Question

A 1300 kg car moving at 20 m/s and a 900 kg car moving at -15 m/s in precisely the opposite directions participate in a head on crash. If we consider this event to be perfectly inelastic collision, what is the speed and direction of the cars after impact?

Answer 1

According to the momentum conservation law, have:

$$m_1v_1 + m_2v_2 = (m_1 + m_2)v$$

where $m_1 = 1300kg, m_2 = 900kg$ are the masses of the cars, $v_1 = 20m/s, v_2 = -15m/s$ are their velocities before the collision, and $v$ is their velocity after the collision. Expressing $v$, obtain:

$$v = \dfrac{m_1v_1 + m_2v_2}{m_1 + m_2}\\ v = \dfrac{1300\cdot 20 - 900\cdot 15}{1300 + 900} \approx 5.7m/s$$

Answer. 5.7 m/s.