Anti-Head-On Collision

The total initial energy is $E_{i} = \frac{1}{2} m_{1} v_1^2 + \frac{1}{2} m_2 v_2^2 = m v^2$ .

To find the total final energy, we use the equation for final energy of an inelastic collision:

$E_f = \frac{1}{2} \frac{m_1^2}{(m_1 + m_2)} v_1^2 + \frac{m_2^2}{(m_1 + m_2)} v_2^2 + 2 \frac{m_1 m_2}{(m_1 + m_2)} v_1 v_2 \text{cos} \theta$

Set $m_1 = m_2 = m$ , $v_1 = v_2 = v$ and $\theta = 0$ , since they are pointing in the same direction. Then $E_f$ simplifies to

$E_f = \frac{1}{2} (\frac{m^2}{2 m} v^2 + \frac{m^2}{2 m} v^2 + 2\frac{m^2}{2 m} v^{2}$ $E_f = m v^2 = E_i$

The answer is 1 .

It is important to note that this problem is a hypothetical limiting case that a real inelastic collision could not reach. If two objects were going in the same direction at the same velocity, they would never collide. However, inelastic collisions of the same mass and almost the same velocity will have final energies very close to the initial energy.