Respuesta :
Answer:
a)  I = √ (2G m³ (1/2r³ - 1/R)), b) I = √ (8 G m³ (1/2r -1/R))
Explanation:
.a) The relation of the Impulse and the moment is
          I = Δp = m [tex]v_{f}[/tex] - m v₀
We can use Newton's second law with force the force of universal attraction
         F = ma
         G m m / r² = m a
         dv / dt = G m / r²
Suppose re the direction where the spheres move is x
         dv/dx  dx/dt = G m / x²
         dv/dx  v = G m / x²
         v dv = G m dx / x²
We integrate
          v² / 2 = Gm (-1 / x)
We evaluate this integra from the lower limit v = 0 for x = R to the upper limit, where the spheres v = v and x = 2r are touched
         v² / 2-0  = G M (-1 / R + 1 / 2r)
         v = √ [2Gm (1 /2r - 1/ R) ]
The impulse on the sphere is
         I = m vf - m v₀
         I = m vf - 0
         I = m √ (2Gm (1 / 2r-1 / R)
         I = √ (2G m³ (1/2r³ - 1/R))
b) during the crash each sphere arrives with a velocity v and leaves with a velocity –v, the same magnitude but opposite direction
           I = m [tex]v_{f}[/tex]- m v₀
           I = m v - m (-v)
           I = 2mv
           I = 2m √ (2Gm (1 / 2r-1 / R)
           I = √ (8 G m³ (1/2r -1/R))