To find the speed of the bullet after it is fired, we can use the principle of conservation of momentum. The momentum before firing equals the momentum after firing.
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Initial momentum (before firing): Since the bullet and the gun are at rest, the total initial momentum is: \[ p_{\text{initial}} = 0 \]
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Final momentum (after firing): The momentum of the bullet and the gun after firing can be expressed as: \[ p_{\text{final}} = (m_{\text{bullet}} \cdot v_{\text{bullet}}) + (m_{\text{gun}} \cdot v_{\text{gun}}) \] where:
- \( m_{\text{bullet}} = 0.005 , \text{kg} \)
- \( v_{\text{bullet}} \) is the speed of the bullet (which we need to find)
- \( m_{\text{gun}} = 0.515 , \text{kg} \)
- \( v_{\text{gun}} = -2.1 , \text{m/s} \) (the negative sign indicates that the gun recoils in the opposite direction)
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Setting the initial and final momentum equal: \[ 0 = (0.005 , \text{kg} \cdot v_{\text{bullet}}) + (0.515 , \text{kg} \cdot -2.1 , \text{m/s}) \]
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Calculating the momentum of the gun: \[ p_{\text{gun}} = 0.515 , \text{kg} \cdot -2.1 , \text{m/s} = -1.0815 , \text{kg m/s} \]
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Plugging this into the momentum equation: \[ 0 = 0.005 v_{\text{bullet}} - 1.0815 \]
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Solving for \( v_{\text{bullet}} \): \[ 0.005 v_{\text{bullet}} = 1.0815 \] \[ v_{\text{bullet}} = \frac{1.0815}{0.005} = 216.3 , \text{m/s} \]
Since the bullet is moving in the opposite direction of the recoil of the gun, it will be negative: \[ v_{\text{bullet}} \approx -216.3 , \text{m/s} \]
Therefore, the closest answer from the options is: a) -216 m/s
2 answers