Elastic collision in two dimensions

Question

Suppose a particle with mass $m_1$ and speed $v_{1i}$ undergoes an elastic collision with stationary particle of mass $m_2$. After the collision, particle of mass $m_1$ moves with speed $v_{1f}$ in a direction of angle $\theta$ above the line it was moving previously. Particle with mass $m_2$ moves with speed $v_{2f}$ in a direction of angle $\phi$ below the line which particle with mass $m_1$ was moving previously. Using equations for conservation of momentum and kinetic energy, how can we prove these two equations

$\frac{v_{1f}}{v_{1i}}=\frac{m_1}{m_1+m_2}[\cos \theta \pm \sqrt{\cos^2 \theta - \frac{m_1^2-m_2^2}{m_1^2}}]$

and

$\frac{\tan(\theta +\phi)}{\tan(\phi)}=\frac{m_1+m_2}{m_1-m_2}$ ?

EDIT. Here is what I've done:

For the first one, set the $xy$ coordinate system so that the positive direction of the $x$ axis points toward the original path of the particle with mass $m_1$. So we have three equations:

$m_1v_{1i}=m_1v_{1f}\cos \theta + m_2v_{2f} \cos \phi$

$0=m_1v_{1f}\sin \theta - m_2v_{2f}\sin \phi$

$m_1v_{1i}^2=m_1v_{1f}^2+m_2v_{2f}^2$.

From the second one, we get:

$v_{2f}=\frac{m_1v_{1f}\sin \theta}{m_2 \sin \phi}$

Plotting this into third equation, we get

$v_{1i}^2=v_{1f}^2(1+\frac{m_1 \sin^2 \theta}{m_2 \sin^2 \phi})$ (1)

From the first equation, we have

$\cos \phi =\frac{m_1(v_{1i}-v_{1f}\cos \theta)}{m_2v_{2f}}$

which after applying the equation we have for $v_2f$ becomes

$\sin^2 \phi = \frac{1}{1+\frac{(v_{1i}-v_{1f}\cos \theta)^2}{\sin^2 \theta \times v_1f^2}}$

Plotting this into equation (1), gives us an equation in terms of $m_1$, $m_2$, $v_{1f}$, $v_{1i}$ and $\theta$, but it is too far from what I expected.

For the second one, assigning the $xy$ coordinate in a way that the positive direction of the $x$ axis points toward the final path of the particle $m_2$, will give us three equations (two for conservation of linear momentum and one for conservation of kinetic energy), but I don't know what to do next.

Answer 1

Since this is a homework problem, I will only provide a sketch of the solution. From the conservation laws, we have the three equations

$$\begin{align} \tag{1} m_1v_{1i} - m_1v_{1f}\cos \theta &= m_2v_{2f} \cos \phi, \\ \tag{2} m_1v_{1f}\sin \theta &= m_2v_{2f}\sin \phi, \\ \tag{3} m_1v_{1i}^2 - m_1v_{1f}^2 &= m_2v_{2f}^2. \end{align}$$

Summing the squares of (1) and (2) eliminates $\phi$. The RHS of the resultant equation contains $v_{2f}^2$ which can be eliminated using (3). Then, one would obtain a quadratic equation in terms of $\frac{v_{1f}}{v_{1i}}$, which can be solved to obtain the desired equation

$$\frac{v_{1f}}{v_{1i}} = \frac{m_1}{m_1+m_2}\left[\cos \theta \pm \sqrt{\cos^2 \theta - \frac{m_1^2-m_2^2}{m_1^2}}\right].$$

For the next equation, we rotate the axes to obtain the angle $\theta+\phi$ more easily. Here, the conservation laws are

$$\begin{align} \tag{4} m_1v_{1i}\cos\phi - m_1v_{1f}\cos(\theta+\phi) &= m_2v_{2f}, \\ \tag{5} m_1v_{1i}\sin \phi &= m_1v_{1f}\sin(\theta+\phi), \\ \tag{6} m_1v_{1i}^2 - m_1v_{1f}^2 &= m_2v_{2f}^2. \end{align}$$ First, we square (4) and use (6) to eliminate $v_{2f}$. Then, we use (5) to eliminate $v_{1i},v_{if}$ from the resultant equation, obtaining an equation in terms of $\phi$ and $\theta+\phi$. Using trigonometric identities, we get an equation in terms of $\tan\phi$ and $\tan(\theta+\phi)$ only. Then, this equation can be rewritten as a quadratic equation in $\frac{\tan\phi}{\tan(\theta+\phi)}$: $$\left[1-\frac{\tan\phi}{\tan(\theta+\phi)} \right]^2=\frac{m_2}{m_1}\left[1-\frac{\tan^2\phi}{\tan^2(\theta+\phi)}\right],$$ which can be solved to obtain the desired equation

$$\frac{\tan(\theta +\phi)}{\tan(\phi)}=\frac{m_1+m_2}{m_1-m_2}.$$