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:
[tex]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex]
where
[tex]m_1[/tex] is the mass of ball 1
[tex]m_2[/tex] is the mass of ball 2
[tex]u_1[/tex] is the initial velocity of ball 1
[tex]u_2[/tex] is the initial velocity of ball 2
[tex]v_1[/tex] is the final velocity of ball 1
[tex]v_2[/tex] is the final velocity of ball 2
The conservation of kinetic energy can be written as
[tex]\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[/tex]
Working together the two equations, it is possible to find two expressions for the final velocities in terms of the initial velocities:
[tex]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[/tex]
In this problem we have:
[tex]m_1 = m_2 = m[/tex] since the mass of the two balls is identical
[tex]u_1=+4.67 m/s[/tex] is the initial velocity of ball 1
[tex]u_2=-7.89 m/s[/tex] is the initial velocity of ball 2
Substituting into the equations, we find the final velocities:
[tex]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[/tex]
Therefore:
- Ball 1 will move backward at 7.89 m/s
- Ball 2 will move forward at 4.67 m/s
