Question

A 0.060-kg tennis ball, moving with a speed of 5.50 m/s has a head-on collision with a 0.090-kg ball initially moving in the same direction at a speed of 3.00 m/s. Assuming a perfectly elastic collision, determine the speed and direction of each ball after the collision?

Short answer

The final speed of the 0.06 kg tennis ball is \(2.5\;{\rm{m/s}}\) , and that of the 0.009 kg tennis ball is \(5\;{\rm{m/s}}\). For the 0.06 kg tennis ball, the direction of the final speed is the same as the initial speed.

Step-by-step solution

Given data

The mass of the first tennis ball is\({m_{\rm{A}}} = 0.06\;{\rm{kg}}\).

The speed of the first tennis ball is\({v_{\rm{A}}} = 5.50\;{\rm{m/s}}\).

The mass of the second tennis ball is\({m_{\rm{B}}} = 0.09\;{\rm{kg}}\).

The speed of the second tennis ball is \({v_{\rm{B}}} = 3.00\;{\rm{m/s}}\).

Relationship between the speeds of tennis balls

The relation for the relative velocity of a perfectly elastic collision is

\(\begin{array}{c}{v_{\rm{A}}} - {v_{\rm{B}}} = - \left( {{{v'}_{\rm{A}}} - {{v'}_{\rm{B}}}} \right)\\5.50\;{\rm{m/s}} - 3\;{\rm{m/s}} = - \left( {{{v'}_{\rm{A}}} - {{v'}_{\rm{B}}}} \right)\\{{v'}_{\rm{B}}} = \left( {2.5\;{\rm{m/s}} + {{v'}_{\rm{A}}}} \right).\end{array}\)…… (i)

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

Calculation of the speed of the first tennis ball after the collision

The relation obtained from the momentum equation is

\({m_{\rm{A}}}{v_{\rm{A}}} + {m_{\rm{B}}}{v_{\rm{B}}} = {m_{\rm{A}}}{v'_{\rm{A}}} + {m_{\rm{B}}}{v'_{\rm{B}}}\).

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

\(\begin{array}{c}{m_{\rm{A}}}{v_{\rm{A}}} + {m_{\rm{B}}}{v_{\rm{B}}} = {m_{\rm{A}}}{{v'}_{\rm{A}}} + {m_{\rm{B}}}\left( {2.5\;{\rm{m/s}} + {{v'}_{\rm{A}}}} \right)\\{{v'}_{\rm{A}}} = \left[ {\frac{{{m_{\rm{A}}}{v_{\rm{A}}} + {m_{\rm{B}}}\left( {{v_{\rm{B}}} - 2.5\;{\rm{m/s}}} \right)}}{{{m_{\rm{A}}} + {m_{\rm{B}}}}}} \right]\end{array}\)

Plugging the values in the above relation, you get

\(\begin{array}{l}{{v'}_{\rm{A}}} = \left( {\frac{{\left( {0.06\;{\rm{kg}}} \right)\left( {5.50\;{\rm{m/s}}} \right) - \left( {0.09\;{\rm{kg}}} \right)\left( {3\;{\rm{m/s}} - 2.50\;{\rm{m/s}}} \right)}}{{0.06\;{\rm{kg}} + 0.09\;{\rm{kg}}}}} \right)\\{{v'}_{\rm{A}}} = 2.5\;{\rm{m/s}}{\rm{.}}\end{array}\)

Calculation of the speed of the second tennis ball after the collision

Substitute the speed of the first tennis ball in equation (i).

\(\begin{array}{l}{{v'}_{\rm{B}}} = \left( {2.50\;{\rm{m/s}} + 2.50\;{\rm{m/s}}} \right)\\{{v'}_{\rm{B}}} = 5\;{\rm{m/s}}\end{array}\)

Thus, \({v'_{\rm{A}}} = 2.5\;{\rm{m/s}}\) is the final speed of the first tennis ball, and \({v'_{\rm{B}}} = 5\;{\rm{m/s}}\) is the final speed of the second tennis ball. Both values are positive, so both the balls will move in the initial direction of the first tennis ball.