(a) The expression for the speed of the larger piece is [tex]\frac{v\sqrt{2} }{3}[/tex].
(b) The numeric value for the speed of the larger piece is 10.84 m/s.
(c) The direction of the larger piece is 45⁰ with respect to y - axis.
Apply the principle of conservation of linear momentum;
m₁v₁ + m₂v₂ + m₃v₃ = 0
where;
mv.i + mv.j + (3m)v₃ = 0
(3m)v₃ = -mv.i - mv.j
3v₃ = - v.i - v.j
[tex]v_3 = \frac{-v_i \ - \ v_j}{3} = \frac{v\ (\sqrt{(-1)^2 + (-1)^2} )}{3} = \frac{v\ \sqrt{2} }{3}[/tex]
[tex]v_3 = \frac{23 \times \sqrt{2} }{3} \\\\v_3 = 10.84 \ m/s[/tex]
tan θ = vj/vi
tan θ = 1/1
tan θ = 1
θ = arc tan(1)
θ = 45⁰
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