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

- Ball 1 will move backward at 7.89 m/s

- Ball 2 will move forward at 4.67 m/s

Step-by-step 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

`(1)/(2)m_1 u_2^2 + (1)/(2)m_2 u_2^2 = (1)/(2)m_1 v_1^2 + (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=(m_1 -m_2)/(m_1 +m_2)u_1+(2m_2)/(m_1+m_2)u_2\\v_2=(2m_1)/(m_1+m_2)u_1 -(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=(2m)/(m+m)u_2=u_2 =-7.89 m/s\\v_2=(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