Answer:
[tex]v_{2f} = \frac{2vm_1}{m_2 + m_1}[/tex]
Explanation:
If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls
Before the collision
[tex]P_i = m_1v[/tex]
After the collision
[tex]P_f = m_1v_{1f} + m_2v_{2f}[/tex]
So using the law of momentum conservation
[tex]P_i = P_f[/tex]
[tex]m_1v = m_1v_{1f} + m_2v_{2f}[/tex]
We can solve for the speed of ball 1 post collision in terms of others:
[tex]v_{1f} = v - v_{2f}\frac{m_2}{m_1}[/tex]
Their kinetic energy is also conserved before and after collision
[tex]m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2[/tex]
[tex]m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2[/tex]
From here we can plug in [tex]v_{1f} = v - v_{2f}\frac{m_2}{m_1}[/tex]
[tex]m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2[/tex]
[tex]m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2[/tex]
[tex]m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2[/tex]
[tex]v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0[/tex]
[tex]v_{2f}(1 + \frac{m_2}{m_1}) = 2v[/tex]
[tex]v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}[/tex]
