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A particle with an initial linear momentum of 3.44 kg · m/s directed along the positive x-axis collides with a second particle, which has an initial linear momentum of 6.88 kg · m/s, directed along the positive y-axis. The final momentum of the first particle is 5.16 kg · m/s, directed 45.0° above the positive x-axis. Find the final momentum of the second particle.

The law of conservation of linear momentum
vector p₁ +vector p₂ =vectorp₁`+vector p₂`
Projections on the axis x and y
p₁⒳ +p₂⒳= p₁`⒳+p₂`⒳
p₁⒴ +p₂⒴= p₁`⒴+p₂`⒴
3.44=5.16•cos45 + p₂`⒳
6.88=5.16•sin45 + p₂`⒴
p₂`⒳ =3.44-5.16•0.707= - 0.17
p₂`⒴=6.88-5.16•0.707= 3.28
p₂=sqrt{ p₂`⒳²+p₂`⒴²}=…
tanβ = p₂`⒴/ p₂`⒳

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