Answer to Question #142203 in Physics for Ranxe

Question

Two object of masses 0.20 kg and 0.10 kg approach each other at 5.0 m/s and 2.5 m/s, respectively. Assuming that the collision is perfectly inelastic, what is the speed of the objects after the collision?

Answer 1

According to the momentum conservation law, the equation of the momentums is:

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

where $m_1 = 0.2kg$, $m_2 = 0.1kg$ are masses of the objects, $v_1 = 5m/s$, $v_2 = 2.5m/s$ are their speeds befor the collistion and $v'$ is their speed after the collision. Since the collision is perfectly inelastic, both objects will move together with the same speed after the collision. Expressing $v'$, obtain:

$$v' = \dfrac{m_1v_1 + m_2v_2 }{m_1+ m_2} = \dfrac{0.2\times 5 + 0.1\times 2.5}{0.2+0.1} \approx 4.17m/s$$

Answer. 4.17 m/s.