Ask Question
20 September, 09:26

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?

+4
Answers (1)
  1. 20 September, 13:14
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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, ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers