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1 January, 14:02

A thread holds a 1.5-kg and a 4.50-kg cart together. After the thread is burned, a compressed spring pushes the carts apart, giving the 1.5 kg cart a speed of 26 cm/s to the left. What is the velocity of the 4.5-kg cart? a. 10 cm/s b. 4.56 cm/s c. 8.67 cm/s d. 7.3 cm/s

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  1. 1 January, 14:15
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    c. 8.67 cm/s

    Explanation:

    From the law of conservation of momentum,

    Total momentum before the thread was burned = Total momentum after was burned

    mu + m'u' = mv + m'v' ... Equation

    Where m = mass of the heavier cart, m' = mass of the lighter cart, u = initial velocity of the bigger cart, u' = initial velocity of the smaller cart, v = final velocity of the bigger cart, v' = final velocity of the smaller cart.

    Note: Both cart where momentarily at rest, as such u = u' = 0. i. e the total momentum before the thread was burn = 0

    And assuming the left is positive,

    We can rewrite equation 1 as

    mv + m'v' = 0 ... Equation 2

    Given: m = 4.5 kg, m' = 1.5 kg, v' = 26 cm/s

    Substitute into equation 2,

    4.5v + 1.5 (26) = 0

    4.5v + 39 = 0

    4.5v = - 39

    v = - 39/4.5

    v = - 8.67 cm/s.

    Note: v is negative because it moves to right.

    Hence the velocity of the 4.5 kg cart = 8.67 cm/s.

    The right option is c. 8.67 cm/s
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