Question

a 1200 kg car travelling at 20 m/s collides with a stationary 1400 kg car. The two cars lock together. Determine the speed of the vehicles immediately after the collision if 53.9% of the initial kinetic energy is converted to heat.

Answer 1

The speed of the vehicles immediately after the collision can be found from the momentum conservation:

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

where $m_1 = 1200kg, m_2 = 1400kg$ are masses of the first and second car respectively, and $v_1 = 20m/s, v_2 = 0m/s$ are the speeds of the first and second car respectively. $v$ is the speed of the vehicles immediately after the collision. Thus, obtain:

$$v = v_1\dfrac{m_1}{m_1+m_2} = 20\cdot \dfrac{1200}{1200+1400} \approx 9.2m/s$$

Answer. 9.2 m/s.