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

`V=4.7km/h`

`theta=61.9`°

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

Step-by-step 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=(m_(2)*v_(oy))/(m_(1)*v_(ox))`

`theta=arctan{(m_(2)*v_(oy))/(m_(1)*v_(ox))}`

`theta=arctan{(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`