A ball of mass 0.440 kg moving east (+x direction) with a speed of \({\bf{3}}{\bf{.80}}\;{{\bf{m}} \mathord{\left/{\vphantom {{\bf{m}} {\bf{s}}}} \right.\\} {\bf{s}}}\) collides head-on with a 0.220-kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?

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

The mass of the first ball is\({m_1} = 0.440\;{\rm{kg}}\).

The initial velocity of the first ball is\({v_{1,{\rm{i}}}} = 3.80\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\).

The mass of the second ball is\({m_2} = 0.220\;{\rm{kg}}\).

The initial velocity of the second ball is \({v_{2,{\rm{i}}}} = 0\).

Since the collision is perfectly elastic, the coefficient of restitution is\(e = 1\).

Consider the two balls as a system. Also, take the direction towards the east positive and the west negative.

Apply the law of conservation of momentum.

\(\begin{array}{c}{m_1}{v_{1,{\rm{i}}}}+{m_2}{v_{2,{\rm{i}}}} = {m_1}{v_{1,{\rm{f}}}} +{m_2}{v_{2,{\rm{f}}}}\\\left({0.440\;{\rm{kg}}}\right)\left({3.80\;{{\rm{m}}\mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right) + \left( {0.220\;{\rm{kg}}}\right)\left( 0 \right) = \left( {0.440\;{\rm{kg}}} \right){v_{1,{\rm{f}}}} +\left({0.220\;{\rm{kg}}}\right){v_{2,{\rm{f}}}}\\\left({0.440\;{\rm{kg}}}\right){v_{1,{\rm{f}}}}+\left({0.220\;{\rm{kg}}}\right){v_{2,{\rm{f}}}}=1.672\;{{{\rm{kg}}\cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}\end{array}\) … (i)

The condition for perfectly elastic collision is as follows:

\(\begin{array}{c}e=\frac{{{v_{2,{\rm{f}}}}{v_{1,f}}}}{{{v_{1,i}}{v_{2,{\rm{i}}}}}}\\1=\frac{{{v_{2,{\rm{f}}}}-{v_{1,f}}}}{{{v_{1,i}}- {v_{2,{\rm{i}}}}}}\\{v_{2,{\rm{f}}}} - {v_{1,f}} = {v_{1,i}} - {v_{2,{\rm{i}}}}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}{v_{2,{\rm{f}}}} - {v_{1,f}} = \left( {3.80\;{{\rm{m}} \mathord{\left/ {\vphantom{{\rm{m}}{\rm{s}}}}\right.\\}{\rm{s}}}}\right)0\\{v_{2,{\rm{f}}}}=\left({3.80\;{{\rm{m}}\mathord{\left/{\vphantom{{\rm{m}}{\rm{s}}}}\right.\\}{\rm{s}}}}\right)+{v_{1,f}}\end{array}\) ……… (ii)

Substitute the value of equation (ii) in equation (i).

\(\begin{array}{c}\left({0.440\;{\rm{kg}}}\right){v_{1,{\rm{f}}}}+\left({0.220\;{\rm{kg}}} \right)\left[ {\left( {3.80\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right) + {v_{1,f}}} \right] = 1.672\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}\\\left( {0.440\;{\rm{kg}}} \right){v_{1,{\rm{f}}}} + \left( {0.220\;{\rm{kg}}} \right){v_{1,f}} = \left( {1.672\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right) - \left( {0.836\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}}{\rm{s}}}}\right.\\}{\rm{s}}}}\right)\\{v_{1,{\rm{f}}}}=1.27\;{{\rm{m}}\mathord{\left/{\vphantom{{\rm{m}}{\rm{s}}}}\right.\\}{\rm{s}}}\end{array}\)

Substitute this value in equation (ii) to get the final velocity of the second ball.

\(\begin{array}{c}{v_{2,{\rm{f}}}} = \left( {3.80\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\}{\rm{s}}}}\right) + {v_{1,f}}\\{v_{2,{\rm{f}}}} = \left( {3.80\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right) + \left( {1.27\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right)\\{v_{2,{\rm{f}}}} = 5.07\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\}{\rm{s}}}\end{array}\)

Thus, the final velocity of the first ball is \(1.27\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\) towards the east, and that of the second ball is \(5.07\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\), also towards the east.