Answer:
 I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s
,  I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s
,  I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s  and I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s
Explanation:
The impulse is equal to the variation of the moment, to apply this relationship to our case, we will assume that initially the mouse was at rest
     I = Δp = m  -m v₀
 -m v₀
     I = m ( -v₀)
  -v₀)
Bold indicates vector quantities, let's calculate the momentum of each mouse in for the x and y axes
We recommend bringing all units to the SI system
Mouse 1.
It has a mass of 22.3 g = 22.3 10⁻³ kg, a final velocity of (v = 0.349 i ^ - 0.301 j ^) m / s with an initial velocity of zero
     Iₓ = m ( - v₀ₓ)
  - v₀ₓ)
     Iₓ = 22.3 10⁻³ (0.349 -0)
     Iₓ = 7.78 10⁻³ J s
     = m (
 = m ( -
  - )
 )
     = 22.3 10⁻³ (-0.301)
 = 22.3 10⁻³ (-0.301)
     = -6.71 10⁻³ J s
 = -6.71 10⁻³ J s
    I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s
Mouse 2
Mass 17.9 g = 17.9 10⁻³ kg
Speed (-0.699 i ^ - 0.815 j ^) m / s
     Iₓ = m ( - v₀ₓ)
  - v₀ₓ)
     Iₓ = 17.9 10⁻³ (-0.699 -0)
     Iₓ = -12.5 10⁻³ J s
      = 17.9 10⁻³ (-0.815 - 0)
 = 17.9 10⁻³ (-0.815 - 0)
      = -14.6 10⁻³ J s
 = -14.6 10⁻³ J s
    I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s
Mouse 3
Mass 19.1 g = 19.1 10⁻³ kg
Speed (0.745i ^ + 0.975 j ^) m / s
     Iₓ = 19.1 10⁻³ (0.745 -0)
     Iₓ = 14.2 10⁻³ J s
      = 19.1 10⁻³(0.975 -0)
 = 19.1 10⁻³(0.975 -0)
      = 18.6 10⁻³ J s
 = 18.6 10⁻³ J s
     I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s
Mouse 4
Mass 10.1 g = 10.1 10⁻³ kg
Speed (-0.905i ^ + 0.717j ^) m / s
     Iₓ = 10.1 10⁻³ (-0.905 -0)
     Iₓ = -9.14 10⁻³ J s
      = 10.1 10⁻³ (0.717 -0)
 = 10.1 10⁻³ (0.717 -0)
      = 7.24 10⁻³ J s
 = 7.24 10⁻³ J s
    I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s