Conservation Laws

Let the $x$ - and $y$ -components of momentum before collision be $p_x$ and $p_y$ respectively. Then $p_x = m_1u_{1x} + m_2 u_{2x} = 2(3) + 3(-7) = - 15$ . Since the collision is elastic, by conservation of momentum, $p_x$ after collision is also $-15$ . Therefore,

$\begin{aligned} m_1v_{1x} + m_2v_{2x} & = -15 \\ 2(-5) + 3v_{2x} & = - 15 \\ \implies v_{2x} & = - \frac 53 \end{aligned}$

Similarly, by conservation of $p_y$ , we have:

$\begin{aligned} m_1v_{1y} + m_2v_{2y} & = m_1u_{1y} + m_2u_{2y} \\ 2v_{1y} + 3v_{2y} & = 2(5) + 3(-1) = 7 \\ \implies v_{2y} & = \frac {7-2v_{1y}}3 & ...(1) \end{aligned}$

Now, consider the conversion of kinetic energy:

$\begin{aligned} \frac 12 m_1 v_1^2 + \frac 12 m_2 v_2^2 & = \frac 12 m_1 u_1^2 + \frac 12 m_2 u_2^2 \\ \frac {m_1}2 \left(v_{1x}^2 + v_{1y}^2\right) + \frac {m_2}2 \left(v_{2x}^2 + v_{2y}^2\right) & = \frac {m_1}2 \left(u_{1x}^2 + u_{1y}^2\right) + \frac {m_2}2 \left(u_{2x}^2 + u_{2y}^2\right) \\ (-5)^3 + v_{1y}^2 + \frac 32 \left(\left(-\frac 53\right)^2 + v_{2y}^2\right) & = 3^2 + 5^2 + \frac 32 \left((-7)^2 + (-1)^2\right) \\ v_{1y}^2 + \frac 32 \color{#3D99F6} v_{2y}^2 & = \frac {479}6 & \small \color{#3D99F6} (1): v_{2y} = \frac {7-2v_{1y}}3 \\ v_{1y}^2 + \frac 32 \color{#3D99F6} \left(\frac {7-2v_{1y}}3\right)^2 & = \frac {479}6 \\ \implies 10v_{1y}^2 - 28v_{1y} - 430 & = 0 \end{aligned}$

Solving the quadratic equation for $v_{1y}$ and checking the respective kinetic energies $E_1$ and $E_2$ :

$\begin{cases} v_{1y} = \color{#3D99F6} \boxed{8.105} & \implies v_{2y} = -3.070 & E_1 = 90.695 & E_2 = 18.305 & E_1:E_2 \approx 5:1 & \small \color{#3D99F6} \text{Accepted.} \\ v_{1y} = -5.305 & \implies v_{2y} = 5.870 & E_1 = 53.145 & E_2 = 55.854 & E_1:E_2 \approx 1:1 & \small \color{#D61F06} \text{Rejected.} \end{cases}$