Question

a typical bullet has a mass of about 0.05 kg and has an initial speed (upon leaving the barrel of the rifle) of about 1,000 m/s. how fast should a medium-sized car (~1,000-kg of mass) be moving in order to have the same momentum?
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next 1:
how is momentum calculated?

options:

• the car should be moving about 0.05 m/s.
• the car will never reach the same momentum as the bullet.
• the car should be moving as fast as the bullet.
• the car should be at rest.

submit, etc.

Explanation:

Step1: Recall momentum formula

Momentum \( p = mv \), where \( m \) is mass and \( v \) is velocity. For the bullet, \( m_b = 0.05 \, \text{kg} \), \( v_b = 1000 \, \text{m/s} \). For the car, \( m_c = 1000 \, \text{kg} \), let \( v_c \) be its velocity. Set \( p_b = p_c \), so \( m_bv_b = m_cv_c \).

Step2: Solve for \( v_c \)

Rearrange the formula: \( v_c=\frac{m_bv_b}{m_c} \). Substitute values: \( m_b = 0.05 \), \( v_b = 1000 \), \( m_c = 1000 \). Then \( v_c=\frac{0.05\times1000}{1000}=\frac{50}{1000}=0.05 \, \text{m/s} \).

Momentum is calculated as \( p = mv \). To have the same momentum as the bullet, the car's momentum (\( m_cv_c \)) must equal the bullet's momentum (\( m_bv_b \)). Solving \( m_bv_b = m_cv_c \) for \( v_c \) gives \( v_c=\frac{m_bv_b}{m_c} \). Substituting \( m_b = 0.05 \, \text{kg} \), \( v_b = 1000 \, \text{m/s} \), and \( m_c = 1000 \, \text{kg} \) yields \( v_c = 0.05 \, \text{m/s} \).

Answer:

The car should be moving about 0.05 m/s.