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?

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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