Questions: Question 11 5 pts A fire arm of mass 7 kg fires a bullet of mass 100 g. The bullet that gains a velocity of 100 m/sec. By how much does the momentum of the fire arm change? (in units of kg m/sec)

Question 11 5 pts A fire arm of mass 7 kg fires a bullet of mass 100 g. The bullet that gains a velocity of 100 m/sec. By how much does the momentum of the fire arm change? (in units of kg m/sec)
Transcript text: Question 11 5 pts A fire arm of mass 7 kg fires a bullet of mass 100 g . The bullet that gains a velocity of 100 $\mathrm{m} / \mathrm{sec}$. By how much does the momentum of the fire arm change? (in units of $\mathrm{kg} \mathrm{m} / \mathrm{sec}$ ) $\square$ Next
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Solution

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Solution Steps

Step 1: Identify the given values

We are given:

  • Mass of the firearm, \( m_f = 7 \, \text{kg} \)
  • Mass of the bullet, \( m_b = 0.1 \, \text{kg} \) (converted from 100 g)
  • Velocity of the bullet, \( v_b = 100 \, \text{m/s} \)
Step 2: Calculate the momentum of the bullet

The momentum of the bullet can be calculated using the formula: \[ p_b = m_b \times v_b \] Substituting the given values: \[ p_b = 0.1 \, \text{kg} \times 100 \, \text{m/s} = 10 \, \text{kg} \cdot \text{m/s} \]

Step 3: Apply the law of conservation of momentum

According to the law of conservation of momentum, the momentum gained by the bullet is equal in magnitude and opposite in direction to the momentum gained by the firearm. Therefore, the change in momentum of the firearm is: \[ \Delta p_f = - p_b \] \[ \Delta p_f = -10 \, \text{kg} \cdot \text{m/s} \]

Final Answer

\[ \boxed{\Delta p_f = -10 \, \text{kg} \cdot \text{m/s}} \]

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