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.

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

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

we need some answer choices, or something like that

explanation: