Question

description: show all you work, including units, on separate paper. follow the \problem solving method\. this final exam is cumulative and covers material from the entire course. instructions: from the list of choices, select the one best answer. multiple attempts not allowed. this test can only be taken once. force completion this test can be saved and resumed later. your answers are saved automatically. question completion status: question numbers 1 - 62, with 40 - 62 visible moving to another question will save this response. question 40 of 62 1.6 points save answer a 2.0 kg ball traveling at 1.0 m/s to the right collides head - on with a 1.0 kg ball traveling at 1.4 m/s toward the left. after the elastic collision, the heavier ball is traveling at 0.60 m/s to the left. what is the final speed of the lighter ball, after the collision? options: 2.0 m/s, 1.8 m/s, 1.2 m/s, 4.6 m/s

Explanation:

Step1: Define Variables and Direction

Let right be positive. Mass of heavier ball \( m_1 = 2.0 \, \text{kg} \), initial velocity \( v_{1i} = 1.0 \, \text{m/s} \). Mass of lighter ball \( m_2 = 1.0 \, \text{kg} \), initial velocity \( v_{2i} = -1.4 \, \text{m/s} \) (left is negative). Final velocity of heavier ball \( v_{1f} = -0.60 \, \text{m/s} \) (left is negative). We use conservation of momentum for elastic collision: \( m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \).

Step2: Substitute Values

Substitute into the momentum equation:
\( (2.0)(1.0) + (1.0)(-1.4) = (2.0)(-0.60) + (1.0)v_{2f} \)

Step3: Calculate Left Side

Left side: \( 2.0 - 1.4 = 0.6 \, \text{kg·m/s} \)

Step4: Calculate Right Side

Right side: \( -1.2 + v_{2f} \)

Step5: Solve for \( v_{2f} \)

Set equal: \( 0.6 = -1.2 + v_{2f} \)
Add 1.2 to both sides: \( v_{2f} = 0.6 + 1.2 = 1.8 \, \text{m/s} \) (positive means right, speed is magnitude 1.8 m/s)

Answer:

1.8 m/s (corresponding to the option "1.8 m/s")