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Oblique Collision

A mass m 1 ...

Question

A mass $m_{1}$ with initial speed $v_{0}$ in the positive $x$ -direction collides with a mass $m_{2} = 2 m_{1}$ which is initially at rest at the origin, as shown in figure. After the collision $m_{1}$ moves off with speed $v_{1} = v_{0} / 2$ in the negative $y$ - direction, and $m_{2}$ moves off with speed $v_{2}$ at angle $θ$ . Determine $tan θ$ , and find $v_{2}$ in terms of $v_{0}$

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Solution

For $X$ components,
The momentum of system before collision is
$P_{m_{1}} = m v_{0}$
$P_{m_{2}} = m (0) = 0$
The momentum of system after collision is
$P_{m_{1}} = 0$
$P_{m_{2}} = (2 m) (v_{2} c o s θ)$
Thus from conservation of momentum, $m v_{0} = 2 m v_{2} c o s θ$

For $Y$ components,
The momentum of system before collision is
$P_{m_{1}} = 0$
$P_{m_{2}} = 0$
The momentum of system after collision is
$P_{m_{1}} = - m \frac{v_{0}}{2}$
$P_{m_{2}} = (2 m) (v_{2} s i n θ)$
Thus from conservation of momentum, $0 = 2 m v_{2} s i n θ - \frac{m v_{0}}{2}$ .

Thus from both the equations,
$v_{2} = \frac{m v_{0}}{2 m c o s θ} = \frac{v_{0}}{2 (\frac{2}{\sqrt{5}})}$
$= \frac{\sqrt{5}}{4} v_{0}$

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