Respuesta :
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.
Using the principle of momentum conservation, the velocity after first impact and final velocity are 4.46 m/s and 0.409 m/s respectively.
Using the principle of conservation of linear momentum, where the initial momentum is equal to the final momentum after impact :
[tex]m_{a}v_{a} = (m_{a} + m_{b})v_{b}[/tex]
[tex](0.03 \times 450) = (0.03 + 3)v_{b}[/tex]
[tex]v_{b} = \frac{13.5}{3.03} = 4.46 m/s[/tex]
B.)
Final velocity of Carrier :
[tex](m_{a}m_{b})v_{b}) = (m_{a} + m_{b} + m_{c})v_{c}[/tex]
[tex](0.03 + 3)4.56 = (3 + 0.3 + 30)v_{c}[/tex]
[tex](3.03)4.46 = (33.3)v_{c}[/tex]
[tex]13.5138 = (33.3)v_{c}[/tex]
[tex]v_{c} = \frac{13.5138}{33.03} = 0.409 m/s[/tex]
Hence, the final velocity of the carrier is 0.409 m/s
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