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An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1 travels up north while m2 travels at a downward angle of θ caused by the collision. In this perfectly elastic collision, find: a) v1'
b) v2'
c) θ

Respuesta :

Answer:

[tex]v"_{1} = v_{1} tan[/tex]Θ

[tex]v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}[/tex]Θ

Θ = [tex]tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )[/tex]

Explanation:

Applying the law of conservation of momentum, we have:

Δ[tex]p_{x = 0}[/tex]

[tex]p_{x} = p"_{x}[/tex]

[tex]m_{1}v_{1} = m_{2}v"_{2} cos[/tex]Θ (Equation 1)

Δ[tex]p_{y} = 0[/tex]

[tex]p_{y} = p"_{y}[/tex]

[tex]0 = m_{1} v"_{1} - m_{2} v"_{2} sin[/tex]Θ (Equation 2)

From Equation 1:

[tex]v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}[/tex]Θ

From Equation 2:

[tex]m_{2} v"_{2}[/tex]sinΘ = [tex]m_{1} v_{1}[/tex]

[tex]v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }[/tex]

Replacing Equation 3 in Equation 4:

[tex]v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}[/tex]

[tex]v"_{1}=v_{1}\frac{sinΘ}{cosΘ}[/tex]

[tex]v"_{1}=v_{1}tan[/tex]Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ=[tex]\frac{v"_{1}}{v_{1}}[/tex]

Θ=[tex]tan^{-1}(\frac{v"_{1}}{v_{1}})[/tex]

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