Question

A 872 g wooden block is initially at rest on a rough horizontal surface when a 15.6 g bullet is fired horizontally into (but does not go through) it. After the impact, the block-bullet combination slides 6.50 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.750, determine the speed of the bullet immediately before impact.

Answer 1

The speed of the bullet before the impact is 556.5 m/s

Step-by-step explanation:

Given;

mass of the wooden block, m₁ = 872 g = 0.872 kg

initial velocity of the wooden block, u = 0

mass of the bullet, m₂ = 15.6 g = 0.0156 kg

distance traveled by the block-bullet system after impact, d = 6.5 m

coefficient of kinetic friction, μ = 0.75

Work done by friction on the system after impact is given by;

W = (m₁ + m₂)μg x d

kinetic energy of the system after impact is given by;

K.E = ¹/₂(m₁ + m₂)v²

Apply work energy theorem;

(m₁ + m₂)μg x d = ¹/₂(m₁ + m₂)v²

μg x d = ¹/₂v²

v² = 2μg x d

v = √(2μg x d)

v = √(2 x 0.75 x 9.8 x 6.5)

v = 9.78 m/s

The speed of the bullet before impact is calculated by applying principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

0.872(0) + 0.0156u₂ = 9.78(0.872 + 0.0156)

0 + 0.0156u₂ = 8.681

u₂ = 8.681 / 0.0156

u₂ = 556.5 m/s

Therefore, the speed of the bullet before the impact is 556.5 m/s