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29 January, 23:43

A 3.0 kg puck slides due east on a horizontal frictionless surface at a constant speed of 4.5 m/s. Then a force of magnitude 6.0 N, directed due north, is applied for 1.5 s. Afterward, a. What is the northward component of the puck's velocity?

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  1. 30 January, 02:04
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    3 m/s

    The northward component of the puck's velocity is 3 m/s

    Explanation:

    Applying the impulse momentum equation;

    Impulse = change in momentum

    Ft = m (∆v)

    ∆v = Ft/m

    F = force = 6.0 N due north

    t = time = 1.5 s

    m = mass = 3.0 kg

    Substituting the values;

    Change in velocity ∆v = (6 * 1.5) / 3.0 = 9/3

    ∆v = 3 m/s due north

    And since the initial northward component of the puck's velocity is zero.

    The final northward component of the puck's velocity is;

    v = 0 + 3 m/s

    v = 3 m/s

    The northward component of the puck's velocity is 3 m/s
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