Question

a 12 - g ball moving 4.0 \\( \frac { m } { s } \\) collides head on and inelastically with a second 12 - g ball initially at rest. after colliding, the balls move together at the speed
\\( \bigcirc 2.0 \frac { m } { s } \\)
\\( \bigcirc 3.0 \frac { m } { s } \\)
\\( \bigcirc 4.0 \frac { m } { s } \\)
\\( \bigcirc 6.0 \frac { m } { s } \\)

Explanation:

Step1: Apply conservation of momentum

Let \( m_1 = 12\,\text{g} \), \( v_1 = 4.0\,\text{m/s} \), \( m_2 = 12\,\text{g} \), \( v_2 = 0\,\text{m/s} \). Combined mass \( M = m_1 + m_2 = 24\,\text{g} \).
Momentum before: \( m_1v_1 + m_2v_2 = 12 \times 4.0 + 12 \times 0 = 48\,\text{g·m/s} \)
Momentum after: \( Mv = 24v \)
Set equal: \( 24v = 48 \)

Step2: Solve for final speed

\( v = \frac{48}{24} = 2.0\,\text{m/s} \)

Answer:

A. 2.0 m/s