Two ice skaters collide on the ice. A 39.6-kg skater moving South at 6.21 m/s collides with a 52.1-kg skater moving East at 4.33 m/s. The two skaters entangle and move together across the ice. Determine the magnitude and direction of their post-collision velocity.

Now using the law of the conservation of linear momentum ( [tex]P_i = P_f[/tex] and knowing that the collision is inelastic we can do the next equations:

[tex]M_{1}V_{1ix}+M_2V_{2ix} = V_{sx}(M_1+M_2)[/tex] (eq. 1)

[tex]M_{1}V_{1iy}+M_2V_{2iy} = V_{sy}(M_1+M_2)[/tex] (eq. 2)

Where [tex]V_{sx}[/tex] and [tex]V_{sy}[/tex] is the velocity of the sistem in x and y after the collision.

Note: the conservation of the linear momentum have to be make once by each axis.

Now, in the (eq. 1) the skater 1 don't have velocity in the axis x, so we can replace [tex]V_{1ix}[/tex] by 0 in the equation and get:

[tex]M_2V_{2ix} = V_{sx}(M_1+M_2)[/tex] (eq. 1)

also, in the (eq. 2) the skater 2 don't have velocity in the axis y, so we can replace [tex]V_{2iy}[/tex] by 0 in the equation and get:

[tex]M_{1}V_{1iy} = V_{sy}(M_1+M_2)[/tex] (eq. 2)

Now, we just replace the data in both equations:

[tex](52.1)(4.33) = V_{sx}(39.6+52.1)[/tex] (eq. 1)

[tex](39.6)(6.21) = V_{sy}(39.6+52.1)[/tex] (eq. 2)

solving for [tex]V_{sx][/tex] and [tex]V_{sy}[/tex] we have:

[tex]V_{sx][/tex] = 2.46 m/s

[tex]V_{sy][/tex] = 2.681 m/s

using the pythagoras theorem we can find the magnitude of the velocity as:

V = [tex]\sqrt{2.46^2+2.681^2}[/tex]

V = 3.6385 m/s

For find the direction we just need to do this;

θ = [tex]tan^{-1}(\frac{2.681}{1.46})[/tex]

θ = 47.46 degrees