Question

8. In the billiard ball room at the basement of Memorial Union, two identical balls are traveling toward each other along a straight line which is chosen to the x-axis. They experience an elastic head-on collision. Ignoring any frictional forces, if one ball m1 moving at a velocity of +4.67 m/s and the other m2 at a velocity of -7.89 m/s, what are the velocity (magnitude and direction) of each ball after the collision?

Answer 1

Answer:

- Ball 1 will move backward at 7.89 m/s

- Ball 2 will move forward at 4.67 m/s

Explanation:

In an elastic collision, both the total momentum and the total kinetic energy of the system are conserved.

The conservation of the momentum can be written as:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where

$m_1$ is the mass of ball 1

$m_2$ is the mass of ball 2

$u_1$ is the initial  velocity of ball 1

$u_2$ is the initial velocity of ball 2

$v_1$ is the final velocity of ball 1

$v_2$ is the final velocity of ball 2

The conservation of kinetic energy can be written as

$\frac{1}{2}m_1 u_2^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Working together the two equations, it is possible to find two expressions for the final velocities in terms of the initial velocities:

$v_1=\frac{m_1 -m_2}{m_1 +m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\v_2=\frac{2m_1}{m_1+m_2}u_1 -\frac{m_1 -m_2}{m_1 +m_2}u_2$

In this problem we have:

$m_1 = m_2 = m$ since the mass of the two balls is identical

$u_1=+4.67 m/s$ is the initial velocity of ball 1

$u_2=-7.89 m/s$ is the initial velocity of ball 2

Substituting into the equations, we find the final velocities:

$v_1=\frac{2m}{m+m}u_2=u_2 =-7.89 m/s\\v_2=\frac{2m}{m+m}u_1=u_1=+4.67 m/s$

Therefore:

- Ball 1 will move backward at 7.89 m/s

- Ball 2 will move forward at 4.67 m/s