A91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity afterward?

The velocity of the players will be 2.88 m/s in the east direction.

Explanation:

Let 'v' be the velocity of the players after collision.

Consider the east direction as positive direction.

Given:

Mass of the first player  is, kg  

Initial velocity of the first player  is, m/s  

Mass of the second player  is, kg  

Initial velocity of the second player is, m/s  

In order to solve this problem we use the law of conservation of momentum which says that the total momentum must be conserved before and after the collision. So we can write:

Solving for v, we get:

Therefore, their velocity after the collision is 2.88 m/s.

The sign of the velocity after collision is positive. So, the players will move in the east direction only after collision.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .