Question

a 6 kg object moving at 8 m/s collides with a 4 kg object moving at 2 m/s. the objects bounce apart. after the collision, the 6 kg object moves at 5 m/s. what is the final velocity of the 4 kg object? a 6.5 m/s b 1 m/s c 2 m/s d 3.33 m/s e 4 m/s

Explanation:

Step1: Apply Conservation of Momentum

The law of conservation of momentum states that the total momentum before a collision equals the total momentum after the collision. The formula is \( m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' \), where \( m_1 = 6 \, \text{kg} \), \( v_1 = 8 \, \text{m/s} \), \( m_2 = 4 \, \text{kg} \), \( v_2 = 2 \, \text{m/s} \), \( v_1' = 5 \, \text{m/s} \), and \( v_2' \) is the final velocity of the 4 kg object.

Step2: Substitute Values into the Formula

First, calculate the total initial momentum: \( (6 \times 8) + (4 \times 2) = 48 + 8 = 56 \, \text{kg·m/s} \).

Then, calculate the total final momentum: \( (6 \times 5) + (4 \times v_2') = 30 + 4v_2' \).

Step3: Solve for \( v_2' \)

Set initial momentum equal to final momentum: \( 56 = 30 + 4v_2' \).

Subtract 30 from both sides: \( 4v_2' = 56 - 30 = 26 \).

Divide both sides by 4: \( v_2' = \frac{26}{4} = 6.5 \, \text{m/s} \).

Answer:

A. 6.5 m/s