Two ice skaters not paying attention collide in a completely inelastic collision, Prior to the collision, skater 1, with a mass of 60 kg, has a velocity of 5.0 km/h eastward, and moves at a right angle to skater 2, who has a mass of 75 kg and a velocity of 7.5 km/h southward. What is the velocity (magnitude and direction) of the skaters after collision? Hint: Define the +x-axis by the initial velocity of skater 1 from west to east and the +y-axis by the initial velocity of skater 2 from north to south

The skaters collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V and an angle respect the axis X.

We need to use the conservation of momentum Law, the total momentum is the same before and after the collision.

In the axis X:

[tex]m_{1}*v_{ox}=(m_{1}+m_{2})Vcos\theta[/tex] (1)

In the axis Y:

[tex]m_{2}*v_{oy}=(m_{1}+m_{2})Vsin\theta[/tex] (2)

We solve the last equations, we divide them:

[tex]tan\theta=\frac{m_{2}*v_{oy}}{m_{1}*v_{ox}}[/tex]

[tex]theta=arctan{\frac{m_{2}*v_{oy}}{m_{1}*v_{ox}}}[/tex]

[tex]theta=arctan{\frac{75*7.5}{60*5}}=61.9[/tex]°

θ is the angle that goes from the positive x axis to the positive y axis

We add the squares of the equations (1) and (2):

[tex]m_{1}^{2}*v_{ox}^{2}+m_{2}^{2}*v_{oy}^{2}=(m_{1}+m_{2})^{2}V^{2}[/tex]

[tex]V=\frac{\sqrt{m_{1}^{2}*v_{ox}^{2}+m_{2}^{2}*v_{oy}^{2}}}{(m_{1}+m_{2})}[/tex]

[tex]V=4.7km/h[/tex]