The speed of the first billiard ball is\({v_{\rm{A}}} = 2\;{\rm{m/s}}\).

The speed of the second billiard ball is\({v_{\rm{B}}} = - 3.60\;{\rm{m/s}}\).

Consider the movement of ball A is in the positive direction, and the movement of ball B is in the negative direction. Obtain a relationship for velocity to find the final velocity after the collision.

The relation for relative velocity is

\(\begin{array}{c}{v_{\rm{A}}} - {v_{\rm{B}}} = - \left( {{{v'}_{\rm{A}}} - {{v'}_{\rm{B}}}} \right)\\2\;{\rm{m/s}} + 3.60\;{\rm{m/s}} = - \left( {{{v'}_{\rm{A}}} - {{v'}_{\rm{B}}}} \right)\\{{v'}_{\rm{B}}} = \left( {5.60\;{\rm{m/s}} + {{v'}_{\rm{A}}}} \right).\end{array}\)… (i)

Here, \({v'_{\rm{A}}}\) is the final velocity of ball A, and \({v'_{\rm{B}}}\) is the final velocity of ball B.

The relation from the momentum equation is

\(\begin{array}{c}m{v_{\rm{A}}} + m{v_{\rm{B}}} = m{{v'}_{\rm{A}}} + m{{v'}_{\rm{B}}}\\{v_{\rm{A}}} + {v_{\rm{B}}} = {{v'}_{\rm{A}}} + {{v'}_{\rm{B}}}.\end{array}\)

Here, m is the mass of the ball.

Substitute the values from equation (i) in the above equation.

\(\begin{array}{c}2\;{\rm{m/s}} + - 3.60\;{\rm{m/s}} = {{v'}_{\rm{A}}} + \left( {5.60\;{\rm{m/s}} + {{v'}_{\rm{A}}}} \right)\\2{{v'}_{\rm{A}}} = - 7.2\;{\rm{m/s}}\\{{v'}_{\rm{A}}} = - 3.6\;{\rm{m/s}}\end{array}\)

Substitute the value of speed in equation (i).

\(\begin{array}{l}{{v'}_{\rm{B}}} = \left( {5.60\;{\rm{m/s}} - 3.6\;{\rm{m/s}}} \right)\\{{v'}_{\rm{B}}} = 2\;{\rm{m/s}}\end{array}\)

Thus, \({v'_{\rm{A}}} = - 3.6\;{\rm{m/s}}\) is the final speed of the first billiard ball whose final direction is opposite to its initial direction, and \({v'_{\rm{B}}} = 2\;{\rm{m/s}}\) is the final speed of the second billiard ball whose final direction is also opposite to its initial direction.