Question

question 39
a 0.50 kg ball traveling at 1.0 m/s to the right collides head-on with a 1.2 kg ball traveling at 0.25 m/s toward the left. after the collision, the lighter ball is traveling at 0.76 m/s to the left. what is the final velocity of the heavier ball, after the collision?
1.4 m/s, right
2.0 m/s, right
4.0 m/s, right
0.48 m/s, right

Explanation:

Step1: Define Variables and Momentum Conservation

Let right be positive. Masses: \( m_1 = 0.50 \, \text{kg} \), \( m_2 = 1.2 \, \text{kg} \). Initial velocities: \( v_{1i} = 1.0 \, \text{m/s} \), \( v_{2i} = -0.25 \, \text{m/s} \) (left is negative). Final velocity of lighter ball: \( v_{1f} = -0.76 \, \text{m/s} \). Use conservation of momentum: \( m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \).

Step2: Plug in Values and Solve for \( v_{2f} \)

Substitute values:
\( (0.50)(1.0) + (1.2)(-0.25) = (0.50)(-0.76) + (1.2)v_{2f} \)
Calculate left side: \( 0.5 - 0.3 = 0.2 \)
Right side: \( -0.38 + 1.2v_{2f} \)
Set equal: \( 0.2 = -0.38 + 1.2v_{2f} \)
Add 0.38 to both sides: \( 0.58 = 1.2v_{2f} \)
Divide by 1.2: \( v_{2f} = \frac{0.58}{1.2} \approx 0.48 \, \text{m/s} \) (positive means right).

Answer:

0.48 m/s, right (Option: 0.48 m/s, right)