Respuesta :
Answer:
a) v₂ = 7 m / s , c) [tex]v_{f}[/tex] = 1.53 m / s
Explanation:
The momentum is defined by
p = mv
In an isolated system in this case formed by the student and the ball the forces that act are forces of action and reaction, so as there are no external forces in impulse, it is conserved in this case
p_student = p_ ball
The student's mass is
M = 44 kg
The mass of the ball
m = 22 kg
a) write the equation in two moments
Initial. Student holds the ball
p₀ = 0 + 0
Final. Throw the ball
Student speed = - 3.5 m / s
[tex]p_{f}[/tex] = M v₁ + m v₂
The moment is preserved
p₀ = [tex]p_{f}[/tex]
0 = M v₁ + m v₂
v₂ = -v₁ M / m
v₂ = - (- 3.5) 44/22
v₂ = 7 m / s
In this case the impulse of the student and the ball is the same.
Due to the mass difference the ball is shot in the opposite direction to the student with a speed of 7 m / s
c) Student in repos and catches the ball, which moves [tex]v_{2i}[/tex]= 4.6 m / s
Initial instant Before catching the ball
p₀ = 0 + m [tex]v_{2i}[/tex]
Final. After catching the ball
[tex]p_{f}[/tex] = (m + M) [tex]v_{f}[/tex]
po = [tex]p_{f}[/tex]
m [tex]v_{2i}[/tex] = (m + M) [tex]m_{f}[/tex]
[tex]v_{f}[/tex] = [tex]v_{2i}[/tex] m / (m + M)
[tex]v_{f}[/tex] = 4.6 22 / (22 + 44)
[tex]v_{f}[/tex] = 1.53 m / s
The momentum of the two is preserved and the student and ball set moves in the same initial direction, but with a lower speed, due to the difference in mass
summary in all cases the momentum is retained.