A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B, which has a mass of 3 kg. After the impact, block B slides on 30-kg carrier C until it impacts the end of the carrier. Knowing the impact between B and C is perfectly plastic and the coefficient of kinetic friction between B and C is 0.2, determine (a) the velocity of the bullet and B after the first impact, (b) the final velocity of the carrier.

(a) The velocity of the bullet and B after the first impact is 4.4554 m/s.

(b) The velocity of the carrier is 0.40872 m/s.

Explanation:

(a) To solve the question, we apply the principle of conservation of linear momentum as follows.

we note that the distance between B and C is 0.5 m

Then we have

Sum of initial momentum = Sum of final momentum

0.03 kg * 450 m/s = (0.03 kg + 3 kg) * v₂

Therefore v₂ = (13.5 kg·m/s) : (3.03 kg) = 4.4554 m/s

The velocity of the bullet and B after the first impact = 4.4554 m/s

(b) The velocity of the carrier is given as follows

Therefore from the conservation of linear momentum we also have

(m₁ + m₂) * v₂ = (m₁ + m₂ + m₃) * v₃

Where:

m₃ = Mass of the carrier = 30 kg

Therefore

(3.03 kg) * (4.4554 m/s) = (3.03 kg+30 kg) * v₃

v₃ = (13.5 kg·m/s) : (33.03 kg) = 0.40872 m/s

The velocity of the carrier = 0.40872 m/s.