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3 December, 10:46

Two masses, MA and MB, with MB = 2MA, are released at the same time and allowed to fall straight down. Neglect air resistance. When we compare their kinetic energies after they have fallen for equal times, we find thata. KB = KA.

b. KB = 2KA.

c. KB = 4KA.

d. KA = 2KB.

e. KA = 4KB.

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  1. 3 December, 11:13
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    KB = 2KA

    Explanation:

    Given

    MA = Mass of object A

    MB = Mass of object B

    Using, K. E = ½mv², where m = mass and v = velocity

    The kinetic energy of A;

    KA = ½MA. V²

    The kinetic energy of B;

    KB = ½MB. V²

    Assuming air resistance is negligible, then velocity of A and B are equal; VA = VB = V

    Given that MB = 2MA

    The kinetic energy of B, becomes

    KB = ½ (2MA). V²

    KB = 2 * ½MA. V²

    By substituton KA = ½MA. V²

    KB becomes

    KB = 2 * KA

    KB = 2KA
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