Question

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

Answer 1

Answer:

$V=4.7km/h$

$theta=61.9$°    

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

Explanation:

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:

$m_{1}*v_{ox}=(m_{1}+m_{2})Vcos\theta$     (1)

In the axis Y:

$m_{2}*v_{oy}=(m_{1}+m_{2})Vsin\theta$      (2)

We solve the last equations, we divide them:

$tan\theta=\frac{m_{2}*v_{oy}}{m_{1}*v_{ox}}$      

$theta=arctan{\frac{m_{2}*v_{oy}}{m_{1}*v_{ox}}}$      

$theta=arctan{\frac{75*7.5}{60*5}}=61.9$°    

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):

$m_{1}^{2}*v_{ox}^{2}+m_{2}^{2}*v_{oy}^{2}=(m_{1}+m_{2})^{2}V^{2}$  

$V=\frac{\sqrt{m_{1}^{2}*v_{ox}^{2}+m_{2}^{2}*v_{oy}^{2}}}{(m_{1}+m_{2})}$

$V=4.7km/h$