Question

a 6 kg object at rest is hit by a 12 kg object moving at 5 m/s. the objects bounce apart. after the collision, the 12 kg object moves at 1 m/s. what is the final velocity of the 6 kg object?
a 6 m/s
b 4 m/s
c 3 m/s
d 8 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 \( m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \), where \( m \) is mass and \( v \) is velocity (initial: \( i \), final: \( f \)).

Given: \( m_1 = 12 \, \text{kg} \), \( v_{1i} = 5 \, \text{m/s} \), \( m_2 = 6 \, \text{kg} \), \( v_{2i} = 0 \, \text{m/s} \) (at rest), \( v_{1f} = 1 \, \text{m/s} \), \( v_{2f} = ? \)

Substitute into the formula:
\( (12)(5) + (6)(0) = (12)(1) + (6)v_{2f} \)

Step2: Solve for \( v_{2f} \)

Calculate left side: \( 60 + 0 = 60 \)
Calculate right side: \( 12 + 6v_{2f} \)
Set equal: \( 60 = 12 + 6v_{2f} \)
Subtract 12: \( 48 = 6v_{2f} \)
Divide by 6: \( v_{2f} = \frac{48}{6} = 8 \, \text{m/s} \)? Wait, no—wait, did I mix up? Wait, no, wait: Wait, the 6 kg object is \( m_2 \). Wait, let's recheck.

Wait, initial momentum: \( 12*5 + 6*0 = 60 \). Final momentum: \( 12*1 + 6*v_{2f} \). So \( 60 = 12 + 6v_{2f} \). Then \( 6v_{2f} = 60 - 12 = 48 \), so \( v_{2f} = 8 \)? But option D is 8 m/s. Wait, but let me check again.

Wait, maybe I made a mistake. Wait, the 12 kg object was moving at 5 m/s, hits 6 kg at rest. After collision, 12 kg moves at 1 m/s. So momentum before: 125 = 60. Momentum after: 121 + 6*v = 12 + 6v. So 60 = 12 + 6v → 6v = 48 → v = 8. So the final velocity of 6 kg is 8 m/s, which is option D.

Answer:

D. 8 m/s