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9 May, 00:51

Ball 1, with a mass of 150 g and traveling at 15 m/s, collides head on with ball 2, which has a mass of 350 g and is initially at rest. A) What are the final velocities of each ball if the collision is perfectly elastic? B) What are the final velocities of each ball if the collision is perfectly inelastic?

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  1. 9 May, 02:55
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    Answer: A. Vb1 = - 6m/s, Vb2 = 9m/s.

    B. Vb1 = 4.5m/s, Vb2 = 4.5m/s

    Explanation: A. In a perfectly elastic collision the objects involved moved with a different velocity after collision. Also kenitic energy is conserved. In other to get the different velocity of ball 1 (Vb1) and ball 2 (Vb2) after collision, we apply the following formula

    Vb1 = {{m1 - m2}*V1}/{m1 + m2}

    Vb2 = {{2m1}*V1}/{m1 + m2}

    Where V1 = velocity of ball 1 before collision=15m/s

    m1 = mass of ball 1

    m2 = mass of ball 2.

    Substituting in the above equation:

    Vb1 = {{0.15kg - 0.35kg}*15}/{0.15 + 0.35}

    = {{-0.2}*15}/0.5

    = - 3/0.5

    = - 6m/s

    Also,

    Vb2 = {{2*0.15kg}*15m/s}/{0.15kg + 0.35kg}

    ={0.3kg*15m/s}/0.5kg

    =4.5/0.5

    =9m/s

    B. In a perfectly inelastic collision the object involved move with a common velocity after collision.

    We have that,

    m1*V1 + m2*v2 = {m1 + m2} V¹

    Where V¹ is the final velocity of both balls after a perfectly inelastic collision.

    NOTE: v2 here is 0m/s since ball 2 was at rest before collision.

    So we have,

    0.15kg*15m/s + 0.35kg*0m/s = {0.15kg + 0.35kg}*V¹

    2.25 kgm/s = 0.5kg * V¹

    Making V¹ subject of formula we have,

    V¹ = 2.25kgm per sec/0.5kg

    =4.5m/s. vb1 = Vb2 = V¹

    NOTE also that the mass of the balls from the question were given in gram hence we converted to kilogram the standard unit for mass by dividing by 1000.
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