Question

part a a special bullet has a mass of about 0.015 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? * some available hints) the car should be moving about 0.015 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

Explanation:

Step1: Recall the principle of conservation of momentum

The momentum \( p \) of an object is given by the formula \( p = mv \), where \( m \) is the mass and \( v \) is the velocity. For the bullet and the car to have the same momentum, \( p_{\text{bullet}} = p_{\text{car}} \), so \( m_{\text{bullet}}v_{\text{bullet}} = m_{\text{car}}v_{\text{car}} \).

Step2: Identify the given values

The mass of the bullet \( m_{\text{bullet}} = 0.015 \, \text{kg} \), the velocity of the bullet \( v_{\text{bullet}} = 1000 \, \text{m/s} \), and the mass of the car \( m_{\text{car}} = 1000 \, \text{kg} \).

Step3: Solve for the velocity of the car

Rearranging the formula \( m_{\text{bullet}}v_{\text{bullet}} = m_{\text{car}}v_{\text{car}} \) for \( v_{\text{car}} \), we get \( v_{\text{car}}=\frac{m_{\text{bullet}}v_{\text{bullet}}}{m_{\text{car}}} \).

Substitute the values: \( v_{\text{car}}=\frac{0.015 \, \text{kg} \times 1000 \, \text{m/s}}{1000 \, \text{kg}} \)

First, calculate the numerator: \( 0.015 \times 1000 = 15 \, \text{kg·m/s} \)

Then, divide by the mass of the car: \( v_{\text{car}}=\frac{15}{1000}= 0.015 \, \text{m/s} \)

Answer:

The car should be moving about 0.015 m/s (the first option: "The car should be moving about 0.015 m/s")