Jerome pitches a baseball of mass 0.15 kg. The ball arrives at home plate with a speed of 45 m/s and is batted straight back to Jerome with a return speed of 65 m/s. What is the magnitude of change in the ball's momentum?Select one:a.16.5 kg⋅m/sb.20 kg⋅m/sc.3 kg⋅m/sd.9.75 kg⋅m/se.6.75 kg⋅m/s
Question
Jerome pitches a baseball of mass 0.15 kg. The ball arrives at home plate with a speed of 45 m/s and is batted straight back to Jerome with a return speed of 65 m/s. What is the magnitude of change in the ball's momentum?Select one:a.16.5 kg⋅m/sb.20 kg⋅m/sc.3 kg⋅m/sd.9.75 kg⋅m/se.6.75 kg⋅m/s
Solution
The momentum of an object is calculated by the product of its mass and velocity. The change in momentum (also known as impulse) is calculated by the difference between the final and initial momentum.
The initial momentum of the ball is the product of its mass and its initial speed. So, the initial momentum (p_initial) is:
p_initial = mass * initial speed = 0.15 kg * 45 m/s = 6.75 kg*m/s
The final momentum of the ball is the product of its mass and its final speed. So, the final momentum (p_final) is:
p_final = mass * final speed = 0.15 kg * 65 m/s = 9.75 kg*m/s
The change in momentum is the difference between the final and initial momentum. So, the change in momentum (Δp) is:
Δp = p_final - p_initial = 9.75 kgm/s - 6.75 kgm/s = 3 kg*m/s
So, the magnitude of the change in the ball's momentum is 3 kgm/s. Therefore, the correct answer is c. 3 kgm/s.
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