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17 December, 09:35

A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction perpendicular to the line between the batter and the pitcher. What is the magnitude of the impulse delivered to the baseball?

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  1. 17 December, 13:31
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    Answer7 NsExplanation

    Given in the question,

    mass of foul ball = 0.140 kg

    initial speed with which ball was hit with the bat = 30 m/s

    final speed = 40 m/s

    According to the scenario the whole scene is making a right angle triangle

    So, to the solve the question we will use pythagorus theorem

    Hypotenuse² = base² + height²

    Here,

    Hypotenuse = Magnitude of impulse

    Base = 1st change of momentum

    height = 2nd change of momentum

    1st impulse (1st change of momentum)

    p = m (1) v (1) = (0.14 kg) (40.0 m/s) = 5.6 kg m / s = 5.6 N s

    2nd impulse (2nd change of momentum)

    p = m (2) v (2) = (0.14 kg) (30.0 m/s) = 4.2 kg m / s = 4.2 N s

    Magnitude of impulse (hypotenuse of triangle)

    impulse² = (5.6) ² + (4.2) ²

    impulse² = 31.36 + 17.64

    impulse² = 49

    impulse² = √49

    impulse = 7.0 N s
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