Answer:
Step-by-step explanation:
We shall apply conservation of momentum along x and y axis.
Let the final momentum of second particle be p₁ along x axis and p₂ along y axis.
Considering momentum along x axis
2 + 0 = 3 cos 45 + p₁
p₁ = 2-2.12 = - 0.12 kg m/s
Considering momentum along y axis
4 + 0 = 3 sin 45 + p₂
p₂ = 4-2.12 = 1.88 kg m/s
Final momentum = √ ( p₁² + p₂² )
=√ ( .12² + 1.88² )
= 1.88 approx