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Sagot :
The magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.
Given that a racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction).
mass m = 42g = 42/1000 = 0.042kg
initial velocity before collision u = 7 m/s
It hits the wall of the court and rebounds to the hitter with a horizontal speed of 7m/s to the left (-x direction). That is,
velocity after collision v = 7 m/s
To calculate the magnitude of the racquetball's change in momentum, we will use the formula below
Change in momentum = Mv - Mu
Since momentum is a vector quantity, we will consider the direction.
Change in momentum = 0.042 x 7 - ( 0.042 x - 7)
Change in momentum = 0.294 + 0.294
Change in momentum = 0.588 kgm/s
Therefore, the magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.
Learn more on momentum here: https://brainly.com/question/402617
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