Question

A 3.00-kg particle has a velocity of 13.00i 2 4.00j2 m/s. (a) Find its x and y components of momentum. (b) Find the magnitude and direction of its momentum.

Answer 1

Let the velocity vector be:

$$\mathbf{v} = (13, 4)\space m/s$$

By definition, the momentum is:

$$\mathbf{p} =m\mathbf{v} = 3\cdot (13, 4) = (39,12)\space kg\cdot m/s$$

Its length:

$$p = \sqrt{p_x^2 + p_y^2} = \sqrt{39^2 + 12^2}\approx 40.8\space kg\cdot m/s$$

The direction:

$$\theta = \arctan{\dfrac{p_y}{p_x}} \approx 17.1\degree$$

Answer. (a) x-component: 39 kg*m/s, y-component: 12 kg*m/s. (b)Magnitude: 40.8 kg*m/s, direction: 17.1 degrees with x-axis.