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Two frisky grasshoppers collide in midair at the top of their respective trajectories and grab onto each other, holding tight thereafter. One is a robust 250 g beast initially moving south at 15.0 cm/s, while the other is a svelte 130 g creature initially moving north at 65.0 cm/s.

Calculate the decrease in kinetic energy that results from the collision in J

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  1. Today, 00:50
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    The decrease in Kinetic energy is 0.0107 Joules

    Explanation:

    Given

    Mass of grasshoppers

    Let m1 = Mass of grasshopper 1

    Let m2 = Mass of grasshopper 2

    Let u1 = initial speed of grasshopper 1

    Let u2 = initial speed of grasshopper 2

    m1 = 250g = 0.25kg

    m2 = 130g = 0.13kg

    u1 = 15cm/s = 0.15m/s

    u2 = 65cm/s = 0.65m/s

    First, we calculate the final velocity of the grasshoppers after collision using conservation of momentum.

    Using

    m1u1 + m2u2 = (m1 + m2) * v

    Where v = final velocity

    By substituton

    0.25 * 0.15 + 0.13 * 0.65 = (0.25 + 0.13) * v

    0.0375 + 0.0845 = 380v

    0.122 = 0.38v

    Make v the subject of formula

    v = 0.122/0.38

    v = 0.321 m/s

    Calculating the Kinetic energies before and after impact.

    Before collision;

    KE = ½m1u1² + ½m2u2²

    KE = ½ (m1u1² + m2u2²)

    By substituton;

    KE = ½ (0.25 * 0.15² + 0.13 * 0.65²)

    KE = 0.030275J

    After collision:

    KE = ½ (m1+m2) v²

    KE = ½ (0.25 + 0.13) * 0.321²

    KE = 0.01957779 J

    Change in kinetic energy = ∆KE

    ∆KE = 0.030275J - 0.01957779J

    ∆KE = 0.01069721J

    ∆KE = 0.0107 J - - - Approximately

    Hence the decrease in Kinetic energy is 0.0107 Joules
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