Respuesta :
Answer:
The velocity of the gun relative to the ground is 19.66 m/s
Explanation:
Given data,
The mass of the gun, M = 15.0 kg
The mass of the bullet, m = 50 g
The velocity of the train, v = 75 km/h
= 20.83 m/s
The velocity of bullet relative to train, V' = 350 m/s
The velocity of bullet relative to ground, V = 350 + 20
= 370 m/s
According to the law of conservation of momentum,
Mv' + mV' = 0
[tex]v' = -\frac{mV'}{M}[/tex]
[tex]v' = -\frac{0.050\times 350}{15}[/tex]
= -1.17 m/s
Therefore, the velocity of the gun with,
v₀ = V + v'
= 20.83 - 1.17
= 19.66 m/s
Hence, the velocity of the gun relative to the ground is 19.66 m/s
The velocity of the gun relative to the ground is 19.73 m/
The given parameters;
- mass of the person, m₁ = 15 kg
- mass of the bullet, m₂ = 50 g = 0.05 kg
- velocity of the train, u₁ = 75 km/h = 20.83 m/s
- velocity of the bullet, u₂ = 350 m/s
Apply the principle of conservation of linear momentum, to determine the velocity of the gun relative to the ground;
[tex]v(m_1 + m_2) = m_1 u_1 + m_2 u_2\\\\20.83(15 + 0.05) = 15u_1 + 0.05(350)\\\\313.49= 15u_1 + 17.5\\\\15u_1 = 295.99\\\\\mu_1 =\frac{295.99}{15} \\\\\mu_1 = 19.73 \ m/s[/tex]
Thus, the velocity of the gun relative to the ground is 19.73 m/s.
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