Answer:
The man's velocity is [tex](2.222,0,0)\frac{m}{s}[/tex]
Explanation:
If we want to solve this exercise, we need to use the Momentum Conservation Principle.
The momentum is a vectorial magnitude define as :
[tex]p=m.v[/tex]
Where ''p'' is the momentum
Where ''m'' is the mass of the object
And where ''v'' is the velocity vector.
For any collision occurring in an isolated system, momentum is conserved. We define an isolated system as a system in which the sumatory of exterior forces is equal to zero.
For this exercise, we can solve it as a collision ⇒
[tex]m1.v1=m2.v2[/tex]
Where ''m1'' is the mass of the ball
Where ''v1'' is the velocity vector of the ball
Where ''m2'' is the mass of the man
And where ''v2'' is the velocity vector of the man after catching the ball.
If we replace all the data in the equation :
[tex](10kg).(20\frac{m}{s})=(90kg).v2[/tex]
[tex]v2=2.222\frac{m}{s}[/tex]
This will be the magnitude of the velocity vector from the man.
If we define as positive the sense in which his friend tosses him the ball (given that we know the direction), the man's velocity will be [tex](2.222,0,0)\frac{m}{s}[/tex]
