Question

question 4
a 1.0 - kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upward. what is the change in momentum of the ball?
hint: assign vector directions
upward is positive +
downward is negative -
12 kg·m/s, downward
12 kg·m/s, upward
24 kg·m/s, downward
24 kg·m/s, upward
zero kg·m/s

Explanation:

Step1: Recall the formula for momentum

Momentum \(p = mv\), where \(m\) is mass and \(v\) is velocity. The change in momentum \(\Delta p=p_{final}-p_{initial}\)

Step2: Assign velocities with signs

Given \(m = 1.0\space kg\), \(v_{initial}=- 12\space m/s\) (downward), \(v_{final}=12\space m/s\) (upward)

Step3: Calculate initial and final momentum

\(p_{initial}=mv_{initial}=(1.0)(-12)=- 12\space kg\cdot m/s\)
\(p_{final}=mv_{final}=(1.0)(12)=12\space kg\cdot m/s\)

Step4: Calculate change in momentum

\(\Delta p=p_{final}-p_{initial}=12-(-12)=24\space kg\cdot m/s\) (positive value indicates upward direction)

Answer:

24 kg·m/s, upward