A 44 kg student on in-line skates is playing with a 22 kg exercise ball, disregarding friction, explain what happens in each situation: The student is holding the ball, and both are at rest. The student then throws the ball horizontally, causing the student to glide back at 3.5m/s. Explain what happens to the ball in part (a) in terms of the momentum of the student and the momentum of the ball The student is initially at rest. The student then catches the ball, which is initially moving to the tight at 4.6 m/s. Explain what happens in part (c) in terms of the momentum of the student and the momentum of the ball

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.