Answer:
(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.
