Question

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description\tshow all you work, including units, on separate paper. follow the \problem solving method\.
\tthis final exam is cumulative and covers material from the entire course.
instructions\tfrom 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.
\tyour answers are saved automatically.
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question 37 \tquestion 37 of 62
a 0.10 kg rubber ball is falling straight down, and just before it strikes the ground, it is traveling at 12 m/s. the ball rebounds upward at 8.0 m/s. the collision with the floor lasts 0.040 s. what is the change in momentum of the ball?
○ 1.2 kg·m/s
○ 0.40 kg·m/s
○ 0.80 kg·m/s
○ 2.0 kg·m/s
\t1.6 points \tsave answer

Explanation:

Step1: Define Momentum and Direction

Momentum is \( p = mv \). Let downward be negative and upward be positive. Initial velocity \( v_i = -12 \, \text{m/s} \) (downward), final velocity \( v_f = 8.0 \, \text{m/s} \) (upward), mass \( m = 0.10 \, \text{kg} \).

Step2: Calculate Initial and Final Momentum

Initial momentum: \( p_i = m v_i = 0.10 \, \text{kg} \times (-12 \, \text{m/s}) = -1.2 \, \text{kg·m/s} \)
Final momentum: \( p_f = m v_f = 0.10 \, \text{kg} \times 8.0 \, \text{m/s} = 0.8 \, \text{kg·m/s} \)

Step3: Calculate Change in Momentum

Change in momentum \( \Delta p = p_f - p_i = 0.8 - (-1.2) = 2.0 \, \text{kg·m/s} \)

Answer:

2.0 kg·m/s (the option with "2.0 kg·m/s")