Two billiard balls have velocities of 2.0 m/s and -1.0 m/s when they meet in an elastic head-on collision. What is the final velocity of the second ball after collision?
A) "-2.0 m/s"
B) "-1.0 m/s"
C) "-0.5 m/s"
D) "+1.0 m/s"

Question

Quizplus · Accepted Answer

Explanation:In an elastic head-on collision, the total momentum and kinetic energy of the system are conserved. The total momentum before the collision is:$$P_i = m_1v_1 + m_2v_2$$where m_1 and m_2 are the masses of the two balls, and v_1 and v_2 are their velocities.The total momentum after the collision is:$$P_f = m_1v_1' + m_2v_2'$$where v_1' and v_2' are the velocities of the two balls after the collision.Since the collision is elastic, the total kinetic energy of the system is also conserved:$$K_i = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$$$$K_f = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2$$Using the conservation of momentum and kinetic energy, we can solve for the final velocities of the two balls. We find that:$$v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1 + \frac{2m_2}{m_1 + m_2}v_2$$$$v_2' = \frac{2m_1}{m_1 + m_2}v_1 + \frac{m_2 - m_1}{m_1 + m_2}v_2$$In this case, the masses of the two balls are equal, so we have:$$v_1' = -v_1$$$$v_2' = v_2$$Therefore, the final velocity of the second ball is 1.0 m/s.

Two Billiard Balls Have Velocities of 2