Question

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

Answer 1

Answer:

$v"_{1} = v_{1} tan$Θ

$v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}$Θ

Θ = $tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )$

Explanation:

Applying the law of conservation of momentum, we have:

Δ$p_{x = 0}$

$p_{x} = p"_{x}$

$m_{1}v_{1} = m_{2}v"_{2} cos$Θ (Equation 1)

Δ$p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$Θ

From Equation 2:

$m_{2} v"_{2}$sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ=$\frac{v"_{1}}{v_{1}}$

Θ=$tan^{-1}(\frac{v"_{1}}{v_{1}})$