² - Fd = 0
½mv² = μmgd
½v² = μgd
d = v² / 2μg

d =2.84²/ (2)0.3(9.81)
d = 1.37 m