Answer:
The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Explanation:
To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.
By definition it is known that the conservation of the moment is given by:

Our values are given by,

As the skater 1 run in x direction, there is not component in Y direction. Then,
Skate 1:


Skate 2:


Then, if we applying the formula in X direction:
m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}
75*5.45-75*1.41=(75+75)v_{fx}
Re-arrange and solving for v_{fx}
v_{fx}=\frac{4.04}{2}
v_{fx}=2.02m/s
Now applying the formula in Y direction:




Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
The velocity with which the jumper leaves the floor is 5.1 m/s.
<h3>
What is the initial velocity of the jumper?</h3>
The initial velocity of the jumper or the velocity with which the jumper leaves the floor is calculated by applying the principle of conservation of energy as shown below.
Kinetic energy of the jumper at the floor = Potential energy of the jumper at the maximum height
¹/₂mv² = mgh
v² = 2gh
v = √2gh
where;
- v is the initial velocity of the jumper on the floor
- h is the maximum height reached by the jumper
- g is acceleration due to gravity
v = √(2 x 9.8 x 1.3)
v = 5.1 m/s
Learn more about initial velocity here: brainly.com/question/19365526
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